This question asks for the rate of heat transfer from a copper fin, which requires the application of heat transfer theory, including the use of specific equations for fin efficiency and heat loss. However, a sample calculation provided deals with finding the initial temperature of a copper piece mixed with water, based on the conservation of energy and specific heat capacities.
Explanation:The subject of this question involves the calculation of heat transfer from a fin made of copper. Given the material's thermal conductivity (k), the fin's dimensions, and the conditions surrounding it (base temperature, air temperature, and convection heat transfer coefficient h), we can find the rate of heat transfer. However, the actual calculation for the rate of heat transfer from this fin isn't provided directly in the context of the question. For such a problem, one would typically employ the fin equations from heat transfer theory, which might involve the use of Biot number, fin efficiency, and heat loss calculations. Instead, to answer a closely related question with given data, we can calculate the initial temperature of a 248-g piece of copper when dropped into 390 mL of water at 22.6°C, where the final temperature is measured as 39.9°C, assuming all heat transfer occurs between the copper and the water. This problem relies on the conservation of energy principle, requiring the application of the specific heat capacities of copper and water
On Earth, froghoppers can jump upward with a takeoff speed of 2.8 m/s. Suppose you took some of the insects to an asteroid. If it is small enough they can jump free of it and escape into space. (a) What is the diameter (in km) of the largest spherical asteroid from which they could jump free? Assume typical asteroid denisty of 2.0 g/cm^3. (b) Suppose that one of the froghoppers jumped horizontally from a small hill on an asteroid. What would the diameter (in km) of the ateroid need to be so that the insect could go into a circular orbit just above the surace?
To develop this problem it is necessary to apply the concepts related to kinetic energy and gravitational potential energy.
By conserving energy we know that
[tex]PE = KE[/tex]
[tex]\frac{1}{2}mv^2=\frac{GMm}{r}[/tex]
Where,
m = mass
v = Velocity
G = Gravitational universal constant
M = Mass of Spherical asteroid
m = mass of object
r = Radius
Mass of a Sphere can be expressed as,
[tex]M= \rho* (\frac{4}{3}\pi r^3 )[/tex]
Replacing we have that,
[tex]\frac{1}{2}*2.8^2 - 6.67*10^{-11}*\frac{4\pi*2000*r^3}{3r} = 0[/tex]
[tex]6.67*10^{-11}*(\frac{4\pi}3{}2000*r^2)=\frac{1}{2}*2.8^2[/tex]
[tex]r = \sqrt{\frac{1}{2}*\frac{2.8^2}{6.67*10^{-11}*2000*4/3*\pi}}[/tex]
[tex]r=2648 m[/tex]
Therefore the diameter is 5296 m.
b) Applying the concept of gravitational force and centripetal force we have to
[tex]F_g = F_c[/tex]
[tex]\frac{G M m}{r^2} = \frac{m v^2}{r}[/tex]
[tex]\frac{G M}{r} = v^2[/tex]
[tex]6.67*10^{-11}*\frac{(\frac{4\pi}{3}2000*r^3)}{r} = 2.8^2[/tex]
[tex]r=3746 m[/tex]
Therefore the diameter would be 7492 m
The diameter of the largest asteroid from which a froghopper could jump free can be calculated using the escape velocity formula. Similarly, for the froghopper to enter a circular orbit, its horizontal speed needs to equal the required orbital speed, calculated using the gravitational field strength at the surface of the asteroid.
Explanation:To find the diameter of the largest spherical asteroid from which a froghopper could jump free, we need to understand the escape speed, the speed necessary to overcome an object's gravitational field. The formula for escape velocity is √(2*G*M/R), where G is the gravitational constant, M is the mass of the asteroid, and R is its radius (half of the diameter).
In this case, we can solve the formula for the radius, R = 2*G*M/v², where v is the takeoff speed of a froghopper (2.8 m/s). The mass of the spherical asteroid can be calculated from the given density (2.0 g/cm³ or 2000 kg/m³) and the volume of a sphere (4/3*π*R³). Substituting the mass into the formula will provide the radius, and hence diameter, of the asteroid.
For a froghopper to enter a circular orbit just above the surface of the asteroid, its horizontal speed must equal the circular orbital speed, which depends on the gravitational field strength at the surface of the asteroid. It's determined by the formula v = √(G*M/R), which can be rearranged to solve for the required diameter of the asteroid.
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A blood transfusion is being set up in an emergency room for an accident victim. Blood has a density of 1060 kg/m3 and a viscosity of 4.00 10-3 Pa·s. The needle being used has a length of 3.0 cm and an inner radius of 0.25 mm. The doctor wishes to use a volume flow rate through the needle of 4.70 10-8 m3/s. What is the distance h above the victim's arm where the level of the blood in the transfusion bottle should be located? As an approximation, assume that the level of the blood in the transfusion bottle and the point where the needle enters the vein in the arm have the same pressure of one atmosphere. (In reality, the pressure in the vein is slightly above atmospheric pressure.)
h =
Answer:
The distance h above the victim's arm where the level of the blood in the transfusion bottle should be located is 0.254 m
Explanation:
given information:
density, ρ = 1060 [tex]kg/m^{3}[/tex]
viscosity, η = 4 x [tex]10^{-3}[/tex] Pa.s
needle length, L = 3 cm = 0.03 m
radius, r = 0.25 mm = 0.00025 m
volume flow rate, Q = 4.70 x [tex]10^{-8}[/tex] [tex]m^{3} /s[/tex]
according to Poiseuille’s law
Q = (π[tex]r^{4}[/tex]ΔP)/8ηL
ΔP = (8QηL)/(π[tex]r^{4}[/tex])
= 8 (4.70 x [tex]10^{-8}[/tex])(4 x [tex]10^{-3}[/tex])(0.03)/(π[tex]0.00025^{4}[/tex])
= 3676.71
Now we can calculate the distance h
ΔP = ρ g h
h = ΔP / ρ g
= 3676.71/(1060)(9.8)
= 0.354 m
An electron moves through a uniform electric field E = (2.00î + 5.40ĵ) V/m and a uniform magnetic field B = 0.400k T. Determine the acceleration of the electron when it has a velocity v = 8.0î m/s. (Give each component in m/s2.)
Answer:[tex]a=1.75\times 10^{21}\left ( 2\hat{i}5.08\hat{j}\right )[/tex]
Explanation:
Given
Electric Field [tex]\vec{E}=2\hat{i}+5.4\hat{j}[/tex]
[tex]\vec{B}=0.4\hat{k}\ T[/tex]
velocity [tex]\vec{v}=8\hat{i} m/s[/tex]
mass of electron [tex]m=9.1\times 10^{-31} kg[/tex]
Force on a charge Particle moving in Magnetic Field
[tex]F=e\left [ \vec{E}+\left ( \vec{v}\times \vec{B}\right )\right ][/tex]
[tex]a=\frac{e\left [ \vec{E}+\left ( \vec{v}\times \vec{B}\right )\right ]}{m}[/tex]
[tex]a=\frac{1.6\times 10^{-19}\left [ 2\hat{i}+5.4\hat{j}+\left ( 8\hat{i}\times 0.4\hat{k}\right )\right ]}{9.1\times 10^{-31}}[/tex]
[tex]a=1.75\times 10^{21}\left ( 2\hat{i}+5.08\hat{j}\right )\ m/s^2[/tex]
The Lorentz force law is used to calculate the force exerted on an electron in an electromagnetic field. This force is then divided by the mass of the electron to calculate the acceleration.
Explanation:The acceleration of an electron moving in both electric and magnetic fields can be found using the Lorentz force law, which states that the force exerted on a charged particle in an electromagnetic field is the vector sum of the electric and magnetic forces. The formula is F = q(E + v × B), where q is the charge of the particle, E is the electric field, v is the velocity of the particle, and B is the magnetic field. The acceleration a is then calculated as the force divided by the mass m of the electron, which is known to be 9.11 x 10^-31 kg.
First, calculate the force F due to the electric field E (since F = qE) and the magnetic field B (since F = qvBsinθ). Using the given values, E = (2.00î + 5.40ĵ) V/m, v = 8.0î m/s, and B = 0.400k T, we find that Fe = qE = -(1.60 x 10^-19 C)(2.00î + 5.40ĵ) N (the charge of an electron is -1.60 x 10^-19 C), and Fb = qvBsinθ = - (1.60 x 10^-19 C)(8.0î m/s)(0.400 T), since the angle θ between v and B is 90 degrees. Then, the total force F is the vector sum of Fe and Fb.
Finally, calculate the acceleration a by dividing the total force F by the mass m of the electron to get a = F/m.
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You drive 4 km at 30 km/h and then another 4 km at 50 km/h. What is your average speed for the whole 8-km trip?
a. Less than 40 km/h.
b. Not enough information.
c. Equal to 40 km/h.
d. More than 40 km/h.
Answer:
option A
Explanation:
given,
drive 4 km at 30 km/h and
another 4 km at 50 km/h
average speed for the whole 8 Km trip = ?
time for the trip of first 4 km
distance = speed x time
[tex]t_1 = \dfrac{d}{s}[/tex]
[tex]t_1 = \dfrac{4}{30}[/tex]
t₁= 0.1333 hr
time from another 4 km trip
[tex]t_2 = \dfrac{d}{s}[/tex]
[tex]t_2= \dfrac{4}{50}[/tex]
t₂ = 0.08 hr
average speed
[tex]s= \dfrac{d}{t}[/tex]
[tex]s= \dfrac{8}{0.1333 + 0.08}[/tex]
[tex]s= \dfrac{8}{0.2133}[/tex]
s = 37.50 Km/hr
Less than 40 Km/h
the correct answer is option A
The pressure inside a bottle of champagne is 2.5 atm higher than the air pressure outside. The neck of the bottle has an cross sectional area of 3.0 cm^2. What is the frictional force on the cork due to the neck of the bottle?
To solve this problem it is necessary to apply the concepts related to Pressure, Strength and Area.
We know by definition that Pressure is the amount of Force expressed per unit area, that is,
[tex]P = \frac{F}{A}[/tex]
Where,
F = Force
A = Cross-sectional Area
The net pressure on the bottle would be given by the difference between the internal pressure and the atmospheric pressure, therefore
[tex]P_{net} = P_{in}-P_{atm}[/tex]
[tex]P_{net} = 2.5atm-1atm[/tex]
[tex]P_{net} = 1.5atm (\frac{101325Pa}{1atm})[/tex]
[tex]P_{net} = 151987.5Pa[/tex]
The given radio is,
[tex]r = 3cm^2[/tex]
Hence the Cross-sectional Area would be
[tex]A= \pi r^2[/tex]
[tex]A = \pi(3*10^{-2})^2[/tex]
[tex]A = 2.827*10^{-3}m^2[/tex]
Applying the equation for Pressure we have that
[tex]P = \frac{F}{A}[/tex]
[tex]151987.5= \frac{F}{2.827*10^{-3}}[/tex]
[tex]F = 429.66N[/tex]
Therefore the frictional force on the cork due to the neck of the bottle is 429.66N.
In a Broadway performance, an 85.0-kg actor swings from a R = 3.90-m-long cable that is horizontal when he starts. At the bottom of his arc, he picks up his 55.0-kg costar in an inelastic collision. What maximum height do they reach after their upward swing?
To solve this problem it is necessary to apply the concepts related to the conservation of the moment and the conservation of energy.
The conservation of energy implies that potential energy is transformed into kinetic energy or vice versa, depending on the process, in this way
[tex]KE = PE[/tex]
[tex]\frac{1}{2}mv^2 = mgh[/tex]
Where,
m = mass
v = velocity
h = height
g = gravitational acceleration
Re-arrange to find v,
[tex]\frac{1}{2}mv^2 = mgh[/tex]
[tex]\frac{1}{2}v^2 = gh[/tex]
[tex]v = \sqrt{2gh_1}[/tex]
Replacing with our values we have that
[tex]v = \sqrt{2gh_1}[/tex]
[tex]v = \sqrt{2(9.8)(3.9)}[/tex]
[tex]v = 8.74m/s[/tex]
At this point we can find the final speed through the conservation of the momentum, that is
[tex]mv = (m+M)V[/tex]
Here,
m = mass of the first person
M = Mass of the second person
v = Initial velocity
V = Final Velocity
Replacing,
[tex]mv = (m+M)V[/tex]
[tex](85)(8.74) = (85+55)V[/tex]
[tex]V = \frac{(85)(8.74) }{(85+55)}[/tex]
[tex]V = 5.30m/s[/tex]
Applying energy conservation again, but in this case using the values of the final state we have to
[tex]\frac{1}{2}(m+M)V^2 = (m+M)gh_2[/tex]
[tex]\frac{1}{2}V^2 = gh_2[/tex]
[tex]h_2 = \frac{V^2}{2g}[/tex]
[tex]h_2 = \frac{5.3^2}{2*9.8}[/tex]
[tex]h_2 = 1.433m[/tex]
Therefore the maximum height that they reach after their upward swing is 1.433m
Find the intensity of a 55 dB sound given I 0=10^-12W/m^2
2.23x10^-6W/m^2
1.01x10^-7W/m^2
3.16x10^-7W/m^2
5.43x10^-6W/m^2
Answer:
3.16 × [tex]10^{-7}[/tex] W/[tex]m^{2}[/tex]
Explanation:
β(dB)=10 × [tex]log_{10}(\frac{I}{I_{0} })[/tex]
[tex]I_{0}[/tex]=[tex]10^{-12}[/tex] W/[tex]m^{2}[/tex]
β=55 dB
Therefore plugging into the equation the values,
55=10 [tex]log_{10}(\frac{I}{ [tex]10^{-12}[/tex]})[/tex]
5.5= [tex]log_{10}(\frac{I}{ [tex]10^{-12}[/tex]})[/tex]
[tex]10^{5.5}[/tex]= [tex]\frac{I}{10^{-12} }[/tex]
316227.76×[tex]10^{-12}[/tex]= I
I= 3.16 × [tex]10^{-7}[/tex] W/[tex]m^{2}[/tex]
A gas sample enclosed in a rigid metal container at room temperature (20.0∘C) has an absolute pressure p1. The container is immersed in hot water until it warms to 40.0∘C. What is the new absolute pressure p2?
The question is from the domain of Physics, where it deals with the behavior of gases subjected to temperature changes, as per Gay-Lussac's Law. It can be solved using the formula P1/T1 = P2/T2 after converting Celsius to Kelvin.
Explanation:The subject of this question is Physics, and it is suitable for students at the High School level. The question deals with the concept of gases and their behavior when subjected to changes in temperature, specifically within the context of the ideal gas law and Gay-Lussac's Law, which states that the pressure of a gas is directly proportional to its temperature when the volume is held constant. To find the new absolute pressure p2 after the gas is heated from 20.0°C to 40.0°C in a rigid container, we can use Gay-Lussac's Law in the form of P1/T1 = P2/T2, assuming the amount of gas and the volume remains constant. Since temperatures must be in Kelvin for gas law calculations, they will be converted first: T1 = 293.15 K and T2 = 313.15 K. The new pressure p2 can then be calculated by multiplying p1 by the ratio of T2 to T1.
When red light shines on a piece of metal, no electrons are released. When the red light is slowly changed to shorter-wavelength light (basically progressing through the rainbow), nothing happens until yellow light shines on the metal, at which point electrons are released from the metal. If this metal is replaced with a metal having a higher work function, which light would have the best chance of releasing electrons from the metal? If this metal is replaced with a metal having a higher work function, which light would have the best chance of releasing electrons from the metal? And please explain why and list any equations if there are any. Thanks!
A. Red.
B. Blue.
C. Yellow would still work fine.
D. We need to know more about the metals involved.
Answer:
B blue
Explanation:
This is an exercise of photoelectric effect where electrons are removed from the metal by the incident radiation, the explanation of this phenomenon was given by Einstein, they assume that the light is formed by quanta and these collide with the electrons promoting their transfer; The equation that describes the process is
[tex]K_{max}[/tex] = h f - Φ
Where[tex]K_{max}[/tex]is the kinetic energy, maximize the electrons started, h the plan constant, f the frequency and Φ the work function of the metal (link energy)
The radiation with the lowest energy that an electron can start
h f = Φ
Let's analyze the energy of the light rays with the wavelength
c = λ f
f = c / λ
λ red = 700 10⁻⁹ m
f = 3 108/700 10⁻⁹
f₁ = 4.29 10¹⁴ Hz
Yellow λ = 600 10⁻⁹ m
f₂ = 3 108/600 10⁻⁹
f₂ = 5 10¹⁴ Hz
λ blue = 450 10⁻⁹ m
f₃ = 6.67 10¹⁴ Hz
In the initial metal with a work function fi1 yellow light was needed, to start the electrons, when analyzing the initial equation if the work function increases Φ₂> Φ₁ more energy is needed to take out the electrons from the atom (E₂>E₁)
light of greater energy (frequency) than yellow is needed so that light is needed to go BLUE
Adisa pulls a 40-N crate up a 5.0-m long inclined plane at a constant velocity. If the plane is inclined at an angle of 37° to the horizontal and there is a constant force of friction of 10 N between the crate and the surface, what is the net change in potential energy of the crate?
To solve this problem we resort to the concept of potential energy which is given by the equation
[tex]PE = mgh[/tex]
Where,
m= mass
g= Gravitational acceleration
h = Height
The height is given in the form of a component, that is, it is given the length that is 5 m and the angle of 37 degrees, therefore the height would be
[tex]h=Lsin\theta[/tex]
[tex]h = 5*sin37[/tex]
[tex]h = 3m[/tex]
Applying the potential energy formula we have to
[tex]PE = mgh[/tex]
[tex]PE = F_g(h)[/tex]
[tex]PE = 40N*3m[/tex]
[tex]PE = 120J[/tex]
Therefore the net change in potential energy of the crate is 120J.
The net change in potential energy of the crate is 148.3 N.m.
Explanation:To calculate the net change in potential energy of the crate, we can use the equation: ∆PE = (mgh) +fric, where ∆PE is the change in potential energy, m is the mass of the crate, g is the acceleration due to gravity, h is the height of the inclined plane, and fric is the work done by friction. In this case, the work done by friction is given by the formula: fric = Ffric * d * cos(θ), where Ffric is the force of friction, d is the length of the inclined plane, and θ is the angle of the inclined plane with respect to the horizontal.
Plugging in the values, we have: ∆PE = (40 N * 5.0 m * sin(37°)) + (10 N * 5.0 m * cos(37°)). Evaluating the trigonometric functions, we get: ∆PE = (107.8 N·m) + (40.5 N·m) = 148.3 N·m.
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The drawing shows a side view of a swimming pool. The pressure at the surface of the water is atmospheric pressure. The pressure at the bottom of the pool is greater because of the weight of water above it. As a storm approaches, the atmospheric pressure drops. What happens to the pressure at the bottom of the pool? A The pressure at the bottom of the pool decreases by exactly the same amount as the atmospheric pressure decreases. B Nothing happens to the pressure at the bottom of the pool. C The pressure at the bottom of the pool increases. D The pressure at the bottom of the pool decreases, but not as much as the atmospheric pressure decreases.
Answer:
A) To true. he pressure at the bottom of the pool decreases by exactly the same amount as the atmospheric pressure decreases
Explanation:
Let us propose the solution of this problem before seeing the final statements. The pressure increases with the depth of raposin due to the weight of water that is above the person and also the pressure exerted by the atmosphere on the entire pool, the equation describing this process is
P =[tex]P_{atm}[/tex] + ρ g y
Where [tex]P_{atm}[/tex] is the atmospheric pressure, ρ the water density, and 'y' the depth measured from the surface.
Let's examine this equation in we see that the total pressure is directly proportional to the atmospheric pressure and depth
Now we can examine the claims
A) To true. State agreement or with the equation above
B) False. Pressure changes with atmospheric pressure
C) False. It's the opposite
D) False. They are directly proportional
A The pressure at the bottom of the pool decreases by exactly the same amount as the atmospheric pressure decreases
Further explanationHydrostatic pressure is pressure caused by the weight of a liquid.
The weight of a liquid is affected by the force of gravity.
If a liquid is placed in a container, the higher the liquid content in the container, the heavier the liquid content is and the greater the liquid pressure at the bottom of the container.
The hydrostatic pressure of a liquid can be formulated:
[tex]\large{\boxed{\bold {P_h ~ = ~ \rho.g.h}}[/tex]
Ph = hydrostatic pressure (N / m², Pa)
ρ = density of liquid (kg / m³)
g = acceleration due to gravity (m / s²)
h = height / depth of liquid surface (m)
If the container is open, then the atmospheric pressure (P₀) can be entered into the equation.
[tex]\large{\boxed {\bold {P_h ~ = ~ P_o ~ + ~ \rho.g.h}}[/tex]
The magnitude of P₀ is usually equal to = 1.01.10⁵ Pascal (Pa = N / m²) = 1 atm
For example submarines :
The deeper a submarine is, the greater the hydrostatic pressure it experiences, so that the hull / wall of the submarine is made thick to withstand that pressure.
As a storm approaches, the atmospheric pressure drops
Because the hydrostatic pressure is proportional to the atmospheric pressure, the pressure at the bottom of the pool will also be reduced
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A rectangular loop (area = 0.15 m2) turns in a uniform magnetic field with B = 0.20 T. At an instant when the angle between the magnetic field and the normal to the plane of the loop is p/2 rads and increasing at the rate of 0.60 rad/s, what is the magnitude of the emf induced in the loop?
Answer:
[tex]\epsilon = 0.018 V[/tex]
Explanation:
given,
Area of rectangular loop = 0.15 m²
Magnetic field = B = 0.20 T
angle between magnetic field and the normal to the plane = π/2
increasing at the rate = 0.60 rad/s
magnitude of emf induced = ?
induce emf through the loop
[tex]\epsilon = -\dfrac{d\phi}{dt}[/tex]
Φ = BA cos θ
[tex]\epsilon = -\dfrac{BAd(cos \theta)}{dt}[/tex]
[tex]\epsilon = -BA (-sin \theta)\dfrac{\theta)}{dt}[/tex]
[tex]\epsilon = BA(sin \theta)\dfrac{\theta)}{dt}[/tex]
now substituting all the given values
[tex]\epsilon = 0.2 \times 0.15 (sin 90^0)\times 0.6[/tex]
[tex]\epsilon = 0.018 V[/tex]
the magnitude of induced emf is equal to [tex]\epsilon = 0.018 V[/tex]
Final answer:
The magnitude of the induced emf in the loop is 0.18 V.
Explanation:
The magnitude of the emf induced in a rectangular loop can be calculated using the formula ε = (dB/dt)(A cos θ), where ε is the induced emf, dB/dt is the rate of change of the magnetic field, A is the area of the loop, and θ is the angle between the magnetic field and the normal to the plane of the loop.
In this case, the area of the loop is given as 0.15 m2 and the magnetic field B = 0.20 T. The angle θ is given as π/2 radians and is increasing at a rate of 0.60 rad/s.
Plugging these values into the formula, we get ε = (0.20 T)(0.15 m2)(0.60 rad/s)(cos π/2) = 0.18 V.
In an oscillating L C circuit, the maximum charge on the capacitor is 1.5 × 10 − 6 C and the maximum current through the inductor is 5.5 mA.
(a) What is the period of the oscillations?
(b) How much time elapses between an instant when the capacitor is uncharged and the next instant when it is fully charged?
Answer:
T= 1.71×10^{-3} sec= 1.71 mili sec
t_{fc}= 4.281×10^{-4} sec or 0.4281 mili sec
Explanation:
First of all we write equation for current oscillation in LC circuits. Note, the maximum current (I_0)= 5.5 mA is the amplitude of this function. Then, we continue to solve for the angular frequency(ω). Afterwards, we calculate the time period T. qo = maximum charge on capacitor. = 1.5× 10 ^− 6 C
a) I(t) = -ωqosin(ωt+φ)
⇒Io= ωqo
⇒ω= Io/qo
also we know that T= 2π/ω
⇒T= [tex]2\pi\frac{q_0}{I_0}[/tex]
now putting the values we get
= [tex]2\pi\frac{1.5\times10^{-6}}{5.5\times10^{-3}}[/tex]
= 1.71×10^{-3} sec
b) note that the time [tex]t_{fc}[/tex] it takes the capacitor to from uncharge to fully charged is one fourth of the period . That is
[tex]t_{fc}= \frac{T}{4}[/tex]
[tex]t_{fc}= \frac{ 1.71\times10^{-3} }{4}[/tex]
t_{fc}= 4.281×10^{-4} sec or 0.4281 mili sec
At resonance, what is impedance of a series RLC circuit?
a. It depends on many other considerations, such as the values of L and C.
b. less than R
c. equal to R
d. larger than R
Which of the following does the quality factor of the circuit depend on? (Select all that apply.)
a. C
b. the width of the resonance
c. L
d. R
Answer:
c) Equal to R
b) the width of the resonance
Explanation: In general the impedance Z of an electrc circuit is:
Z = R + jX
Now when the circuit is capacitive, the above mentioned relation become Z = R + 1/jwc where ( w = 2πf )
And when the circuit is inductive Z becomes
Z = R + j wl
Resonance condition implies that the reactance created by capacitors are equal at the inductances produced by inductors, in other words the circuit will behaves as it were resistive.
The impedance will be equal to R
The Q factor is[
Q = X/R in which X is the module of either the capacitive or inductive reactance.
Q has an inverse relation with the band width.
A meter stick is at rest on frictionless surface. A hockey puck is going towards the 30cm mark on the stick and is traveling perpendicular to the stick. After the collision the puck is deflected 30 degrees from original path and is traveling half its original speed.
~Mass of a meter stick = 0.05 kg~Mass of hockey puck = 0.17 kg~Initial speed of the hockey puck = 9 m/sA) Choosing an origin at the starting position of the meter stick's center of mass, what is the angular momentum of the hockey puck before the collision.B) What is the angular momentum of the hockey puck after the collision? (Use same origin.)C) What is the velocity (direction and speed) of the stick's center of mass after the collision?D) What is the angular velocity of the stick (assume it will rotate about its center of mass)?
Answer:
A)[tex]0.306k[/tex]
B)[tex]0.1325k[/tex]
C)[tex]v_{s_{f}} =7.65(-i)+17.35(-j)[/tex]
D)[tex]w=41.64 rads^{-1}[/tex]
Explanation:
Given:
hockey puck is moving towards 30cm mark perpendicular to the stick[tex]m_{s} = 0.05kg[/tex][tex]m_{h} =0.17kg[/tex][tex]v_{h_{i} } = 9 ms^{-1}[/tex]after collision the puck is deflected 30°[tex]v_{h_{f} } =4.5 ms^{-1}[/tex]To find the initial angular momentum about origin which is the 50th mark of the metre scale (It's COM) : angular momentum [tex]L[/tex]=[tex]mv[/tex]x[tex]r[/tex] where, v - is the velocity of the puck perpendicular to the radial vector
r - is the radius vector
∴
A) [tex]L_{i} = m_{h}r*v_{h_{i} }\\L_{i}= 0.17*\frac{50-30}{100} (-i)*9(-j)\\L_{i} = 0.306 k[/tex]
B) after collision , it moves 30° from original path;
and it's speed = [tex]\frac{9}{2} =4.5ms^{-1}[/tex];
∴the perpendicular velocity [tex]v_{per}[/tex] = 4.5[tex]cos30[/tex] = [tex]2.25\sqrt{3}[/tex][tex]ms^{-1}[/tex]
⇒[tex]L=m_{h}r*v=0.17*0.2(-i)*2.25\sqrt{3}(-j) \\L= 0.1325 k[/tex]
C) since the net external force on the system is zero , the total momentum of the system can be conserved .
thus ,
[tex]m_{s}v_{s_{i}}+m_{h}v_{h_{i}}=m_{s}v_{s_{f}}+m_hv_{h_{f}}\\0+0.17*9(-j)=0.05*v_{s_{f}}+0.17*(2.25(i)+2.25\sqrt{3}(-j))\\[/tex]
solving this we get,
⇒[tex]v_{s_{f}} =7.65(-i)+17.35(-j)[/tex]
D) since there is no external torque about the system ,the angular momentum can be conserved.
[tex]L_{h_{i}}= L_{h_{f}} + Iw[/tex]
where ,
[tex]w[/tex] is the angular velocity of the stick.
[tex]I[/tex] is the moment of inertia of the stick about COM :
[tex]I =\frac{m_{s}l^{2}}{12} \\m_{s}=0.05kg\\l=1m\\I = 0.004167 kgm^{2}[/tex]
∴
⇒[tex]0.306k=0.1325k+0.004167w\\w=41.64 rads^{-1}[/tex]
A) The angular momentum of the hockey puck before the collision is 0.459 kg.m^2/s. B) The angular velocity of the puck after the collision is 15 rad/s. C) The velocity of the stick's center of mass after the collision is 3.475 m/s.
Explanation:A) The angular momentum of an object is given by the equation:
L = Iω
where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
Since the meter stick is at rest, its angular velocity is zero, and therefore its angular momentum is also zero.
On the other hand, the angular momentum of the hockey puck before the collision can be calculated using the equation:
L = mvr
where m is the mass of the puck, v is its linear velocity, and r is the distance from its axis of rotation.
Using the given values, the angular momentum of the hockey puck before the collision is 0.17 kg * 9 m/s * 0.3m = 0.459 kg.m^2/s.
B) After the collision, the puck is deflected at an angle of 30 degrees and its linear velocity is halved.
The angular velocity of the puck after the collision can be calculated using the equation:
ω = v'/r'
where v' is the new linear velocity of the puck and r' is the new distance from its axis of rotation.
Since the linear velocity of the puck is halved, its new linear velocity is 4.5 m/s.
Assuming the radius of rotation remains the same, the angular velocity of the puck after the collision is 4.5 m/s / 0.3m = 15 rad/s.
C) The velocity of the stick's center of mass after the collision can be calculated using the conservation of linear momentum.
Before the collision, the momentum of the system is zero, since the meter stick is at rest.
After the collision, the momentum of the system is the linear momentum of the puck, which can be calculated using the equation:
p = mv
where m is the mass of the puck and v is its linear velocity.
Using the given values, the linear momentum of the puck after the collision is 0.17 kg * 4.5 m/s = 0.765 kg.m/s.
Since the meter stick and the puck move together after the collision, their velocities are the same and equal to 0.765 kg.m/s divided by the total mass of the system, which is 0.05 kg + 0.17 kg = 0.22 kg.
Therefore, the velocity of the stick's center of mass after the collision is 0.765 kg.m/s / 0.22 kg = 3.475 m/s.
D) The angular velocity of the stick can be calculated using the equation:
ω = v/r
where v is the linear velocity of the stick's center of mass and r is the radius of the stick.
Using the given values, the linear velocity of the stick's center of mass after the collision is 3.475 m/s.
Assuming the radius remains the same, the angular velocity of the stick is 3.475 m/s / 0.3m = 11.583 rad/s.
The water stream strikes the inclined surface of the cart. Determine the power produced by the stream if, due to rolling friction, the cart moves to the right with a constant velocity of 2 m/s. The discharge from the 50-mm-diameter nozzle is 0.04 m3/s. One-fourth of the discharge flows down the incline, and three-fourths flows up the incline. Assume steady flow, all within the vertical plane.
Answer
given,
constant speed of cart on right side = 2 m/s
diameter of nozzle = 50 mm = 0.05 m
discharge flow through nozzle = 0.04 m³
One-fourth of the discharge flows down the incline
three-fourths flows up the incline
Power = ?
Normal force i.e. Fn acting on the cart
[tex]F_n = \rho A (v - u)^2 sin \theta[/tex]
v is the velocity of jet
Q = A V
[tex]v = \dfrac{0.04}{\dfrac{\pi}{4}d^2}[/tex]
[tex]v= \dfrac{0.04}{\dfrac{\pi}{4}\times 0.05^2}[/tex]
v = 20.37 m/s
u be the speed of cart assuming it to be u = 2 m/s
angle angle of inclination be 60°
now,
[tex]F_n = 1000 \times \dfrac{\pi}{4}\times 0.05^2\times (20.37 - 2)^2 sin 60^0[/tex]
F n = 2295 N
now force along x direction
[tex]F_x = F_n sin 60^0[/tex]
[tex]F_x = 2295 \times sin 60^0[/tex]
[tex]F_x = 1987.52\ N[/tex]
Power of the cart
P = F x v
P = 1987.52 x 20.37
P = 40485 watt
P = 40.5 kW
The power produced by the stream is; 40.5 kW
What is the power produced?We are given;
Constant speed of cart on right side; u = 2 m/s
Diameter of nozzle; d = 50 mm = 0.05 m
Discharge flow through nozzle; Q = 0.04 m³/s
Formula for the normal force acting on the cart is;
Fₙ = ρA(v - u)²sin θ
Let us find v the speed of the jet from;
Q = Av
v = Q/A
v = 0.04/(π * 0.05²/4)
v = 20.37 m/s
From online research about this question, the angle of inclination is 60°. Thus;
Fₙ = ρA(v - u)²sin θ
Fₙ = 1000 * (π * 0.05²/4) * (20.37 - 2)sin 60
Fₙ = 2295 N
Force along the x-direction will be;
Fₓ = Fₙ * sin 60
Fₓ = 2295 * sin 60
Fₓ = 1987.52 N
Power of the Cart is;
P = Fₓ * v
P = 1987.52 * 20.37
P = 40.5 kW
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A manometer that is filled with water (rho=1000 kg/m3) is connected to a wind tunnel to measure the pressure inside the test section. The other manometer is open to the atmosphere and the atmospheric pressure is 100 kPa. The difference in the manometer level is 100 mm. Determine the absolute pressure in the test section in kPa.
Answer:
99.019 kPa
Explanation:
[tex]P_a[/tex] = Atmospheric pressure = 100 kPa
[tex]\rho[/tex] = Density of water in manometer = 1000 kg/m³
g = Acceleration due to gravity = 9.81 m/s²
h = Difference in height = 100 mm
As the pressure in the test section is lower than the pressure of the atmosphere it will be subtracted.
The absolute pressure in the test section is given by
[tex]P_{ab}=P_a-\rho gh\\\Rightarrow P_{ab}=100\times 10^3-1000\times 9.81\times 100\times 10^{-3}\\\Rightarrow P_{ab}=99019\ Pa=99.019\ kPa[/tex]
The absolute pressure in the test section is 99.019 kPa
Observations of the Crab Nebula taken over several decades show that gas blobs that are now 100 arcseconds from the center of the nebula are moving away from the center by about 0.11 arcsecond per year. Use that information to estimate the year in which the explosion ought to have been observed.
Answer:
Estimation: year 1110.
Explanation:
We need to know how much time it takes to move 100 arcseconds if it moves at 0.11 arcsecond per year. Similarly to any velocity equation [tex]v=\frac{d}{t}[/tex], where in our case the distances are angular, we will obtain the time by doing:
[tex]t=\frac{d}{v}=\frac{100arc}{0.11arc/year}=909 years[/tex]
Which, considering from 2019, the explosion ought to have been observed around 1110 (in reality it was observed by Chinese astronomers in 1054).
The explosion of the Crab Nebula was estimated to have been observed around 1054 AD based on the given information.
Explanation:To estimate the year in which the explosion of the Crab Nebula was observed, we can use the information provided. The gas blobs that are now 100 arcseconds from the center of the nebula are moving away from the center by about 0.11 arcsecond per year. By dividing the distance of 100 arcseconds by the rate of motion of 0.11 arcsecond per year, we can estimate that it would have taken approximately 909 years for the gas blobs to reach their current position. Therefore, the explosion of the Crab Nebula would have been observed around 1054 AD.
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A student is doing a lab experiment in which she uses diffraction to measure the width of a shaft of her hair. She shines a 530 nm green laser pointer on a single hair, which produces a diffraction pattern on a screen 1.2 m away. The width of the central maximum of the pattern is 14 mm. a. What is the thickness of the hair? (91µm) b. If she chose a wider shaft of hair, how would this change the width of the central maximum? (b): For a single slit, we know that a wider slit gives a narrower pattern; the same is true for the hair.
Answer:
.09 mm
Explanation:
In case of diffraction by single slit with width a the width of central maxima is given below
width = 2 λD/a
where λ is wavelength of light , D is distance of screen , a is width o slit
substituting the given values
14 x 10⁻³ = [tex]\frac{2\times530\times10^{-9}\times1.2}{a}[/tex]
a = .09 mm
b) If a is greater , width of central maxima will be less wide.
A string that is 3.6 m long is tied between two posts and plucked. The string produces a wave that has a frequency of 320 Hz and travels with a speed of 192 m/s. How many full wavelengths of the wave fit on the string?
To solve this problem it is necessary to apply the concepts related to wavelength depending on the frequency and speed. Mathematically, the wavelength can be expressed as
[tex]\lambda = \frac{v}{f}[/tex]
Where,
v = Velocity
f = Frequency,
Our values are given as
L = 3.6m
v= 192m/s
f= 320Hz
Replacing we have that
[tex]\lambda = \frac{192}{320}[/tex]
[tex]\lambda = 0.6m[/tex]
The total number of 'wavelengths' that will be in the string will be subject to the total length over the size of each of these undulations, that is,
[tex]N = \frac{L}{\lambda}[/tex]
[tex]N = \frac{3.6}{0.6}[/tex]
[tex]N = 6[/tex]
Therefore the number of wavelengths of the wave fit on the string is 6.
The most distant galaxies observed to date have a redshift of approximately z =10. Part A How does the wavelength of light we observe from those galaxies compare with its original wavelength when it was emitted? Express your answer as an integer.
To solve the problem it is necessary to apply the concepts related to wavelength and error calculation.
The calculation of the error is usually defined by
[tex]z = \frac{\lambda_o-\lambda_r}{\lambda_r}[/tex]
Where
[tex]\lambda_o =[/tex] Wavelength from the observer
[tex]\lambda_r =[/tex] Wavelength from the rest
The previous formula is exactly equal to,
[tex]z= \frac{\lambda_o}{\lambda_r}-1[/tex]
[tex](z+1) = \frac{\lambda_o}{\lambda_r}[/tex]
[tex]\frac{\lambda_o}{\lambda_r}=(10+1)[/tex]
[tex]\lambda_o=11 \lambda_r[/tex]
Therefore the observed wavelength will be 11 times longer that the emitted wavelength.
If we make the comparison for the ultraviolet region (from 10nm to 400nm) we get that,
[tex]\lambda_o=11 \lambda_r[/tex]
[tex]\lambda_o=11 (\frac{10nm*400nm}}{2})[/tex]
[tex]\lambda_o = 2255nm[/tex]
That is the infrared region of electromagnetic spectra.
The wavelength of light we observe from these galaxies compares to its original wavelength when it was emitted, a function of the relative velocity between the emitting source and the observing receiver.
The redshiftThe redshift (z) corresponds to a change in the way in which the frequency of light photon waves is observed in the spectroscope as a function of the relative velocity between the emitting source and the observing receiver (a measure of distance, of according to cosmological expansion, based on recessional velocity).
A decrease in frequency occurs as the distance from the source galaxy increases, causing the light ray to be captured as a standard color. The deviation or distance measurement is represented by the letter z.
With this information, we can conclude that the wavelength of light we observe from these galaxies compares to its original wavelength when it was emitted, a function of the relative velocity between the emitting source and the observing receiver.
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A 155 kg satellite is orbiting on a circular orbit 5995 km above the Earth's surface. Determine the speed of the satellite. (The mass of the Earth is 5.97×1024 kg, and the radius of the Earth is 6370 km.)
Answer:[tex]5.67\times 10^{3} m/s[/tex]
Explanation:
Given
mass of satellite [tex]m=155 kg[/tex]
Satellite is orbiting 5995 km above Earth surface
mass of Earth [tex]M=5.97\times 10^{24} kg[/tex]
Radius of Earth [tex]R=6370 km[/tex]
here [tex]r=R+5995=6370+5995=12,365 km[/tex]
Gravitational Force will Provide the centripetal Force
[tex]\frac{GMm}{r^2}=\frac{mv^2}{r}[/tex]
[tex]v=\sqrt{\frac{GM}{r}}[/tex]
[tex]v=\sqrt{\frac{6.67\times 10^{-11}\times 5.97\times 10^{24}}{12365\times 10^{3}}}[/tex]
[tex]v=\sqrt{32.203\times 10^6}[/tex]
[tex]v=5.67\times 10^3m/s=5674.7 km/s[/tex]
A kettle bell is a piece of gym equipment used in workouts. The bottom part of the kettle bell is a sphere made completely out of iron. What is the weight of just the sphere if it has a radius is 3.4 inches and iron weights 0.28 pounds per cubic inch?
Answer:204.9 N
Explanation:
Given
Radius [tex]r=3.4 in.[/tex]
Density of iron weight [tex]\rho =0.28 pound/in.^3[/tex]
Volume of sphere[tex]=\frac{4\pi r^3}{3}[/tex]
[tex]V=\frac{4\pi (3.4)^3}{3}[/tex]
[tex]V=164.65\ in.^3[/tex]
mass of sphere [tex]m=\rho \cdot V[/tex]
[tex]m=0.28\times 164.65=46.10 Pounds\approx 20.91 kg[/tex]
[tex]weight=mg=20.91\times 9.81=204.91\ N\ or\ 46.063353\ Pound-Force [/tex]
The weigh of the sphere depend on its mass. The weigh of the sphere is 204.722 N.
What is the mass?Mass is defined as the amount of matter contained in a physical body.
Given that the radius r of the sphere is 3.4 inches and the density [tex]\rho[/tex] of the iron is 0.28 pounds per cubic inch.
The volume of the sphere is calculated as below.
[tex]V = \dfrac {4}{3}\times \pi \times r^3[/tex]
[tex]V = \dfrac {4}{3} \times 3.14\times (3.4)^3[/tex]
[tex]V = 164.55 \;\rm cube\;Inches[/tex]
Now the mass of the sphere is calculated as given below.
[tex]m = \rho \times V[/tex]
[tex]m = 0.28 \times 164.55[/tex]
[tex]m = 46.074 \;\rm Pounds[/tex]
We know that 1 pound is equal to 0.453592 kg. So,
[tex]m = 46.074 \times 0.453592 \;\rm kg[/tex]
[tex]m = 20.89 \;\rm kg[/tex]
The weight of the sphere is given below.
[tex]W = mg[/tex]
Where g is the gravitational acceleration.
[tex]W = 20.89 \times 9.8[/tex]
[tex]W = 204.722 \;\rm N[/tex]
Hence we can conclude that the weigh of the sphere is 204.722 N.
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Suppose the person, whose mass is m, is being held up against the wall with an angular velocity of ω ′ = 2 ωmin.The magnitude of the frictional force between the person and the wall is 1. F = 1 5 m g . 2. F = 1 4 m g . 3. F = m g . 4. F = 1 2 m g . 5. F = 4 m g . 6. F = 2 m g . 7. F = 1 3 m g . 8. F = 5 m g . 9. F = 3 m g .
Answer:
3. F = m g
Explanation:
The centripetal force the person is exerting on the wall is;
[tex]f = \frac{mv^{2}}{r}[/tex]
The maximum force due to static friction is f = µ R ,
where R is the normal force exerted by the wall on the person. For the person to be held up against the wall, the maximal friction force must be larger than the force of gravity mg.
The actual force is now equal to or less than the maximum µR
f ≤ fmax = µR
When the angular velocity increase beyond the minimum angular velocity ωmin that is required to hold the person against the wall, the frictional force does to increase any more. Therefore, f still equals mg regardless of the maximum frictional force.
The magnitude of the frictional force between the person and the wall can be determined by considering Newton's laws of motion. According to Newton's second law, the net force on an object is equal to its mass multiplied by its acceleration. In this case, the person is being held against the wall with an angular velocity, which implies a centripetal acceleration. The frictional force between the person and the wall acts as the centripetal force, so we can equate the two.
Explanation:The magnitude of the frictional force between the person and the wall can be determined by considering Newton's laws of motion. According to Newton's second law, the net force on an object is equal to its mass multiplied by its acceleration. In this case, the person is being held against the wall with an angular velocity, which implies a centripetal acceleration. The frictional force between the person and the wall acts as the centripetal force, so we can equate the two:
Ffriction = mω²r
Where Ffriction is the frictional force, m is the mass of the person, ω is the angular velocity, and r is the distance between the person and the axis of rotation (which is the radius of the cylinder in this problem). Therefore, the correct answer is option 9, F = 3mg.
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A spaceship of mass 8600 kg is returning to Earth with its engine turned off. Consider only the gravitational field of Earth. Let R be the distance from the center of Earth. In moving from position R1 = 7300 km to position R1 = 6700 km the kinetic energy of the spaceship increases by _____.
The kinetic energy of the spaceship increases by approximately 0.379 × 10^9 Joules as it moves from R1 = 7300 km to R2 = 6700 km in the gravitational field of Earth.
Explanation:To find the change in kinetic energy of the spaceship as it moves from position R1 = 7300 km to position R1 = 6700 km in the gravitational field of Earth, we can use the gravitational potential energy and the fact that the mechanical energy is conserved. The change in kinetic energy is equal to the negative of the change in potential energy.
The gravitational potential energy (U) is given by the formula:
U = -G * (m1 * m2) / R
Where:
- G is the universal gravitational constant (approximately 6.674 × 10^(-11) N·(m/kg)^2).
- m1 is the mass of the Earth (approximately 5.972 × 10^24 kg).
- m2 is the mass of the spaceship (8600 kg in this case).
- R is the distance from the centre of the Earth.
Now, let's calculate the potential energy at R1 = 7300 km and R2 = 6700 km:
For R1 = 7300 km:
R1 = 7300 km = 7,300,000 meters
U1 = -G * (m1 * m2) / R1
U1 = -6.674 × 10^(-11) N·(m/kg)^2 * (5.972 × 10^24 kg * 8600 kg) / 7,300,000 m
U1 ≈ -5.125 × 10^9 J
For R2 = 6700 km:
R2 = 6700 km = 6,700,000 meters
U2 = -G * (m1 * m2) / R2
U2 = -6.674 × 10^(-11) N·(m/kg)^2 * (5.972 × 10^24 kg * 8600 kg) / 6,700,000 m
U2 ≈ -5.504 × 10^9 J
Now, let's find the change in potential energy (ΔU) as the spaceship moves from R1 to R2:
ΔU = U2 - U1
ΔU ≈ (-5.504 × 10^9 J) - (-5.125 × 10^9 J)
ΔU ≈ -0.379 × 10^9 J
So, the change in potential energy is approximately -0.379 × 10^9 Joules.
The change in kinetic energy is equal to the negative of the change in potential energy, so:
Change in kinetic energy = -ΔU
Change in kinetic energy ≈ 0.379 × 10^9 J
Therefore, the kinetic energy of the spaceship increases by approximately 0.379 × 10^9 Joules as it moves from R1 = 7300 km to R2 = 6700 km in the gravitational field of Earth.
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A 3.49 x 10-2-kg block is resting on a horizontal frictionless surface and is attached to a horizontal spring whose spring constant is 111 N/m. The block is shoved parallel to the spring axis and is given an initial speed of 10.3 m/s, while the spring is initially unstrained. What is the amplitude of the resulting simple harmonic motion?
Answer:
18.2 cm
Explanation:
The kinetic energy of the block
= .5 x 3.49 x 10⁻² x 10.3²
= 1.85 J
If x be the compression created in the spring due to this kinetic energy
1.85 = 1/2 k x²
= .5 x 111 x²
x² = 3.3 x 10⁻²
x = 0.182 m .
= 18.2 cm
This will be the amplitude of oscillation under SHM.
An object of mass 0.50 kg is transported to the surface of Planet X where the object's weight is measured to be 20 N. The radius of the planet is 4.0 ´ 106 m. What free fall accerleration will the 0.5 kg object experience when at the surface of Planet X. What is the mass of Planet X? (G = 6.67 ´ 10-11 N × m2/kg2)
Answer:
Mass of the planet, [tex]M=9.59\times 10^{24}\ kg[/tex]
Explanation:
Given that,
Mass of the object, m = 0.5 kg
Weight of the object, W = 20 N
Radius of the planet, [tex]r=4\times 10^6\ m[/tex]
The weight of an object is given by :
W = mg
g is the acceleration due to gravity on the surface of Planet
[tex]g=\dfrac{W}{m}[/tex]
[tex]g=\dfrac{20}{0.5}[/tex]
[tex]g=40\ m/s^2[/tex]
The expression for g is given by :
[tex]g=\dfrac{GM}{r^2}[/tex]
[tex]M=\dfrac{r^2g}{G}[/tex]
[tex]M=\dfrac{(4\times 10^6)^2\times 40}{6.67\times 10^{-11}}[/tex]
[tex]M=9.59\times 10^{24}\ kg[/tex]
So, the mass of the planet X is [tex]9.59\times 10^{24}\ kg[/tex]. Hence, this is the required solution.
The free-fall acceleration (g) on Planet X is 40 m/s². The estimated mass of Planet X, derived using Newton's law of universal gravitation and provided data, is 9.57 * 10⁵⁴ kg.
Explanation:The free-fall acceleration, or the acceleration due to gravity, can be found by using the equation F (force which is the weight of the object) = m (mass) * g (acceleration due to gravity). Given the weight as 20 N and the mass is 0.50 kg, we can rearrange the equation to find g = F/m = 20 N /0.50 kg = 40 m/s².
To determine the mass of Planet X (M), we use Newton's law of universal gravitation: F = G * (m * M) / r², where F is the force (the weight of the object), G is the universal gravitational constant, m is the object's mass, and r is the radius of the planet. We already know g (acceleration due to gravity), which replaces F/m in the equation, thus becoming g = G * M / r². After rearranging the formula to solve for M, M = g*r²/G = 40 m/s² * (4.0 * 10⁶ m)² / (6.67 * 10^-11 N*m²/kg²) = 9.57 * 10⁵⁴ kg.
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A major difference between AC and DC electricity is in its usefulness to effectively supply electricity to large groups. The best explanation for the difference is which of these statements?
A) DC current can be used in transformers while AC current cannot.
B) AC current can be used in transformers while DC current cannot.
C) AC current fluctuates too much to be used effectively in power generation.
D) DC current fluctuates too much to be used effectively in power generation.
Answer:
B) AC current can be used in transformers while DC current cannot.
Explanation:
AC current is used for long distance transportation of electrical energy with amplifying the voltage and reducing the current in the conductors so that there is minimum loss of energy in the form of heat.
The relation between heat energy and electrical current is given by Joule's law as:
[tex]Q=i^2.R.t[/tex]
where:
Q = amount of heat energy generatedi = current in the conductorR = resistance of the conductort = time for which the current flowsThe above mentioned variation of current and voltage in the transmission lines of AC is achieved by a device called transformer. It consist of a rectangular shaped steel core common to two of the insulated wire winding of which one side acts as input (primary coil) and the other one acts as secondary (output).
AC i.e. alternating current can only be used in the transformers because they have the continuously varying amplitude which in turn provides the change in flux with respect to the time inducing an emf in the nearby secondary coil of the transformer without any physically moving parts.
The alternating nature of AC current aloows changes in magnet fields to be used to transform electricity and make it more efficient to transport over long distances. AC current can be used in transfromers while DC current cannot.
A 145.0 km long electric power line made out of copper is installed during the Spring months at an average temperature of 13.7 °C. What is the change in the length of the power line on a hot Summer day, when the temperature is 42.0 °C? The coefficient of linear expansion for copper is 1.62×10-5 1/C°. Use positive sign for an increase in length and negative for a decrease.
Answer:
66.4767 m
Explanation:
[tex]L_0[/tex] = Original length of power line = 145 km
[tex]\alpha[/tex] = Coefficient of linear expansion = [tex]1.62\times 10^{-5}\ /^{\circ}C[/tex]
Initial temperature = 13.7°C
Final temperature = 42°C
Change in length of a material is given by
[tex]\Delta L=\alpha L_0\Delta T\\\Rightarrow \Delta L=1.62\times 10^{-5}\times 145000\times (42-13.7)\\\Rightarrow \Delta=66.4767\ m[/tex]
The change in length will be 66.4767 m
|| Climbing ropes stretch when they catch a falling climber, thus increasing the time it takes the climber to come to rest and reducing the force on the climber. In one standardized test of ropes, an 80 kg mass falls 4.8 m before being caught by a 2.5-m-long rope. If the net force on the mass must be kept below 11 kN, what is the minimum time for the mass to come to rest at the end of the fall
To solve this problem it is necessary to apply the concepts related to Newton's second law and the kinematic equations of movement description.
Newton's second law is defined as
[tex]F = ma[/tex]
Where,
m = mass
a = acceleration
From this equation we can figure the acceleration out, then
[tex]a = \frac{F}{m}[/tex]
[tex]a = \frac{11*10^3}{80}[/tex]
[tex]a = 137.5m/s[/tex]
From the cinematic equations of motion we know that
[tex]v_f^2-v_i^2 = 2ax[/tex]
Where,
[tex]v_f =[/tex]Final velocity
[tex]v_i =[/tex]Initial velocity
a = acceleration
x = displacement
There is not Final velocity and the acceleration is equal to the gravity, then
[tex]v_f^2-v_i^2 = 2ax[/tex]
[tex]0-v_i^2 = 2(-g)x[/tex]
[tex]v_i =\sqrt{2gx}[/tex]
[tex]v_i = \sqrt{2*9.8*4.8}[/tex]
[tex]v_i = 9.69m/s[/tex]
From the equation of motion where acceleration is equal to the velocity in function of time we have
[tex]a = \frac{v_i}{t}[/tex]
[tex]t = \frac{v_i}{a}[/tex]
[tex]t =\frac{9.69}{137.5}[/tex]
[tex]t = 0.0705s[/tex]
Therefore the time required is 0.0705s
The minimum time for the mass to come to rest at the end of the fall is about 0.071 s
[tex]\texttt{ }[/tex]
Further explanationNewton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.
[tex]\large {\boxed {F = ma }[/tex]
F = Force ( Newton )
m = Object's Mass ( kg )
a = Acceleration ( m )
[tex]\texttt{ }[/tex]
[tex]\large {\boxed {F = \Delta (mv) \div t }[/tex]
F = Force ( Newton )
m = Object's Mass ( kg )
v = Velocity of Object ( m/s )
t = Time Taken ( s )
Let us now tackle the problem !
[tex]\texttt{ }[/tex]
Given:
mass of climber = m = 80 kg
height of fall = h = 4.8 m
net force = ∑F = 11 k N = 11000 N
Asked:
minimum time = t = ?
Solution:
FIrstly , we could find the initial velocity of climber as he caught by the rope:
[tex]v^2 = u^2 + 2gh[/tex]
[tex]v^2 = 0^2 + 2(9.8)(4.8)[/tex]
[tex]v^2 = 94.08[/tex]
[tex]v = \frac{28}{3}\sqrt{5} \texttt{ m/s}[/tex]
[tex]\texttt{ }[/tex]
Next , we will use Newton's Law of Motion to calculate the minimum time:
[tex]\Sigma F = \Delta p \div t[/tex]
[tex]\Sigma F = m \Delta v \div t[/tex]
[tex]11000 = 80 (\frac{28}{3}\sqrt{5}) \div t[/tex]
[tex]t = 80 (\frac{28}{3}\sqrt{5}) \div 11000[/tex]
[tex]t \approx 0.071 \texttt{ s}[/tex]
[tex]\texttt{ }[/tex]
Learn moreImpacts of Gravity : https://brainly.com/question/5330244Effect of Earth’s Gravity on Objects : https://brainly.com/question/8844454The Acceleration Due To Gravity : https://brainly.com/question/4189441Newton's Law of Motion: https://brainly.com/question/10431582Example of Newton's Law: https://brainly.com/question/498822[tex]\texttt{ }[/tex]
Answer detailsGrade: High School
Subject: Physics
Chapter: Dynamics