Answer:
v > 133.5 m/s
Explanation:
Let's analyze this problem a little, park run a complete circle we must know the speed of the system at the top of the circle.
Let's start by using the concepts of energy to find the velocity at the top of the circle
Initial. Top circle
Em₀ = K + U = ½ m v² + m g y
If we place the reference system at the bottom of the cycle y = 2R = L
Em₀ = ½ m v² + m g y
final. Low circle
[tex]Em_{f}[/tex] = K = ½ m v₁²
Emo = [tex]Em_{f}[/tex]
½ m v² + m g y = 1/2 m v₁²
v₁² = v² + (2g L)
v₁² = v² + 2 g L
The smallest value that v can have is zero, with this value the bullet + block system reaches this point and falls, with any other value exceeding it and completing the circle. Let's calculate for this minimum speed point
v₁ = √2g L
We already have the speed system at the bottom we can use the moment
Starting point before crashing
p₀ = m v₀
End point after collision at the bottom of the circle
[tex]p_{f}[/tex] = (m + M) v₁
The system is formed by the two bodies and therefore the forces to last before the crash are internal and the moment is conserved
p₀ = [tex]p_{f}[/tex]
m v₀ = (m + M) v₁
v₀ = (m + M) / m v₁
Let's replace
v₀ = (1+ M / m) √ 2g L
Let's reduce to the SI system
m = 40 g (kg / 1000g) = 0.040 kg
Let's calculate
v₀ = (1 + 1.35 / 0.040) RA (2 9.8 0.753)
v₀ = 34.75 3.8417
v₀ = 133.5 m / s
the velocity must be greater than this value
v > 133.5 m/s
A 0.750kg block is attached to a spring with spring constant 13.5N/m . While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 32.0cm/s .
What are :
A) The amplitude of the subsequent oscillations?
Express your answer with the appropriate units.
B) The block's speed at the point where x= 0.750 A?
Express your answer with the appropriate units.
Answer:
A)A=0.075 m
B)v= 0.21 m/s
Explanation:
Given that
m = 0.75 kg
K= 13.5 N
The natural frequency of the block given as
[tex]\omega =\sqrt{\dfrac{K}{m}}[/tex]
The maximum speed v given as
[tex]v=\omega A[/tex]
A=Amplitude
[tex]v=\sqrt{\dfrac{K}{m}}\times A[/tex]
[tex]0.32=\sqrt{\dfrac{13.5}{0.75}}\times A[/tex]
A=0.075 m
A= 0.75 cm
The speed at distance x
[tex]v=\omega \sqrt{A^2-x^2}[/tex]
[tex]v=\sqrt{\dfrac{K}{m}}\times \sqrt{A^2-x^2}[/tex]
[tex]v=\sqrt{\dfrac{13.5}{0.75}}\times \sqrt{0.075^2-(0.075\times 0.75)^2}[/tex]
v= 0.21 m/s
The amplitude of the subsequent oscillations can be calculated using the formula for potential energy and the initial kinetic energy. The block's speed at a given point can be found using the conservation of mechanical energy and the equation for kinetic energy.
Explanation:To determine the amplitude of the subsequent oscillations, we need to first calculate the potential energy stored in the spring when the block is hit. Since the block is moving at a speed of 32.0 cm/s, we can calculate its kinetic energy. We can then equate this kinetic energy to the potential energy of the spring, using the formula potential energy = (1/2) * k * A^2, where k is the spring constant and A is the amplitude. Rearranging the formula, we find that the amplitude is equal to the square root of (2 * kinetic energy / k).
To find the block's speed when x = 0.750 A, we can use the conservation of mechanical energy. At the maximum displacement, all the energy is converted between potential and kinetic energy. At this point, the potential energy is zero, so the kinetic energy is equal to the initial kinetic energy. Using the formula for kinetic energy, we can find the speed when x = 0.750 A using the equation kinetic energy = (1/2) * k * (x^2 - A^2) and solving for the speed.
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A white dwarf has a greater mass than either a red dwarf or a brown dwarf. Yet a white dwarf has a smaller radius than either a red dwarf or a brown dwarf. Explain why, in terms of the types of pressure that keep the different kinds of dwarfs from collapsing under their own gravity.
Explanation:
Red dwarf and brown dwarf masses are less than a typical white dwarf mass measuring around 1.2 solar masses. But it's only a few kilometers of the radius. This is precisely because there is no force to overcome the contraction due to gravity. There is a constant battle between the external force of fusion (who wants to expand the star) and inward pressure because of gravity (who wants to compact the star) of regular stars on the main sequence. There remains a balance between these two forces as long as the star remains on the celestial equator.
Red dwarfs are helped by the nuclear fusion force, but brown dwarfs were not large enough to cause the fusion of hydrogen, they are massive enough to generate sufficient energy in the core by fusing deuterium to sustain their volume. However as soon as the star runs out of hydrogen to burn it weakens the force of the external fusion and gravity starts to compact the center of the star. The contraction heats up the core into more massive stars and helium fusion begins, rendering the star once again stable. However this helium fusion does not occur in stars with masses below 1.44Mo. Tightness persists for such stars until the star's gasses degenerate.
In a sample of 18-karat gold, 75 percent of the total mass is pure gold, while the rest is typically 16 percent silver and 9 percent copper. If the density of pure gold is rhogold=19.3g/cm3, while the densitites of silver and copper are respectively rhosilver=10.5g/cm3 and rhocopper=8.90g/cm3, what is the overall density rho18kt of this alloy of 18-karat gold?
Answer:
15.57 g/cm³
Explanation:
[tex]\rho_g[/tex] = Density of gold = 19.3 g/cm³
[tex]\rho_s[/tex] = Density of silver = 10.5 g/cm³
[tex]\rho_c[/tex] = Density of copper = 8.9 g/cm³
Assuming total mass as 1000 g
Volume of gold
[tex]V_g=\frac{0.75\times 1000}{19.3}\\\Rightarrow V_g=38.86\ cm^3[/tex]
Volume of silver
[tex]V_g=\frac{0.16\times 1000}{10.5}\\\Rightarrow V_g=15.238\ cm^3[/tex]
Volume of copper
[tex]V_c=\frac{0.09\times 1000}{8.9}\\\Rightarrow V_c=10.11\ cm^3[/tex]
Density of the alloy would be
[tex]\rho=\frac{M}{V_g+V_s+V_c}\\\Rightarrow \rho=\frac{1000}{38.86+15.238+10.11}\\\Rightarrow \rho=15.57\ g/cm^3[/tex]
The overall density of this alloy is 15.57 g/cm³
The overall density of an 18-karat gold alloy, which consists of 75% gold, 16% silver, and 9% copper, is approximately 15.2 grams per cubic centimeter given the known densities of the pure metals.
Explanation:The density of an alloy (a mixture of metals) is calculated by finding the volume-weighted average of the densities of the constituent metals. In the case of 18-karat gold, we have a mixture of 75% gold, 16% silver, and 9% copper. Using the given densities of these metals, we can calculate the overall density of the alloy as follows:
rho18kt = 0.75*rhogold+0.16*rhosilver+0.09*rhocopper = 0.75*19.3g/cm³+0.16*10.5g/cm³+0.09*8.90g/cm³ = 15.2g/cm³.
Therefore, the overall density of the 18-karat gold alloy, rho18kt, is approximately 15.2g/cm³.
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calculate the density of a neutron star with a radius 1.05 x10^4 m, assuming the mass is distributed uniformly. Treat the neutron star as a giant ucleaus and consider the mass of a nucleon 1.675 x 10^-27 kg. Your answer should be in the form of N x 10^17 kg/m^3. Enter onlt the number N with teo decimal places, do not enter unit.
To develop this problem it is necessary to apply the concepts related to the proportion of a neutron star referring to the sun and density as a function of mass and volume.
Mathematically it can be expressed as
[tex]\rho = \frac{m}{V}[/tex]
Where
m = Mass (Neutron at this case)
V = Volume
The mass of the neutron star is 1.4times to that of the mass of the sun
The volume of a sphere is determined by the equation
[tex]V = \frac{4}{3}\pi R^3[/tex]
Replacing at the equation we have that
[tex]\rho = \frac{1.4m_{sun}}{\frac{4}{3}\pi R^3}[/tex]
[tex]\rho = \frac{1.4(1.989*10^{30})}{\frac{4}{3}\pi (1.05*10^4)^3}[/tex]
[tex]\rho = 5.75*10^{17}kg/m^3[/tex]
Therefore the density of a neutron star is [tex] 5.75*10^{17}kg/m^3[/tex]
A child is sitting on the seat of a swing with ropes 5 m long. Her father pulls the swing back until the ropes make a 30o angle with the vertical and then releases the swing. If air resistance is neglected, what is the speed of the child at the bottom of the arc of the swing when the ropes are vertical?
Answer:
v = 3.7 m/s
Explanation:
As the swing starts from rest, if we choose the lowest point of the trajectory to be the zero reference level for gravitational potential energy, and if we neglect air resistance, we can apply energy conservation as follows:
m. g. h = 1/2 m v²
The only unknown (let alone the speed) in the equation , is the height from which the swing is released.
At this point, the ropes make a 30⁰ angle with the vertical, so we can obtain the vertical length at this point as L cos 30⁰, appying simply cos definition.
As the height we are looking for is the difference respect from the vertical length L, we can simply write as follows:
h = L - Lcos 30⁰ = 5m -5m. 0.866 = 4.3 m
Replacing in the energy conservation equation, and solving for v, we get:
v = √2.g.(L-Lcos30⁰) = √2.9.8 m/s². 4.3 m =3.7 m/s
17. (a) Will the electric field strength between two parallel conducting plates exceed the breakdown strength for air ( 3.0×106 V/m ) if the plates are separated by 2.00 mm and 5.0×103 V a potential difference of is applied? (b) How close together can the plates be with this applied voltage?
Answer:
Explanation:
Distance between plates d = 2 x 10⁻³m
Potential diff applied = 5 x 10³ V
Electric field = Potential diff applied / d
= 5 x 10³ / 2 x 10⁻³
= 2.5 x 10⁶ V/m
This is less than breakdown strength for air 3.0×10⁶ V/m
b ) Let the plates be at a separation of d .so
5 x 10³ / d = 3.0×10⁶ ( break down voltage )
d = 5 x 10³ / 3.0×10⁶
= 1.67 x 10⁻³ m
= 1.67 mm.
A box with a weight of 50 N rests on a horizontal surface. A person pulls horizontally on it with a force of 10 N and it does not move. To start it moving, a second person pulls vertically upward on the box. If the coefficient of static friction is 0.4, what is the smallest vertical force for which the box moves?
Answer:
25 N
Explanation:
Given that
Weight ,mg = 50 N
m=Mass of the box
g=acceleration due to gravity
Horizontal force F= 10 N
Coefficient of friction ,μ = 0.4
Lets take vertical force = R N
In vertical direction
R + N = mg
N= mg - R
The friction force Fr
Fr= μ N
Fr= μ ( mg - R)
To start the motion
F > Fr
10 > 0.4 ( 50 - R )
25 > 50 - R
R > 50 - 25
R > 25 N
Therefore minimum force R= 25 N
In the given problem, the minimum vertical force required to set the box into motion is greater than 15N. This force reduces the normal force and consequently the frictional force such that it becomes less than the horizontally applied force.
Explanation:In this physical situation, you need to understand the role of static friction and normal force in setting the box into motion. The frictional force is calculated by multiplying the coefficient of static friction (0.4 in this case) and normal force. The normal force on the box is the weight of the box minus the upward force applied. Initially, the upward force is zero, so the frictional force is 0.4 * 50N = 20N. This is greater than the 10N horizontal force applied, so the box does not move.
To make the box move, the vertical force has to reduce the normal force such that the frictional force (which is now less due to the decreased normal force) becomes less than the applied horizontal force (10N). Let's consider the vertical upward force needed as F. Hence, the new frictional force will be 0.4 * (50N - F) and should be less than 10N for the box to move. Solving this inequality, the minimum F needed is >15N, anything above this will make the box move.
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Be sure to answer all parts. In a future hydrogen-fuel economy, the cheapest source of H2 will certainly be water. It takes 467 kJ to produce 1 mol of H atoms from water. What is the frequency, wavelength, and minimum energy of a photon that can free an H atom from water? Enter your answers in scientific notation. Frequency = 7.05 × 10 -32 s−1 Wavelength = 4.26 × 10 -25 m Minimum energy = × 10 kJ/photon
Answer:
[tex]7.7549\times 10{19}\ J[/tex]
[tex]1.17037\times 10^{15}\ Hz[/tex]
[tex]2.56329\times 10^{-7}\ m[/tex]
Explanation:
c = Speed of light = [tex]3\times 10^8\ m/s[/tex]
[tex]N_A[/tex] = Avogadro's number = [tex]6.022\times 10^{23}[/tex]
[tex]\nu[/tex] = Frequency
[tex]\lambda[/tex] = Wavelength
The minimum energy is given by
[tex]E=\frac{1\ mol}{N_A}\\\Rightarrow E=467\times 10^{3}\times \frac{1}{6.022\times 10^{23}}\\\Rightarrow E=7.7549\times 10{19}\ J[/tex]
The minimum energy is [tex]7.7549\times 10{19}\ J[/tex]
The energy of a photon is given by
[tex]E=h\nu\\\Rightarrow \nu=\frac{E}{h}\\\Rightarrow \nu=\frac{467\times 10^{3}\frac{1}{6.022\times 10^{23}}}{6.626\times 10^{-34}}\\\Rightarrow \nu=1.17037\times 10^{15}\ Hz[/tex]
The frequency of the photon is [tex]1.17037\times 10^{15}\ Hz[/tex]
Wavelength is given by
[tex]\lambda=\frac{c}{\nu}\\\Rightarrow \lambda=\frac{3\times 10^8}{1.17037\times 10^{15}}\\\Rightarrow \lambda=2.56329\times 10^{-7}\ m[/tex]
The wavelength is [tex]2.56329\times 10^{-7}\ m[/tex]
In some applications of ultrasound, such as its use on cranial tissues, large reflections from the surrounding bones can produce standing waves. This is of concern because the large pressure amplitude in an antinode can damage tissues. For a frequency of 1.0 MHz, what is the distance between antinodes in tissue?
a. 0.38 mm
b. 0.75 mm
c. 1.5 mm
d. 3.0 mm
Answer:
b. 0.75 mm
Explanation:
The distance between antinodes d is half the wavelength [tex]\lambda[/tex]. We can obtain the wavelength with the formula [tex]v=\lambda f[/tex], where f is the frequency given ([tex]f=1MHz=1\times10^6Hz[/tex]) and v is the speed of sound in body tissues (v=1540m/s), so putting all together we have:
[tex]d=\frac{\lambda}{2}=\frac{v}{2f}=\frac{1540m/s}{2(1\times10^6Hz)}=0.00077m=0.77mm[/tex]
which is very close to the 0.75mm option.
Future space stations will create an artificial gravity by rotating. Consider a cylindrical space station of 380 m diameter rotating about its axis. Astronauts walk on the inside surface of the space station. What rotation period will provide "normal" gravity? On a space walk on the outside of the space station how much gravity would they experience?
Answer:
27.66 s
Explanation:
Space station creates artificial gravity by rotational movement about its axis .
The object inside also move in circular motion creating centrifugal force which creates acceleration in them .
centrifugal acceleration = ω² R where ω is angular velocity and R is radius of the cylindrical space station .
R = 380 /2 = 190 m
Given
ω² R = g = 9.8
ω² = 9.8 / R
= 9.8 / 190
= 5.15x 10⁻²
ω = 2.27 x 10⁻¹
= .227 rad / s
2π / T = .227 ( T is time period of rotation )
T = 2π / .227
= 27.66 s .
outside of the space station they will experience zero acceleration , because they are rotating around the earth.
A CD player rotates at a variable speed so that a laser can scan pits and lands on the disk’s bottom surface at a constant tangential speed of 1.2 m/s. The disk has a moment of inertia of 1.2 x 10-4 kg m2and the music is first detected when the laser is located 15 mm from the disk’s center. Assuming the disk started from rest, find the work done by the motor during this start-up.
Answer:
W = 0.384 J
Explanation:
Work and energy in the rotational movement are related
W = ΔK = [tex]K_{f}[/tex] - K₀
W = ½ I [tex]w_{f}[/tex]² - 1 /2 I w₀²
where W isthe work, I is the moment of inertia and w angular velocity
With the system part of the rest the initial angular speed is zero (w₀ = 0)
The angular and linear quantities are related
v = w r
w = v / r
Let's replace
W = ½ I (v / r)²
Let's calculate
W = ½ 1.2 10⁻⁴ 1.2² / (1.5 10⁻²)²
W = 0.384 J
W = 38.4 J
You and your friends find a rope that hangs down 16m from a high tree branch right at the edge of a river. You find that you can run, grab the rope, swing out over the river, and drop into the water. You run at 2.0 m/s and grab the rope, launching yourself out over the water. How long must you hang on if you want to drop into the water at the greatest possible distance from the edge?
Answer:
[tex]t = 2\ s[/tex]
Explanation:
given,
length of the rope = 16 m
speed of the man = 2 m/s
using the formula of time period
[tex]T =2 \pi \sqrt{\dfrac{L}{g}}[/tex]
[tex]T =2 \pi \sqrt{\dfrac{16}{9.8}}[/tex]
[tex]T = 8.028\ s[/tex]
To cover the maximum distance you need to leave the when the rope is shows maximum displacement.
To reach the displacement time to leave the rope is one fourth of the time period.
[tex]t = \dfrac{T}{4}[/tex]
[tex]t = \dfrac{8.03}{4}[/tex]
[tex]t = 2\ s[/tex]
The time period of pendulum is time taken by it to complete one cycle of swing left to right and right to left.
The total time taken to hang on if you want to drop into the water at the greatest possible distance from the edge is 2 seconds.
What is time period of pendulum?The time period of pendulum is time taken by it to complete one cycle of swing left to right and right to left.
It can be given as,
[tex]T=2\pi\sqrt{\dfrac{T}{g}}[/tex]
Here, [tex]g[/tex] is the gravitational force of Earth.
Given information-
Total length of the rope is 16 m.
The speed of the man is 2.0 m/s.
Let the time of to swing by rope both side is [tex]T[/tex]. Thus put the values in the above formula as,
[tex]T=2\pi\sqrt{\dfrac{16}{9.8}}[/tex]
[tex]T=8.028 \rm s[/tex]
Now the greatest possible distance from the edge will be at the greatest displacement.
Thus the time to cove the greatest possible distance will be one forth (1/4) of the total time. Thus,
[tex]t=\dfrac{8.028 }{4}\\t=2\rm s[/tex]
Hence, the total time taken to hang on if you want to drop into the water at the greatest possible distance from the edge is 2 seconds.
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You wish to boil 1.9 kg of water, which has a specific heat capacity of 4186 J/kg-K. The water is initially at room temperature (293 K). Water boils at 373 K.
1) How much energy must be added to the water by heating it in order for it to start boiling?
Answer:
6.36 105 J
Explanation:
In calorimetry all the energy given to a system is converted to heat and the equation for heat is
Q = m [tex]c_{e}[/tex] ΔT
The temperature can be in degrees Celsius or Kelvin since the interval between them is the same, substitute and calculate
Q = 1.9 4186 (373-293)
Q = 6.36 105 J
This heat equals the energy supplied
An alien spaceship traveling at 0.600c toward the Earth launches a landing craft with an advance guard of purchasing agents and physics teachers. The lander travels in the same direction with a speed of 0.800c relative to the mother ship. As observed on the Earth, the spaceship is 0.200 ly from the Earth when the lander is launched.(a) What speed do the Earth observers measure for the approaching lander?(b) What is the distance to the Earth at the time of lander launch, as observed by the aliens?
Answer:
Explanation:
Let the velocity of lander with respect to earth be v .
In relativistic mechanism the expression for relative velocity is
v_r = [tex]\frac{v -u}{1-\frac{uv}{c^2}}[/tex]
Given u = .6c , v_r = .8 c
Substituting the values
.8c = [tex]\frac{v -0.6c}{1-\frac{.6c\times v}{c^2} }[/tex]
.8c-.48v = v - .6c
v = .946c
b )
Distance in terms of time = .2 ly
In relativistic mechanism , expression for relativistic time is given by the following relation
t = [tex]\frac{t_0}{\sqrt{1-\frac{v^2}{c^2} }}[/tex]
Substituting v = .946c
t₀ = .2
t = [tex]\frac{.2}{\sqrt{1-\frac{0.946\times .946c^2}{c^2}}}[/tex]
.2 / √.1050
= .62 ly
distance to the Earth at the time of lander launch, as observed by the aliens will be .62 ly.
Final answer:
The speed of the lander as observed on Earth is 0.865c. The distance to the Earth at the time of lander launch, as observed by the aliens, is 0 ly, which is an unreasonable result.
Explanation:
(a) To find the speed of the lander as observed on Earth, we need to use the relativistic velocity addition formula. The formula is given by:
v' = (v1 + v2)/(1 + (v1*v2)/c^2)
Substituting in the values, where v1 = 0.800c and v2 = 0.600c, we get:
v' = (0.800c + 0.600c)/(1 + (0.800c*0.600c)/c^2) = 1.280c/1.480 = 0.865c
So, the speed of the lander as observed on Earth is 0.865c.
(b) To find the distance to the Earth at the time of lander launch, as observed by the aliens, we can use the time dilation formula. The formula is given by:
t' = t/sqrt(1 - (v^2/c^2))
Where t' is the time measured by the aliens, t is the time measured on Earth, v is the velocity of the spaceship relative to Earth, and c is the speed of light.
In this case, t' = 0 (since the lander is launched at the same time as observed on Earth), t = 0.200 ly (given in the question), v = 0.600c (velocity of the spaceship relative to Earth), and c is the speed of light. Substituting the values, we get:
0 = 0.200/sqrt(1 - (0.600c)^2/c^2)
0 = 0.200/sqrt(1 - 0.360)
0 = 0.200/sqrt(0.640)
0 = 0.200/0.8
0 = 0.25
Since 0 = 0.25 is not possible, this result is unreasonable. It suggests that there is a discrepancy in the calculations or assumptions made.
A 1.40 mH inductor and a 1.00 µF capacitor are connected in series. The current in the circuit is described by I = 14.0 t, where t is in seconds and I is in amperes. The capacitor initially has no charge.
(a) Determine the voltage across the inductor as a function of time. mV
(b) Determine the voltage across the capacitor as a function of time. (V/s2) t2
(c) Determine the time when the energy stored in the capacitor first exceeds that in the inductor.
Answer:
Explanation:
Inductance L = 1.4 x 10⁻³ H
Capacitance C = 1 x 10⁻⁶ F
a )
current I = 14 .0 t
dI / dt = 14
voltage across inductor
= L dI / dt
= 1.4 x 10⁻³ x 14
= 19.6 x 10⁻³ V
= 19.6 mV
It does not depend upon time because it is constant at 19.6 mV.
b )
Voltage across capacitor
V = ∫ dq / C
= 1 / C ∫ I dt
= 1 / C ∫ 14 t dt
1 / C x 14 t² / 2
= 7 t² / C
= 7 t² / 1 x 10⁻⁶
c ) Let after time t energy stored in capacitor becomes equal the energy stored in capacitance
energy stored in inductor
= 1/2 L I²
energy stored in capacitor
= 1/2 CV²
After time t
1/2 L I² = 1/2 CV²
L I² = CV²
L x ( 14 t )² = C x ( 7 t² / C )²
L x 196 t² = 49 t⁴ / C
t² = CL x 196 / 49
t = 74.8 μ s
After 74.8 μ s energy stored in capacitor exceeds that of inductor.
Motor oil , with a viscosity of 0 . 250 Ns / m2 , is flowing through a tube that has a radius of 5 . 00 mm and is 25 . 0 cm long . The drop in pressure is 300 kPa . What is the volume of oil flowing through the tube per unit time ?
Answer:
1.1775 x 10^-3 m^3 /s
Explanation:
viscosity, η = 0.250 Ns/m^2
radius, r = 5 mm = 5 x 10^-3 m
length, l = 25 cm = 0.25 m
Pressure, P = 300 kPa = 300000 Pa
According to the Poisuellie's formula
Volume flow per unit time is
[tex]V=\frac{\pi \times Pr^{4}}{8\eta l}[/tex]
[tex]V=\frac{3.14 \times 300000\times \left ( 5\times 10^{-3} \right )^{4}}{8\times 0.250\times 0.25}[/tex]
V = 1.1775 x 10^-3 m^3 /s
Thus, the volume of oil flowing per second is 1.1775 x 10^-3 m^3 /s.
An ideal parallel-plate capacitor consists of a set of two parallel plates of area Separated by a very small distance 푑. This capacitor is connected to a battery that charges the capacitor such that the energy stored in the capacitor is 푈'. Now the battery is disconnected and the separation between the plates is doubled, how much energy is stored in the capacitor?
Answer:
doubled the initial value
Explanation:
Let the area of plates be A and the separation between them is d.
Let V be the potential difference of the battery.
The energy stored in the capacitor is given by
U = Q^2/2C ...(1)
Now the battery is disconnected, it means the charge is constant.
the separation between the plates is doubled.
The capacitance of the parallel plate capacitor is inversely proportional to the distance between the plates.
C' = C/2
the new energy stored
U' = Q^2 / 2C'
U' = Q^2/C = 2 U
The energy stored in the capacitor is doubled the initial amount.
(a) Triply charged uranium-235 and uranium-238 ions are being separated in a mass spectrometer. (The much rarer uranium-235 is used as reactor fuel.) The masses of the ions are 3.90x10—25 kg and 3.95x10—25 kg , respectively, and they travel at 3.00x105 m/s in a 0.250-T field. What is the separation between their paths when they hit a target after traversing a semicircle? (b) Discuss whether this distance between their paths seems to be big enough to be practical in the separation of uranium-235 from uranium-238.
Answer:
(a) 2.5 cm
(b) Yes
Solution:
As per the question:
Mass of Uranium-235 ion, m = [tex]3.95\times 10^{- 25}\ kg[/tex]
Mass of Uranium- 238, m' = [tex]3.90\times 10^{- 25}\ kg[/tex]
Velocity, v = [tex]3.00\times 10^{5}\ m/s[/tex]
Magnetic field, B = 0.250 T
q = 3e
Now,
To calculate the path separation while traversing a semi-circle:
[tex]\Delta x = 2(R_{U_{35}} - 2R_{U_{38}})[/tex]
The radius of the ion in a magnetic field is given by:
R = [tex]\frac{mv}{qB}[/tex]
[tex]\Delta x = 2(R_{U_{35}} - 2R_{U_{38}})[/tex]
[tex]\Delta x = 2(\frac{mv}{qB} - \frac{m'v}{qB})[/tex]
[tex]\Delta x = 2(\frac{m - m'}{qB}v)[/tex]
Now,
By putting suitable values in the above eqn:
[tex]\Delta x = 2(\frac{3.95\times 10^{- 25} - 3.90\times 10^{- 25}}{3\times 1.6\times 10^{- 19}\times 0.250}\times 3.00\times 10^{5}) = 2.5\ cm[/tex]
[tex]\Delta x = 1.25\ cm[/tex]
(b) Since the order of the distance is in cm, thus clearly this distance is sufficiently large enough in practical for the separation of the two uranium isotopes.
In a mass spectrometer, uranium-235 and uranium-238 ions can be separated based on their masses and velocities. The separation distance between their paths can be determined using the equation: d = mv/(qB). The distance between their paths is practical for the separation of uranium-235 from uranium-238.
Explanation:In a mass spectrometer, triply charged uranium-235 and uranium-238 ions are separated based on their masses and velocities.
The separation of their paths when they hit a target after traversing a semicircle can be calculated using the equation:
d = mv/(qB)
where d is the separation, m is the mass of the ion, v is the velocity, q is the charge, and B is the magnetic field.
The distance between their paths seems to be big enough to be practical in the separation of uranium-235 from uranium-238, as even a small separation can result in significant enrichment over multiple passes through the mass spectrometer.
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Time to be a crash investigator! Two cars are in an accident: Car A was driving due east when Car B, driving due North, ran a stop sign and collided with his car, causing an inelastic collision (the cars stuck together). The skid marks from the accident go off at a 66 degree angle above the horizontal (East). A traffic camera witnessed Car A going 30 mph prior to the crash, and both cars traveling 29.5 mph after the crash. Unfortunately, the camera was angled so it did not see Car B prior to the crash.
Car A has a mass of 1000 kg, and Car B has a mass of 1500 kg.
The driver of Car B claims that he was driving the 30 mph speed limit, and didn’t see the stop sign because of foggy weather. Driver A claims Driver B was speeding and that is why he missed the stop sign.
How fast was Car B traveling prior to the crash?
Answer:
The carriage speed B was 67.4 mph
Explanation:
This is an exercise for the moment, that as a vector we must look for the solution of each axis (x, y). We define a system formed by the two cars, for this system the forces during the crash are internal, so the moment is preserved.
The data they give is the car more A m = 1000kg and its speed is v1₁₀ = 30 mph i^ and the mass of the car B M = 1500 kg
Let's write the moment for each axis
X axis
p₀ₓ = [tex]p_{fx}[/tex]
m v₁ₓ + 0 = (m + M) vₓ
Y axis
poy = [tex]p_{fy}[/tex]
0 + M [tex]v_{2y}[/tex] = (m + M) [tex]v_{y}[/tex]
Let's look for the components of the final velocity with trigonometry
sin 66 = [tex]v_{y}[/tex] / v
cos 66 = vₓ / v
[tex]v_{y}[/tex] = v sin 66
vₓ = v cos 66
We substitute and write the system of equations
m v₁ₓ = (m + M) v cos 66
M [tex]v_{2y}[/tex] = (m + M) v sin66
From the first equation
v = m / (m + M) v₁ₓ / cos 66
v = 1000 / (1000 + 1500) 30 / cos 66
v = 29.5 mph
From the second equation
[tex]v_{2y}[/tex] = (m + M)/m v sin 66
[tex]v_{2y}[/tex] = (1000 + 1500) /1000 29.5 sin 66
[tex]v_{2y}[/tex] = 67.4 mph
The carriage speed B was 67.4 mph
An angry rhino with a mass of 2600 kg charges directly toward you with a speed of 3.70 m/s. Before you start running, as a distraction, you throw a 0.180-kg rubber ball directly at the rhino with a speed of 8.39 m/s. Determine the speed of the ball after it bounces back elastically toward you.
Answer:
vf₂ = 15.79 m/s
Explanation:
Theory of collisions
Linear momentum is a vector magnitude (same direction of the velocity) and its magnitude is calculated like this:
p=m*v
where
p:Linear momentum
m: mass
v:velocity
There are 3 cases of collisions : elastic, inelastic and plastic.
For the three cases the total linear momentum quantity is conserved:
P₀ = Pf Formula (1)
P₀ :Initial linear momentum quantity
Pf : Final linear momentum quantity
Data
m₁ = 2600 kg : mass of the rhino
m₂= 0.18 kg : mass of the ball
v₀₁ = 3.70 m/s : initial velocity of the rhino
v₀₂= - 8.39 m/s, initial velocity of the ball
Problem development
We appy the formula (1):
P₀ = Pf
m₁*v₀₁ + m₂*v₀₂ = m₁*vf₁ + m₂*vf₂
(2600)*(3.7) + (0.18)*(- 8.39) = (2600)*vf₁ +(0.18)*vf₂
9620 -1.5102 = (2600)*vf₁ +(0.18)*vf₂
9618.4898 = (2600)*vf₁ +(0.18)*vf₂ Equation (1)
Because the shock is elastic, the coefficient of elastic restitution (e) is equal to 1.
[tex]e = \frac{v_{f2}-v_{f1} }{v_{o1}-v_{o2}}[/tex]
1*( v₀₁- v₀₂ ) = (vf₂ -vf₁)
( 3.7 -( -8.39 ) = (vf₂ -vf₁)
12.09 = (vf₂ -vf₁)
vf₂ = vf₁ + 12.09 Equation (2)
We replace Equation (2) in the Equation (1)
9618.4898 = (2600)*vf₁ +(0.18)*vf₂
9618.4898 = (2600)*vf₁ +(0.18)* (vf₁ + 12.09)
9618.4898 = (2600)*vf₁ +(0.18)*vf₁ + (0.18)(12.09)
9618.4898 = (2600)*vf₁ +(0.18)*vf₁ + 2.1762
9618.4898 -2.1762 = (2600.18)*vf₁
9616.3136 = (2600.18)*vf₁
vf₁ = (9616.3136) / (2600.18)
vf₁ = 3.698 m/s : Final velocity of the rhino
We replace vf₁ = 3.698 m/s in the Equation (2)
vf₂ = vf₁ + 12.09
vf₂ = 3.698 + 12.09
vf₂ = 15.79 m/s : Final velocity of the ball
The speed of the ball after it bounces back elastically toward you is 8.39 m/s.
To solve this problem, we can use the principle of conservation of momentum and the fact that the collision is elastic. In an elastic collision, both momentum and kinetic energy are conserved.
Using the conservation of momentum:
[tex]\[ m_r v_{r,i} + m_b v_{b,i} = m_r v_{r,f} + m_b v_{b,f} \][/tex]
Since the rhino is much more massive than the ball, its velocity will change very little after the collision. This assumption is valid in the limit where the mass of one object is much larger than the other, which is the case here.
With this assumption, the conservation of momentum simplifies to:
[tex]\[ m_b v_{b,i} = m_b v_{b,f} \][/tex]
Since the mass of the ball cancels out, we find that:
[tex]\[ v_{b,f} = v_{b,i} \][/tex]
This means that the speed of the ball after the collision is the same as before the collision, which is 8.39 m/s, but in the opposite direction.
Additionally, in an elastic collision, the relative speed of approach is equal to the relative speed of separation. Since the rhino's speed is much greater than the ball's speed, the ball will essentially rebound with the same speed it had before the collision but in the opposite direction.
The hollow tube is pivoted about a horizontal axis through point O and is made to rotate in the vertical plane with a constant counterclockwise angular velocity = 2.9 rad/sec. If a 0.15-lb particle is sliding in the tube toward O with a velocity of 4.8 ft/sec relative to the tube when the position θ = 36° is passed, calculate the magnitude N of the normal force exerted by the wall of the tube on the particle at this instant.
To solve the problem it is necessary to apply the concepts related to Newton's second law, as well as to the sum of forces in this type of bodies.
According to the description I make a diagram that allows a better understanding of the problem.
Performing sum of forces in the angular direction in which it is inclined we have to
[tex]\sum F = ma_{\theta}[/tex]
[tex]N - W cos\theta = ma_{\theta}[/tex]
[tex]N = ma_{\theta}+Wcos\theta[/tex]
Tangential acceleration can be expressed as
[tex]a_{\theta} = (r\ddot{\theta}+2\dot{r}\dot{\theta})[/tex]
Our values are given by,
[tex]\dot{\theta} = 2.9rad/s[/tex]
[tex]m = 0.15 lb[/tex]
[tex]\theta = 36\°[/tex]
[tex]v = 4.8ft/s[/tex]
Substituting [tex]\ddot{\theta}=0rad/s^2 , \dot{r}=-4.8ft/s, \dot{\theta}=2.9 rad/s[/tex]
[tex]a_{\theta} = r*0+2*(4.8*2.9)\\a_{\theta}=27.84ft/s^2[/tex]
At the same time we acan calculate the mass of the particle, then
W = mg
Where,
W = Weight of the particle
m = mass
g = acceleration due to gravity
[tex]0.15lb = m(32.2ft/s^2)[/tex]
[tex]m = 4.66*10^{-3}Lb[/tex]
Now using our first equation we have that
[tex]N = ma_{\theta}+Wcos\theta[/tex]
[tex]N = (4.66*10^{-3})(27.84)+0.2cos36[/tex]
[tex]N = 0.2914Lb[/tex]
Therefore the normal force exerted by the wall of the tube on the particle at this instant is 0.2914Lb
The magnitude of the normal force [tex]\( N \)[/tex] exerted by the wall of the tube on the particle at the instant when the angle [tex]\( \theta = 36^\circ \)[/tex] is approximately 0.249 lb
To determine the normal force [tex]\( N \)[/tex]exerted by the wall of the tube on the particle, we analyze the forces acting on the particle in a rotating reference frame. Here’s the step-by-step process:
1. Identify the given data:
- Angular velocity [tex]\( \omega = 2.9 \)[/tex] rad/sec
- Particle weight [tex]\( W = 0.15 \)[/tex] lb
- Relative velocity towards O [tex]\( v_r = 4.8 \)[/tex] ft/sec
- Angle [tex]\( \theta = 36^\circ \)[/tex]
2. **Convert the weight to mass:**
-[tex]\( W = mg \)[/tex]
-[tex]\( m = \frac{W}{g} \)[/tex]
- Using[tex]\( g = 32.2 \text{ ft/sec}^2 \)[/tex]:
- [tex]\( m = \frac{0.15 \text{ lb}}{32.2 \text{ ft/sec}^2} = 0.00466 \text{ slugs} \)[/tex]
3. Calculate the distance ( r ):
- Since [tex]\( v_r = 4.8 \)[/tex] ft/sec is towards O and the tube is rotating with [tex]\( \omega \),[/tex] the distance [tex]\( r \)[/tex] can be found using [tex]\( r = \frac{v_r}{\omega} \)[/tex]:
- [tex]\( r = \frac{4.8 \text{ ft/sec}}{2.9 \text{ rad/sec}} = 1.655 \text{ ft} \)[/tex]
4. **Determine the forces:**
- Centrifugal force [tex]\( F_c = m \omega^2 r \)[/tex]:
[tex]\[ F_c = 0.00466 \text{ slugs} \times (2.9 \text{ rad/sec})^2 \times 1.655 \text{ ft} = 0.0638 \text{ lb} \][/tex]
- Component of gravitational force in the radial direction [tex]\( F_{g, r} = W \cos \theta \)[/tex]:
[tex]\[ F_{g, r} = 0.15 \text{ lb} \times \cos 36^\circ = 0.1214 \text{ lb} \][/tex]
5. Calculate the normal force [tex]\( N \)[/tex]:
- Normal force [tex]\( N \)[/tex] must balance the radial forces:
[tex]\[ N = F_c + F_{g, r} + m \cdot (\omega^2 r) \][/tex]
[tex]\[ N = 0.0638 \text{ lb} + 0.1214 \text{ lb} + 0.0638 \text{ lb} \][/tex]
[tex]\[ N = 0.249 \text{ lb} \][/tex]
Write the ground-state electron configurations of the following ions. (a) Li+ (b) N3− (c) In3+ (Use the noble gas core electron configuration in your answer.) (d) Tl+ (Use the noble gas core electron configuration in your answer.)
Li+ [He]
N³- [Ne]
In³+ [Kr] 4d10
Tl+ [Xe] 4f14 5d10 6S2
The ground-state electron configuration of the ions Li+, N3−, In3+, and Tl+ are [He], [Ne], [Kr]4d10, and [Xe]4f145d106s26p1 respectively. This notation suggests these ions have similar electronic structures to the noble gases and additional electrons in certain cases.
Explanation:The ground-state electron configurations of the ions Li+, N3−, In3+, and Tl+ can be described using the noble gas core electron configuration. The noble gas core is essentially the electron configuration of the closest noble gas with less atomic number than the atom we are considering.
(a) Li+ has lost an electron compared to neutral Lithium. Its electron configuration becomes [He] - it resembles helium, a noble gas.
(b) N3− has gained three electrons compared to neutral Nitrogen and its electron configuration becomes [Ne] - it resembles neon, a noble gas.
(c) In3+ has lost three electrons compared to neutral Indium. Its electron configuration becomes [Kr]4d10 - core is like Kr (krypton), a noble gas, plus 10 electrons added in the d orbital.
(d) Tl+ has lost one electron compared to neutral Thallium and its electron configuration is [Xe]4f145d106s26p1 - core is like Xe (xenon), a noble gas, plus additional electrons.
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Force F = − + ( 8.00 N i 6.00 N j ) ( ) acts on a particle with position vector r = + (3.00 m i 4.00 m j ) ( ) .
What are
(a) the torque on the particle about the origin, in unit-vector notation
(b) the angle between the directions of r and F ?
Explanation:
Given that,
Force, [tex]F=((-8i)+6j)\ N[/tex]
Position of the particle, [tex]r=(3i+4j)\ m[/tex]
(a) The toque on a particle about the origin is given by :
[tex]\tau=F\times r[/tex]
[tex]\tau=((-8i)+6j) \times (3i+4j)[/tex]
Taking the cross product of above two vectors, we get the value of torque as :
[tex]\tau=(0+0-50k)\ N-m[/tex]
(b) Let [tex]\theta[/tex] is the angle between r and F. The angle between two vectors is given by :
[tex]cos\theta=\dfrac{r.F}{|r|.|F|}[/tex]
[tex]cos\theta=\dfrac{(3i+4j).((-8i)+6j)}{(\sqrt{3^2+4^2} ).(\sqrt{8^2+6^2}) }[/tex]
[tex]cos\theta=\dfrac{0}{50}[/tex]
[tex]\theta=90^{\circ}[/tex]
Your friend is catching a falling basketball after it has passed through the basket. Her hands move straight down while catching the ball. It takes about 0.10 s for the player to lower her hands to stop the ball. Assume the mass of the ball is 0.60 kg, and that the ball has fallen a vertical distance of 1.2 m before reaching the player's hand. Determine the average force that her hands exert on the ball while catching it.
To determine the average force exerted by her hands on the ball while catching it, we can use the equation F = m * a. First, we find the initial velocity at the top of the ball's trajectory. Then, we calculate the acceleration using the change in velocity and time. Finally, we can find the force using the mass and acceleration.
Explanation:To determine the average force that her hands exert on the ball while catching it, we can use the equation F = m * a, where F is the force, m is the mass of the ball, and a is the acceleration. First, we need to find the acceleration of the ball. We can use the equation a = Δv / Δt, where Δv is the change in velocity and Δt is the change in time. Since the ball stops in 0.10 s, the change in velocity is equal to the initial velocity, which is the velocity at the top of the ball's trajectory. Using the equation v = u + a * t, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken, we can find the initial velocity at the top of the trajectory. The initial velocity can be determined using the equation v² = u² + 2as, where s is the vertical distance traveled. Substituting the known values, we can find the initial velocity. Once we have the initial velocity, we can use it to calculate the acceleration, and finally the force.
Let's go step by step. First, we need to find the initial velocity at the top of the trajectory. Using the equation v² = u² + 2as, where v is the final velocity (0 m/s at the top), u is the initial velocity, a is the acceleration (-9.8 m/s²), and s is the vertical distance traveled (1.2 m), we can solve for u: 0 = u² + 2(-9.8 m/s²)(1.2 m). Solving for u, we get u = 3.13 m/s. Now that we have the value of the initial velocity, we can find the acceleration using the equation a = (v - u) / t, where v is the final velocity (0 m/s), u is the initial velocity (3.13 m/s), and t is the time taken (0.10 s). Substituting the values, we get a = (0 m/s - 3.13 m/s) / 0.10 s = -31.3 m/s². Finally, we can calculate the force using the equation F = m * a, where m is the mass of the ball (0.60 kg) and a is the acceleration (-31.3 m/s²). Substituting the values, we get F = (0.60 kg)(-31.3 m/s²) = -18.78 N. Since force is a vector quantity, it is important to note that we have considered the negative sign in the calculation, which indicates that the force is directed upwards.
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The force exerted is approximately 29.1 N in the opposite direction of the ball's motion.
Determining the Average Force Exerted on a Falling Basketball
To find the average force exerted by the player's hands while catching the falling basketball, we need to understand a few key physics concepts involving motion and force.
Steps to Calculate the Average Force:
Calculate the velocity of the ball just before it is caught: Using the equation for free fall, we can determine the final velocity (v) of the ball just before it reaches the player's hands:
[tex]v^2 = u^2[/tex]+ 2gh, where u = initial velocity (0 m/s), g = acceleration due to gravity (9.8 m/s²), and h = height (1.2 m)
[tex]v^2[/tex] = 0 + 2(9.8)(1.2)
[tex]v^2[/tex]= 23.52
v = [tex]\sqrt{23.52}[/tex]≈ 4.85 m/s
Calculate the change in momentum: The change in momentum, or impulse, occurs as the ball is caught and brought to a stop in the player's hands. The ball's mass is 0.60 kg.
Impulse = change in momentum = m x (v - u)
Impulse = 0.60 kg x (0 m/s - 4.85 m/s)
Impulse = -2.91 kg·m/s
Determine the average force: We use the equation for impulse, Impulse = F_avg x Δt, where Δt = time duration (0.10 s).
-2.91 kg·m/s = F_avg x 0.10 s
F_avg = -2.91 kg·m/s / 0.10 s
F_avg ≈ -29.1 N
The negative sign indicates that the force is exerted in the opposite direction of the ball's motion, which is typical for a decelerating force.
A 15.5 kg mass vibrates in simple harmonic motion with a frequency of 9.73 Hz. It has a maximum displacement from equilibrium of +14.6 cm at time, t = 0.00 s. The displacement from equilibrium of the mass at time, t = 1.25 s is_______?
The displacement of a 15.5 kg mass undergoing simple harmonic motion with a frequency of 9.73 Hz, at a point in time 1.25 s after it was at its maximum displacement of 14.6 cm, is found to be -14.1 cm.
Explanation:The displacement of a mass undergoing simple harmonic motion at any given point of time can be found using the formula x(t) = A cos(wt + φ), where 'A' is the amplitude (maximum displacement), 'w' is the angular frequency, and 'φ' is the phase constant. Given that the maximum displacement or amplitude 'A' is 14.6 cm (or 0.146 m), the frequency 'f' is 9.73 Hz, and the phase constant φ = 0 (as the displacement is maximum at t = 0), the angular frequency 'w' can be calculated as 2πf, which equals approximately 61.1 rad/s. Substituting all these values into the formula, we find that the displacement at time t = 1.25 s is x(t) = 0.146 cos(61.1*1.25 + 0) = -0.141 m, or -14.1 cm. Note that the negative sign indicates that the displacement is in the opposite direction of the initial maximum displacement.
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The displacement from equilibrium at time t = 1.25s is -0.042m.
Explanation:The displacement from equilibrium of the mass at time t = 1.25 s can be calculated using the formula for simple harmonic motion. The displacement at any given time t is given by the equation x = A * cos(2πft), where A is the amplitude and f is the frequency. In this case, the amplitude is 0.146 m and the frequency is 9.73 Hz. Plugging in the values, we get x = 0.146 * cos(2π * 9.73 * 1.25), which gives us x = -0.042 m.
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Consider a motor that exerts a constant torque of 25.0N \cdot m to a horizontal platform whose moment of inertia is 50.0kg \cdot m^2 . Assume that the platform is initially at rest and the torque is applied for 12.0rotations . Neglect friction.
Part A ) How much work W does the motor do on the platform during this process?
Enter your answer in joules to four significant figures.
W =
1885
\rm J
Part B ) What is the rotational kinetic energy of the platform K_rot,f at the end of the process described above?
Enter your answer in joules to four significant figures.
K_rot,f =
1885
\rm J
Part C ) What is the angular velocity omega_f of the platform at the end of this process?
Enter your answer in radians per second to three significant figures.
omega_f =
8.68
{\rm rad / s}
Part D ) How long \Delta t does it take for the motor to do the work done on the platform calculated in Part A?
Enter your answer in seconds to three significant figures.
\Delta t =
17.4
\rm s
Part E ) What is the average power P_avg delivered by the motor in the situation above?
Enter your answer in watts to three significant figures.
P_avg = 109 \rm W
Part F ) Note that the instantaneous power P delivered by the motor is directly proportional to omega, so P increases as the platform spins faster and faster. How does the instantaneous power P_f being delivered by the motor at the time t_{\rm f} compare to the average power P_avg calculated in Part E?
Note that the instantaneous power delivered by the motor is directly proportional to , so increases as the platform spins faster and faster. How does the instantaneous power being delivered by the motor at the time compare to the average power calculated in Part E?
P = P_{\rm avg}
P = 2 * P_{\rm avg}
P = P_{\rm avg} / 2
none of the above
Answer:
A) W = 1885 J , B) [tex]K_{f}[/tex] = 1885 J , C) w = 8.68 rad / s , D) t = 8,687 s , E) P = 109 W F) P = 2 [tex]P_{rms}[/tex]
Explanation:
Part A The work in the rotational movement is
W = τ θ
Let's look at the rotated angle
θ = 12.0 rot (2pi rad / 1rot) = 75.398 rad
W = 25.0 75.40
W = 1885 J
Part B Let's use the relationship between work and kinetic energy
W = ΔK = Kf - Ko
As the body leaves the rest w₀ = 0 ⇒ K₀ = 0
W = [tex]K_{f}[/tex] -0
[tex]K_{f}[/tex] = 1885 J
Part C The formula for kinetic energy is
K = ½ I w²
w² = 2k / I
w = √ (2 1885/50)
w = 8.68 rad / s
Part D The power in the rotational movement
P = τ w
P = 25 8.68
P = 217 W
P = W / t
t = W / P
t = 1885/217
t = 8,687 s
Part E At average power is
P = τ ([tex]w_{f}[/tex] -w₀)/ 2
We look for angular velocity with kinematics
[tex]w_{f}[/tex = w₀ + α t
τ = I α
α = τ / I
α = 25/50
α = 0.5 rad / s²
calculate
P = 25 (0.5 8.687)
P = 108.6 W
P = 109 W
Part F
The average power is
[tex]P_{rms}[/tex] = τ ([tex]w_{f}[/tex] -w₀) /
The instant power is
P = τ w
The difference is that in one case the angular velocity is instantaneous and between averages
P / [tex]P_{rms}[/tex] = τ w / (τ ([tex]w_{f}[/tex]-w₀) / 2)
P / [tex]P_{rms}[/tex]= 2 w / Δw
For this case w₀ = o
p / [tex]P_{rms}[/tex] = 2
The motor exerts rotational motion to do 1885 Joules of work on the platform. The final angular velocity of the platform is 8.68 rad/s. The time it takes to do this work is 17.4 seconds. The average power delivered by the motor is 109 Watts and the instantaneous power is twice the average power.
Explanation:A motor exerts a constant torque on a horizontal platform and we need to determine the work done, the rotational kinetic energy, the angular velocity, the time it takes, the average power, and compare the instantaneous power to the average power.
Firstly, the work done by the motor is calculated using the formula W = Torque x angular displacement. The angular displacement for 12.0 rotations will equals to 12.0 x 2π radians. So, W = 25.0 N.m x 12 x 2π rad = 1885 J. Hence, the motor does 1885 Joules of work on the platform.
For the rotational kinetic energy, since there is no friction, all of the work done on the platform is converted into kinetic energy, so, K_rot,f = 1885 J.
The final angular velocity ωf can be found from the rotational kinetic energy and the moment of inertia by the relation K_rot,f = 1/2 I ωf^2. From this, we can find ωf = √(2K_rot,f / I)= 8.68 rad/s.
The time it takes Δt to do the work can be calculated using Δt = angular displacement / average angular velocity. Hence, Δt= (12 x 2π) / ((0 + ωf) / 2)= 17.4 s.
The average power P_avg is given by the total work done divided by the total time, which gives P_avg= W / Δt = 109 Watts.
On the final part, the instantaneous power Pf is proportional to the final angular velocity. As Pf = Torque x ωf, we get Pf = 2 x P_avg. So, the instantaneous power being delivered by the motor at the end is twice the average power calculated before.
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Although the evidence is weak, there has been concern in recent years over possible health effects from the magnetic fields generated by electric transmission lines. A typical high-voltage transmission line is 20 m above the ground and carries a 200 A current at a potential of 110 kV.What is the magnetic field strength on the ground directly under such a transmission line?
The concept necessary to develop this exercise is that of a magnetic field on a surface. The magnetic field is a tool to describe how a magnetic force is distributed in the space around and within something magnetic.
It can be defined as,
[tex]B = \frac{\mu_0 I}{2\pi R}[/tex]
Where,
R= The distance from the point
I = Current
[tex]\mu_0 =[/tex] Permeability constant in free space
Our values are given as,
[tex]\mu_0 = 4\pi * 10^{-7}H/m[/tex]
[tex]I = 200 A[/tex]
[tex]R = 20m[/tex]
Replacing ,
[tex]B = \frac{(4\pi*10^{-7})(200)}{2\pi 20}[/tex]
[tex]B = 2*10^{-6}T[/tex]
Therefore the magnetic field strength on the ground directly under such a transmission line is [tex]2*10^{-6}T[/tex]
The magnetic field strength on the ground directly under a typical high-voltage transmission line carrying a 200 A current at a potential of 110 kV, which is 20 m above the ground, is 2 μT. This is calculated using Ampere's Law. Current evidence does not conclusively support any health hazards associated with such exposure.
Explanation:To answer your question about the strength of the magnetic field on the ground directly under a typical high-voltage transmission line carrying a 200 A current at a potential of 110 kV, we should first understand the basic physics principle. The magnetic field created by an electric current in a long straight wire forms concentric circles around the wire, and the strength B of the magnetic field is given by Ampere's law.
Ampere's law states that B = μ₀I / 2πr, where B is the magnetic field strength, μ₀ is the permeability of free space (4π × 10^-7 T m/A), I is the current, and r is the distance from the wire. In this case, the transmission line is 20 m above the ground, so r = 20 m. The current I = 200 A. Substituting these values into the equation, we get B = (4π × 10^-7 T m/A * 200 A) / (2π * 20 m) = 2 × 10^-6 T or 2 μT.
So, the strength of the magnetic field on the ground directly under the transmission line is 2 μT, which is less than a tenth of Earth's admittedly weak magnetic field. While there's an ongoing controversy regarding potential health hazards associated with exposure to these electromagnetic fields (E-fields), current evidence does not conclusively support these claims.
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An ice skater is spinning at 5.2 rev/s and has a moment of inertia of 0.32 kg * m2.
Calculate the angular momentum, in kilogram meters squared per second, of the ice skater spinning at 5.2 rev/s.
Suppose instead he keeps his arms in and allows friction of the ice to slow him to 2.75 rev/s. What is the magnitude of the average torque that was exerted, in N * m, if this takes 12 s?
Explanation:
The angular momentum is given by the moment of inertia, multiplied by the angular speed of the rotating body:
[tex]L=I\omega[/tex]
The angular speed is given by:
[tex]\omega=2\pi f\\\omega=2\pi 5.2\frac{rev}{s}\\\omega=32.67\frac{rad}{s}[/tex]
Now, we calculate the angular momentum:
[tex]L=0.32kg\cdot m^2(32.67\frac{rad}{s})\\L=10.45\frac{kg\cdot m^2}{s}[/tex]
The average torque is defined as:
[tex]\tau=I\alpha[/tex]
[tex]\alpha[/tex] is the angular acceleration, which is defined as:
[tex]\alpha=\frac{\omega_f-\omega_0}{t}[/tex]
We have to calculate [tex]\omega_f[/tex]:
[tex]\omega_f=2\pi (2.75\frac{rad}{s})\\\omega_f=17.28\frac{rad}{s}[/tex]
Now, we calculate the angular acceleration:
[tex]\alpha=\frac{17.28\frac{rad}{s}-32.67\frac{rad}{s}}{12s}\\\alpha=-1.28\frac{rad}{s^2}[/tex]
Finally, we can know the average torque:
[tex]\tau=0.32kg\cdot m^2(-1.28\frac{rad}{s^2})\\\tau=-0.41N\cdot m[/tex]
(a) The angular momentum of the skater is 10.45 kgm²/s
(b) The magnitude of the average torque that was exerted, is 0.41 Nm.
Angular momentum of the ice skaterThe angular momentum of the skater is calculated as follows;
L = Iω
where;
ω is angular speed (rad/s)ω = 5.2 rev/s x 2π rad = 32.67 rad/s
L = 0.32 x 32.67
L = 10.45 kgm²/s
Angular acceleration of the skaterThe angular acceleration is calculated as follows;
[tex]\alpha= \frac{\omega _f - \omega _i}{t}[/tex]
ωf is the final angular speed = 2.75 rev/s x 2π rad = 17.28 rad/s[tex]\alpha = \frac{17.28 -32.67 }{12} \\\\\alpha = -1.28 \ rad/s^2[/tex]
Average torqueThe magnitude of the average torque that was exerted, is calculated as;
τ = Iα
τ = 0.32 x (1.28)
τ = 0.41 Nm.
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A spaceship in distress sends out two escape pods in opposite directions. One travels at a speed v1 = + 0.70 c in one direction, and the other travels at a speed v2 = − 0.76 c in the other direction, as observed from the spaceship. Part A What speed does the first escape pod measure for the second escape pod? Express your answer using two significant figures.
To develop this problem it is necessary to apply the related concepts at relative speed.
When an observer perceives the relative speed of a second observer, the function is described,
[tex]v' = \frac{v_1-v_2}{1-\frac{v_1v_2}{c^2}}[/tex]
Where,
[tex]v_1[/tex] = The velocity of the first escape pod
[tex]v_2[/tex] = The velocity of the second escape pod
c = Speed of light
v' = Speed of the first escape pod relative to the second escape pod.
Our values are given as,
[tex]v_1[/tex]= 0.7c
[tex]v_2[/tex]= -0.76c
Replacing we have,
[tex]v' = \frac{v_1-v_2}{1-\frac{v_1v_2}{c^2}}[/tex]
[tex]v' = \frac{0.7c-(-0.76c)}{1-\frac{(0.7c)(-0.76c)}{(3*10^8)^2}}[/tex]
[tex]v' = \frac{0.7c-(-0.76c)}{1-\frac{(0.7c)(-0.76c)}{(3*10^8)^2}}[/tex]
[tex]v' = 2.85*10^8m/s[/tex]
Therefore the speed of the first escape pod measure for the second escape pod is [tex]v' = 2.85*10^8m/s[/tex]
The relative speed of the second escape pod as measured by the first escape pod, using relativistic velocity addition, is approximately -0.04c (or just 0.04c considering the magnitude), where c is the speed of light.
Explanation:The question involves calculating the relative speed of one escape pod as observed by the other escape pod in a scenario where they are moving in opposite directions. To do this, we must use the formula for relativistic velocity addition. The formula is as follows:
V = (v1 + v2) / (1 + v1*v2/c²),
where V is the relative velocity as measured by one escape pod, v1 and v2 are the velocities of the escape pods, and c is the speed of light. For this particular problem:
v1 = +0.70c (the velocity of the first escape pod as observed from the spaceship)v2 = -0.76c (the velocity of the second escape pod as observed from the spaceship, note the negative sign because it is in the opposite direction)We substitute the values into the relativistic velocity addition formula to find the relative speed:
V = (0.70c - 0.76c) / (1 - 0.70*(-0.76)c²/c²)
Doing the calculations:
V = -0.06c / (1 + 0.532)c²/c²)
V = -0.06c / 1.532
V = -0.039c ≈ -0.04c
The negative sign indicates that the second escape pod is moving in the direction opposite to the first pod as measured by the first pod. It's important to keep in mind that this result is only an approximation, rounded to two significant figures as per the question's request.
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(a) Steve exerts a steady force of magnitude 210N (about 47 lb.) on the stalled car in fig. 6.3 as he pushesit a distance of 18 m. The care also has a flat tire, so to make the car track straight Steve must push at an angle of 30 degrees to the direction of motion. How much work does Steve do? (b) refer to book.
Answer:
3 300 J
Explanation:
Work done, mathematically speaking, is the distance moved multiplied by the force exerted in the only in direction of motion. In this question therefore we will multiply the distance by the component of the force that moved the car forward
W = Fdcosθ
= 210 N * 18 * Cos 30°
= 3 300 J
The work done by Steve while pushing the car 18 meters at an angle of 30 degrees, applying a force of 210N, is calculated to be 3465 joules using the formula: Work = Force x distance x cos(angle).
Explanation:The subject of this question is Physics, specifically the concept of work done by a force. In physics, work is defined as the product of the component of the force in the direction of the displacement and the magnitude of this displacement.
For Steve's situation, we can calculate the work done using the formula: Work = Force x distance x cos(angle). Given the parameters: Force = 210N, Distance = 18m, and Angle = 30 degrees, we get:
Work = 210N x 18m x cos(30) = 3465 J
So, Steve does 3465 Joules of work to push the car 18 meters under these conditions. This example illustrates how the angle at which force is applied can affect the work output, a key principle in physics.
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