Answer:
The weight at a distance 4R from the center of earth is 10.37 N.
Explanation:
Given that,
Weight = 166 N
Distance = 4R
Let m be the mass of the object.
We know that,
Mass of earth [tex]M_{e}=5.98\times10^{24}\ kg[/tex]
Gravitational constant[tex]G = 6.67\times10^{-11}\ N-m^2/kg^2[/tex]
Radius of earth [tex]R = 6.38\times10^{6}\ m[/tex]
We need to calculate the weight at a distance 4 R from the center of earth
Using formula of gravitational force
[tex]W = \dfrac{GmM_{e}}{R^2}[/tex]
Put the value in to the formula
[tex]166=\dfrac{6.67\times10^{-11}\times m\times5.98\times10^{24}}{(6.38\times10^{6})^2}[/tex]
[tex]m=\dfrac{166\times(6.38\times10^{6})^2}{6.67\times10^{-11}\times5.98\times10^{24}}[/tex]
[tex]m=16.94 kg[/tex]
Now, Again using formula of gravitational
[tex]W=\dfrac{6.67\times10^{-11}\times 16.94\times5.98\times10^{24}}{(4\times6.38\times10^{6})^2}[/tex]
[tex]W=10.37 N[/tex]
Hence, The weight at a distance 4R from the center of earth is 10.37 N.
Frequency and velocity of a particle in simple harmonic motion Problem in that the amplitude is 0.24 in. and the maximum acceleration is 225 ft/s maximum velocity of the particle ist/s, and the frequency of its motion s z Print 20r
Answer:
velocity maximum = 21.6 ft /s
frequency = 16.88 Hz
Explanation:
Amplitude, A = 0.24 in = 0.02 ft
maximum acceleration, a = 225 ft/s\
The formula for maximum acceleration is
a = ω² A
225 = ω² x 0.02
ω² = 11250
ω = 106.06 rad/s
Maximum velocity, v = ω A
v = 106.06 x 0.02 = 2.1 ft/s
Let f be the frequency
ω = 2 x 3.14 x f
f = 106.06 / (2 x 3.14) = 16.88 Hz
velocity maximum = 21.6 ft /s
frequency = 16.88 Hz
Amplitude, A = 0.24 in = 0.02 ft
maximum acceleration, a = 225 ft/s
The formula for maximum acceleration is
a = ω² A
225 = ω² x 0.02
ω² = 11250
ω = 106.06 rad/s
Maximum velocity, v = ω A
v = 106.06 x 0.02 = 2.1 ft/s
Let f be the frequency
ω = 2 x 3.14 x f
f = 106.06 / (2 x 3.14) = 16.88 Hz
One billiard ball is shot east at 2.00 m/s. A second, identical billiard ball is shot west at 1.00 m/s. The balls have a glancing collision, not a head-on collision, deflecting the second ball by 90° and sending it north at 1.41 m/s. What are the speed and direction of the first ball after the collision?
Answer:
Velocity is 1.73 m/s along 54.65° south of east.
Explanation:
Let unknown velocity be v, mass of billiard ball be m and east direction be positive x axis.
Here momentum is conserved.
Initial momentum = Final momentum
Initial momentum = m x 2i + m x (-1)i = m i
Final momentum = m x v + m x 1.41 j = mv + 1.41 m j
Comparing
mi = mv + 1.41 m j
v = i - 1.41 j
Magnitude of velocity
[tex]v=\sqrt{1^2+(-1.41)^2}=1.73m/s[/tex]
Direction,
[tex]\theta =tan^{-1}\left ( \frac{-1.41}{1}\right )=-54.65^0[/tex]
Velocity is 1.73 m/s along 54.65° south of east.
Using the law of conservation of momentum, one can deduce the speed and direction of the first ball after the collision. It's found to be traveling east at 1 m/s.
Explanation:The scenario described is an example of a two-dimensional collision. In such collision, the law of conservation of momentum applies both in the east-west direction (x-axis) and the north-south direction (y-axis).
With the given velocities before the collision, the total momentum in the x-axis before the collision is: momentum_east = mass*velocity = m*2.00 m/s, and momentum_west = m*(-1.00) m/s. Therefore, total momentum in the x-axis before the collision = m*2.00 m/s + m*(-1.00) m/s = m m/s.
After the collision, the first ball keeps moving in the east-west direction (since the second ball is deflected north), but we don't know its velocity, let's call it v1. Applying the conservation of momentum after the collision in the x-direction, we get: total momentum = m*v1 + 0 (since the second ball no longer moves in the east-west direction) = m m/s. From this, we can solve for v1 and find that v1 = 1 m/s east. Thus, the first billiard ball is traveling east at 1 m/s after the collision.
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A uniform plank is 8-m long and weighs 99 N. It is balanced on a sawhorse at its center. An additional 182 N weight is now placed on the left end of the plank. To keep the plank balanced, it must be moved what distance (in m) to the right? Round your answer to the nearest tenth.
Answer:
It must be moved by 2.59 m to the right
Explanation:
To keep the plank in uniform state, the moment at sawhorse = 0
Let y be the distance of saw horse from left of plank.
Weight per unit length of plank [tex]=\frac{99}{8}=12.375N/m[/tex]
Taking moment about saw horse
Moment due to left portion,
182 x y + 12.375 x y x 0.5 y = 182 y + 6.1875 y²
Moment due to right portion,
12.375 x (8-y) x 0.5 (8-y) = 396 + 6.1875 y² - 99y
So to keep balance these moments should be same
182 y + 6.1875 y² = 396 + 6.1875 y² - 99y
281 y = 396
y = 1.41 m
Distance moved to right = 0.5 x 8 - 1.41 = 2.59 m
So it must be moved by 2.59 m to the right
You are in downtown Chicago (where streets run N-S and E-W). You started from 600 N. Michigan Avenue, and walked 3 blocks toward north, 4 blocks toward west, and 1 block toward north to a train station. Length of one block is 100 [m]. What is the magnitude and direction of your displacement from the start position?
Answer:
Displacement is 565.69 m at 45° west of north
Explanation:
Let north represent positive y axis and east represent positive x axis.
We have journey started from 600 N. Michigan Avenue, and walked 3 blocks toward north, 4 blocks toward west, and 1 block toward north to a train station.
3 blocks toward north = 300 j m
4 blocks toward west = -400 i m
1 blocks toward north = 100 j m
Total displacement = -400 i + 400 j m
Magnitude
[tex]s=\sqrt{(-400)^2+400^2}=565.69m[/tex]
Direction,
[tex]\theta =tan^{-1}\left ( \frac{400}{-400}\right )=135^0[/tex]
Direction is 45° west of north.
Displacement is 565.69 m at 45° west of north
A length change - 0.18 m will occur for an object that is L- 80 m long, If the coeffcient of thermal expansion is12 x 106/C and if the original temperature is 83 C, find the final temperature.
Answer:
The final temperature is 270.5°C
Explanation:
Given that,
Change length = 0.18 m
Length = 80 m
Coefficient of thermal expansion [tex]\alpha=12\times10^{-6}\ C^{\circ}[/tex]
Temperature = 83°C
We need to calculate the final temperature
Using formula thermal expansion
[tex]\delta l=l\alpha\times\delta t[/tex]
Where, [tex]\Delta l[/tex] =change in length
[tex]\Delta t[/tex] = change in temperature
[tex]\alpha[/tex] = coefficient of thermal expansion
Put the value into the formula
[tex]0.18=80\times12\tmes10^{6}\times(T_{f}-T_{i})[/tex]
[tex]T_{f}-83=\dfrac{0.18}{80\times12\times10^{-6}}[/tex]
[tex]T_{f}=187.5+83=270.5^{\circ} C[/tex]
Hence, The final temperature is 270.5°C
The crankshaft in a race car goes from rest to 3000 rpm in 3.0 s . (a) What is the crankshaft's angular acceleration?
(b) How many revolutions does it make while reaching 3000 rpm ?
Answer:
75 rotations
Explanation:
f0 = 0, f = 3000 rpm = 50 rps, t = 3 s
(a) use first equation of motion for rotational motion
w = w0 + α t
2 x 3.14 x 50 = 0 + α x 3
α = 104.67 rad/s^2
(b) Let θ be the angular displacement
use second equation of motion for rotational motion
θ = w0 t + 1/2 α t^2
θ = 0 + 0.5 x 104.67 x 3 x 3
θ = 471.015 rad
The angle turn in one rotation is 2 π radian.
Number of rotation = 471.015 / (2 x 3.14) = 75 rotations
When ________ is constant, the enthalpy change of a process equal to the amount of heat transferred into or out of the system? pressure and volume temperature temperature and volume pressure volume
Answer:
temperature
Explanation:
When temperature is constant, the enthalpy change of a process equal to the amount of heat transferred into or out of the system.
When pressure is constant, the enthalpy change of a process equal to the amount of heat transferred into or out of the system
What is enthalpy?Enthalpy, the sum of the internal energy and the product of the pressure and volume of a thermodynamic system. Enthalpy is an energy-like property or state function—it has the dimensions of energy (and is thus measured in units of joules or ergs),
and its value is determined entirely by the temperature, pressure, and composition of the system and not by its history. In symbols, the enthalpy, H, equals the sum of the internal energy, E, and the product of the pressure, P, and volume, V, of the system: H = E + PV.
According to the law of energy conservation, the change in internal energy is equal to the heat transferred to, less the work done by, the system.
If the only work done is a change of volume at constant pressure, the enthalpy change is exactly equal to the heat transferred to the system
Hence when pressure is constant, the enthalpy change of a process equal to the amount of heat transferred into or out of the system
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Laura, Jeff, and Amy are in the same work group; they work well with other people in the work group but they don’t get along. This type of conflict can best be described as ________.
A train whistle has a sound intensity level of 70. dB, and a library has a sound intensity level of about 40. dB. How many times greater is the sound intensity of the train whistle than that of the library?
Answer:
The sound intensity of train is 1000 times greater than that of the library.
Explanation:
We have expression for sound intensity level,
[tex]L=10log_{10}\left ( \frac{I}{I_0}\right )[/tex]
A train whistle has a sound intensity level of 70 dB
We have
[tex]70=10log_{10}\left ( \frac{I_1}{I_0}\right )[/tex]
A library has a sound intensity level of about 40 dB
We also have
[tex]40=10log_{10}\left ( \frac{I_2}{I_0}\right )[/tex]
Dividing both equations
[tex]\frac{70}{40}=\frac{10log_{10}\left ( \frac{I_1}{I_0}\right )}{10log_{10}\left ( \frac{I_2}{I_0}\right )}\\\\\frac{7}{4}=\frac{log_{10}\left ( \frac{I_1}{I_0}\right )}{log_{10}\left ( \frac{I_2}{I_0}\right )}\\\\10^7\frac{I_2}{I_0}=10^4\frac{I_1}{I_0}\\\\\frac{I_1}{I_2}=10^3=1000[/tex]
The sound intensity of train is 1000 times greater than that of the library.
Suppose that you wish to construct a simple ac generator having an output of 12 V maximum when rotated at 60 Hz. A uniform magnetic field of 0.050 T is available. If the area of the rotating coil is 100 cm2, how many turns do you need?
Answer:
The number of turns is 64.
Explanation:
Given that,
Magnetic field = 0.050 T
Area of coil = 100 cm²
Frequency = 60 Hz
Output voltage emf= 12 V
We need to calculate the number of turns
Using formula of induced emf
[tex]emf =NAB\omega[/tex]
[tex]N=\dfrac{emf}{A\times B\times2\pi f}[/tex]
[tex]N=\dfrac{12}{0.01\times0.050\times2\times3.14\times60}[/tex]
[tex]N =63.6 = 64\ turns[/tex]
Hence, The number of turns is 64.
Answer:
You need 63.66 turns.
Explanation:
The number of turns of a magnetic field is given by the following formula:
[tex]N = \frac{V}{S*T*2\pi f}[/tex]
In which N is the number of turns, V is the maximum output voltage, S is the area of the rotating coil, in square meters and T is the measure of the magnetic field and f is the frequency.
In this problem, we have that:
Suppose that you wish to construct a simple ac generator having an output of 12 V maximum when rotated at 60 Hz. This means that [tex]V = 12[/tex] and [tex]f = 60[/tex].
A uniform magnetic field of 0.050 T is available. This means that [tex]T = 0.050[/tex].
If the area of the rotating coil is 100 cm2, how many turns do you need?
This means that [tex]S = 0.01[/tex]m². So:
[tex]N = \frac{V}{S*T*2\pi f}[/tex]
[tex]N = \frac{12}{0.01*0.05*120\pi}[/tex]
[tex]N = 63.66[/tex]
You need 63.66 turns.
Newton's Law of Gravitation says that the magnitude F of the force exerted by a body of mass m on a body of mass M is F = GmM r2 where G is the gravitational constant and r is the distance between the bodies. (a) Find dF/dr. dF dr = − 2GmM r3 What is the meaning of dF/dr? dF/dr represents the rate of change of the mass with respect to the distance between the bodies. dF/dr represents the amount of force per distance. dF/dr represents the rate of change of the mass with respect to the force. dF/dr represents the rate of change of the distance between the bodies with respect to the force. dF/dr represents the rate of change of the force with respect to the distance between the bodies. What does the minus sign indicate? The minus sign indicates that as the distance between the bodies increases, the magnitude of the force decreases. The minus sign indicates that the bodies are being forced in the negative direction. The minus sign indicates that the force between the bodies is decreasing. The minus sign indicates that as the distance between the bodies increases, the magnitude of the force increases. The minus sign indicates that as the distance between the bodies decreases, the magnitude of the force remains constant. (b) Suppose it is known that the earth attracts an object with a force that decreases at the rate of 4 N/km when r = 20,000 km. How fast does this force change when r = 10,000 km? N/km
Answer 1 (a):
According to Newton's Law of Gravitation, the Gravity Force is:
[tex]F=\frac{GMm}{{r}^{2}}[/tex] (1)
This expression can also be written as:
[tex]F=GMm{r}^{-2}[/tex] (2)
If we derive this force [tex]F[/tex] respect to the distance [tex]r[/tex] between the two masses:
[tex]\frac{dF}{dr}dFdr=\frac{d}{dr}(GMm{r}^{-2})dr[/tex] (3)
Taking into account [tex]GMm[/tex] are constants:
[tex]\frac{dF}{dr}dFdr=-2GMm{r}^{-3}[/tex] (4)
Or
[tex]\frac{dF}{dr}dFdr=-2\frac{GMm}{{r}^{3}}[/tex] (5)
Answer 2 (a): dF/dr represents the rate of change of the force with respect to the distance between the bodies.
In other words, this means how much does the Gravity Force changes with the distance between the two bodies.
More precisely this change is inversely proportional to the distance elevated to the cubic exponent.
As the distance increases, the Force decreases.
Answer 3 (a): The minus sign indicates that the bodies are being forced in the negative direction.This is because Gravity is an attractive force, as well as, a central conservative force.
This means it does not depend on time, and both bodies are mutually attracted to each other.
Answer 4 (b): [tex]X=-32N/km[/tex]
In the first answer we already found the decrease rate of the Gravity force respect to the distance, being its unit [tex]N/km[/tex]:
[tex]\frac{dF}{dr}dFdr=-2\frac{GMm}{{r}^{3}}[/tex] (5)
We have a force that decreases with a rate 1 [tex]\frac{dF_{1}}{dr}dFdr=4N/km[/tex] when [tex]r=20000km[/tex]:
[tex]4N/km=-2\frac{GMm}{{(20000km)}^{3}}[/tex] (6)
Isolating [tex]-2GMm[/tex]:
[tex]-2GMm=(4N/km)({(20000km)}^{3})[/tex] (7)
In addition, we have another force that decreases with a rate 2 [tex]\frac{dF_{2}}{dr}dFdr=X[/tex] when [tex]r=10000km[/tex]:
[tex]XN/km=-2\frac{GMm}{{(10000km)}^{3}}[/tex] (8)
Isolating [tex]-2GMm[/tex]:
[tex]-2GMm=X({(10000km)}^{3})[/tex] (9)
Making (7)=(9):
[tex](4N/km)({(20000km)}^{3})=X({(10000km)}^{3}[/tex] (10)
Then isolating [tex]X[/tex]:
[tex]X=\frac{4N/km)({(20000km)}^{3}}{{(10000km)}^{3}}[/tex]
Solving and taking into account the units, we finally have:
[tex]X=-32N/km[/tex]>>>>This is how fast this force changes when [tex]r=10000 km[/tex]
Torque is dependent on the angle between the force applied and the length of the level arm. When is the torque maximum? Not enough information. θ = 90 θ = 0 θ = 45
Answer:
90 degree
Explanation:
According to the formula of torque
torque = force x displacement x Sine of angle between force and displacement
So, for the maximum torque, the value of Sin theta should be maximum.
the maximum value of Sin theta is 1.
that means the value of theta is 90 degree.
Two particles, one with charge -5.45 uC and one with charge 3.55 uC, are 4.34 cm apart. What is the magnitude of the force that one particle exerts on the other?
Answer:
[tex]F = 92.45 N[/tex]
Explanation:
As we know that the force between two charge particles is given by
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
here we know that
[tex]q_1 = 3.55 \mu C[/tex]
[tex]q_2 = 5.45 \mu C[/tex]
now the distance between the two charges is
r = 4.34 cm
now from the formula of electrostatic force we will have
[tex]F = \frac{(9\times 10^9)(3.55 \mu C)(5.45 \mu C)}{0.0434^2}[/tex]
[tex]F = 92.45 N[/tex]
An object that is thrown straight up falls back to Earth. This is one-dimensional motion. (a) When is its velocity zero? (b) Does its velocity change direction? (c) Does the acceleration due to gravity have the same sign on the way up as on the way down?
Answer:
a) The velocity is zero when the object is at the highest vertical point (peak). b) Yes, at the peak, the velocity for an instant gets to zero and then the direction of motion and the velocity will be downwards.
C) No, the direction of ‘g’ will be opposite in direction when the object is on the way down when compared to what it was during the upward motion.
Explanation:
a) Since gravity is acting on the object and trying to pull it down to earth, the ball will slow down to a velocity of zero when the ball starts its free fall downward.
b) When the direction of the object changes, its velocity will also change direction.
c) When the ball has a positive velocity, gravity is pulling it in the opposite direction, so it has a negative acceleration trying to slow it down. When the ball's velocity is negative (downward free fall), the acceleration will be positive at the instant when the ball hits the ground. The only way to stop the ball from falling is a positive force upward, which results in a positive acceleration when the ball hits the ground.
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(a) The velocity of the object will be zero at the maximum height
(b) The velocity of the object changes direction.
(c) The acceleration due to gravity will not have the same sign on the way up as on the way down.
(A) The velocity of the object decreases as the object ascends upwards and eventually becomes zero at the maximum height. As the object descends downwards, the velocity increases and becomes maximum before the object hits the ground.
(B) The velocity of the object is always directed in the direction of the motion of the object. As the object moves upward, the direction of the object is upward and as the object moves downwards the direction of the object is downwards. Thus, the velocity of the object changes direction.
(c) The upward motion of the object is opposite direction to acceleration due to gravity and the sign of acceleration due to gravity becomes negative.
As the object moves downwards, it will be moving in the same direction as the acceleration due to gravity. The sign of acceleration due to gravity is positive.
Thus, the acceleration due to gravity will not have the same sign on the way up as on the way down.
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What was the speed of a space shuttle that orbited Earth at an altitude of 1482 km?
Answer:
v = 7121.3 m/s
Explanation:
As we know that the centripetal force for the space shuttle is due to gravitational force of earth due to which it will rotate in circular path with constant speed
so here we will have
[tex]\frac{mv^2}{r} = \frac{GMm}{r^2}[/tex]
here we know that
r = orbital radius = 6370 km + 1482 km
[tex]r = 7.852 \times 10^6 m[/tex]
also we know that
[tex]M = 5.97 \times 10^{24} kg[/tex]
now we will have
[tex]v^2 = \frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{7.852 \times 10^6}[/tex]
[tex]v^2 = 5.07 \times 10^7[/tex]
[tex]v = 7121.3 m/s[/tex]
The speed of the space shuttle that orbited the earth at an altitude of 1482km will be [tex]V=7121.3\dfrac{m}{s}[/tex]
What will be the speed of a space shuttle that orbited Earth at an altitude of 1482 km?
As we know that the centripetal force for the space shuttle is due to the gravitational force of the earth due to which it will rotate in a circular path with constant speed
so here we will have
[tex]\dfrac{mv^2}{r} = \dfrac{GMm}{r^2}[/tex]
[tex]V^2=\dfrac{GM}{r}[/tex]
here we know that
[tex]\rm r= orbital \ radius =6370+1482=7852 \ km[/tex]
[tex]r=7.852\times10^6\ m[/tex]
mass of earth [tex]M=5.97\times10^{24}\ kg[/tex]
Gravitational constant [tex]G=6.67\times10^{-11}[/tex]
By putting all the values we get
[tex]V^2=\dfrac{(6.67\times10^{-11} )(5.97\times10^{24})}{7.852\times10^{6}}[/tex]
[tex]V^2=5.07\times10^7[/tex]
[tex]V=7121.3 \dfrac{m}{s}[/tex]
Thus the speed of the space shuttle that orbited the earth at an altitude of 1482km will be [tex]V=7121.3\dfrac{m}{s}[/tex]
A spring of equilibrium length L1 and spring constant k1 hangs from the ceiling. Mass m1 is suspended from its lower end. Then a second spring, with equilibrium length L2 and spring constant k2, is hung from the bottom of m1. Mass m2 is suspended from this second spring. How far is m2 below the ceiling?
Answer:
[tex]d = L_1 + L_2 + \frac{(m_1 + m_2)g}{K_1} + \frac{m_2g}{K_2}[/tex]
Explanation:
For first spring the total extension is given as
[tex]F_{net} = K_1 x_1[/tex]
here net force is due to weight of both masses
[tex](m_1 + m_2)g = K_1 x_1[/tex]
so extension of first spring is
[tex]x_1 = \frac{(m_1 + m_2)g}{K_1}[/tex]
now similarly the extension in second spring is only due to the weight of mass m2
so here we will have
[tex]m_2g = K_2 x_2[/tex]
[tex]x_2 =\frac{m_2g}{K_2}[/tex]
so the total distance from the ceiling for mass m2 is given as
[tex]d = L_1 + L_2 + \frac{(m_1 + m_2)g}{K_1} + \frac{m_2g}{K_2}[/tex]
The total distance of m₂ from the ceiling is [tex]d = L_1 + L_2 + \frac{g(m_1 + m_2)}{k_1} + \frac{m_2g}{k_2}[/tex]
The given parameters;
equilibrium of length of the spring, = L₁spring constant, = K₁mass on the spring = m₁length of the first spring, = L₂spring constant = K₂mass on the second spring, = m₂The extension of the first spring due to mass m₁ and m₂ is calculated as;
[tex]F = kx\\\\ g(m_1+m_2) = k_1x_1\\\\x_1 = \frac{g(m_1+m_2)}{k_1}[/tex]
The extension of the second spring due to mass (m₂) is calculated as;
[tex]m_2g = k_2x_2\\\\x_2 = \frac{m_2g}{k_2}[/tex]
The total distance of m₂ from the ceiling is calculated as follows;
[tex]d = L_1 + L_2 + x_1 + x_2\\\\d = L_1 + L_2 + \frac{g(m_1 + m_2)}{k_1} + \frac{m_2g}{k_2}[/tex]
Thus, the total distance of m₂ from the ceiling is [tex]d = L_1 + L_2 + \frac{g(m_1 + m_2)}{k_1} + \frac{m_2g}{k_2}[/tex]
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A parallel-plate capacitor initially has a potential difference of 400 V and is then disconnected from the charging battery. If the plate spacing is now doubled (without changing Q), what is the new value of the voltage?
The capacitance of a capacitor is the ratio of the stored charge to its potential difference, i.e.
C = Q/ΔV
C is the capacitance
Q is the stored charge
ΔV is the potential difference
Rearrange the equation:
ΔV = Q/C
We also know the capacitance of a parallel-plate capacitor is given by:
C = κε₀A/d
C is the capacitance
κ is the capacitor's dielectric constant
ε₀ is the electric constant
A is the area of the plates
d is the plate separation
If we substitute C:
ΔV = Qd/(κε₀A)
We assume the stored charge and the area of the plates don't change. Then if we double the plate spacing, i.e. we double the value of d, then the potential difference ΔV is also doubled.
Chameleons can rapidly project their very long tongues to catch nearby insects. The tongue of the tiny Rosette-nosed chameleon has the highest acceleration of a body part of any amniote (reptile, bird, or mammal) ever measured. In a somewhat simplified model of its tongue motion, the tongue, starting from rest, first undergoes a constant-acceleration phase with an astounding magnitude of 2500 m/s^2. This acceleration brings the tongue up to a final speed of 5.0 m/s. It continues at this speed for 22 m/s until it hits its target. What is the total distance traveled by the chameleon's tongue?
Answer:
0.115 m
Explanation:
Consider the motion of tongue during acceleration :
v₀ = initial velocity of the tongue = 0 m/s
v = final velocity of the tongue = 5 m/s
a = acceleration = 2500 m/s²
d = distance traveled during acceleration phase
Using the equation
v² = v₀² + 2 a d
5² = 0² + 2 (2500) d
d = 0.005 m
Consider the motion of tongue after it attains constant speed
d' = distance traveled during constant velocity
v = constant velocity = 5 m/s
t = time of travel = 22 ms = 0.022 s
using the equation
d' = v t
d' = 5 x 0.022
d' = 0.11 m
D = Total distance traveled by tongue
Total distance traveled by tongue is given as
D = d + d'
D = 0.005 + 0.11
D = 0.115 m
A motorcycle moving at 13.2 m/s increases in speed to 25.7 m/s over a period of 8.6 s. If the motorcycle's mass is 352 kg, what net force in Newtons acts on the motorcycle?
Answer:
510.4 N
Explanation:
u = 13.2 m /s, v = 25.7 m/s, t = 8.6 s, m = 352 kg
Use first equation of motion
v = u + a t
a = (25.7 - 13.2) / 8.6 = 1.45 m/s^2
Use Newton's second law
F = m a = 352 x 1.45 = 510.4 N
What accelerating potential is needed to produce electrons of wavelength 5.20 nm? V- 10.5563 Previous Answers Reguest Answer Submit Incorrect; Try Again; 5 attempts remaining ▼ Part B What would be the energy of photons having the same wavelength as these electrons? eV Submit Request Answer
The accelerating potential needed for electrons of a particular wavelength can be found by rearranging the de Broglie wavelength equation. The energy of photons of the same wavelength can be found using Planck's equation, by substituting the given wavelength.
Explanation:The accelerating potential needed to produce electrons of a wavelength (de Broglie wavelength) can be calculated using the equation: λ = h / √(2mVq), where h is Planck's constant, m is the mass of the electron, V is the accelerating voltage, and q is the charge of the electron. For the wavelength of 5.20 nm, one can rearrange and solve for V to get V = h² / (2mqλ²).
Now, for the energy of photons having the same wavelength as the electrons, we can use Planck's equation, E = hv = hc/λ. Here, 'h' is Planck's constant, 'v' is the frequency of light, 'c' is the speed of light and 'λ' is the wavelength. Substituting λ = 5.20 nm, we get the energy of photons in terms of electron-volts (eV).
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Three stars, each with the mass of our sun, form an equilateral triangle with sides 1.0×1012m long. (This triangle would just about fit within the orbit of Jupiter.) The triangle has to rotate, because otherwise the stars would crash together in the center. What is the period of rotation?
Answer:
period of rotation is 9.9843 years
Explanation:
Given data
sides = 1.0×10^12 m
to find out
period of rotation
solution
first we use the gravititional formula i.e
F(g) = 2 F cos 30
here F = G × mass(sun)² / radius²
F = 6.67 × [tex]10^{-11}[/tex] × (1.99× [tex]10^{30}[/tex])² / (1× [tex]10^{30}[/tex])²
so
F(g) = 2 × 6.67 × [tex]10^{-11}[/tex] × (1.99× [tex]10^{30}[/tex])² / (1× [tex]10^{30}[/tex])² × (1.73/2)
F(g) = 4.57 × [tex]10^{26}[/tex] N
and
by the law of sines
r/sin30 = s/sin120
so r will be
r = 1.0×10^12 (0.5/.87)
r = 5.75 × [tex]10^{11}[/tex] m
we know that gravititional force = centripetal force
so here
centripetal force = mv² /r
centripetal force = 1.99× [tex]10^{30}[/tex] v² / 5.75 × [tex]10^{11}[/tex]
so
4.57 × [tex]10^{26}[/tex] = mv² /r
4.57 × [tex]10^{26}[/tex] = 1.99× [tex]10^{30}[/tex] v² / 5.75 × [tex]10^{11}[/tex]
v² = 4.57 × [tex]10^{26}[/tex] × 5.75 × [tex]10^{11} / 1.99× [tex]10^{30}[/tex]
and
v = 11480 m/sec
and we know period of rotation formula
t = d /v
t = 2 [tex]\pi[/tex] r /v
t = 2 × [tex]\pi[/tex] × 5.75 × [tex]10^{11} / 11480
t = 314 × [tex]10^{5} / 31536000 year
t = 9.9843 year
so period of rotation is 9.9843 years
The period of rotation for the three stars can be calculated using Kepler's Third Law and recognizing that the stars form a circular orbital pattern. The relevant distance is from the center of the mass distribution (center of the equilateral triangle) to any of the stars and all three stars have identical masses.
Explanation:This question relates to astrophysics and can be solved using Kepler's Third Law that provides a relationship between the period of revolution (P), semimajor axis (D), and the total mass (M) of a binary system: D³ = (M₁ + M₂)P². However, since we are dealing with three stars of equal mass (each equivalent to the mass of our sun), they form a circular orbital pattern. We use the formula P = √((4π²/G)(r³/3M)), where r is the distance from the center of the mass distribution to any of the stars.
Firstly, we must understand that in an equilateral triangle, the distance from the center to a vertex is r = √3/2 * side. So, the distance from the center to any star is 0.866 * 1.0×10¹²m. The gravitational constant, G, is approximately 6.67 * 10^-11 N(m/kg)².
Substituting these values into our formula, we can calculate the period of rotation of the stars. One should note that all three stars have identical masses which simplifies the calculation.
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The electric field of a charge is defined by the force on: An electron A probe charge A proton. A source charge.
Answer:
A probe charge
Explanation:
As we know that electric field intensity is the force experienced by the probe charge which is placed in the electric field region
Here we can say it as
[tex]E = \frac{F}{q}[/tex]
so here that probe charge should be very small so that it will not disturb the electric field in the space.
If the probe charge is of large magnitude then the field will get disturbed and the intensity which is to be measured is different from its actual value.
Also the sign of the probe charge is taken to be positive.
so correct answer here will be
A probe charge
Final answer:
The electric field of a charge is defined by the force experienced by a probe charge placed within the electric field, so the correct option will be D probe charge
Explanation:
The electric field of a charge is defined by the force on a probe charge. An electric field is a vector field that affects the space surrounding a source charge and exerts a force on any test charge (probe charge) within that region. The direction of the electric field represents the direction a positive probe charge would move if placed within the field.
It's calculated as the force per unit charge exerted on a test charge and can be represented mathematically by the equation F = qE, where F is the force experienced by the test charge, q represents the charge of the test charge, and E is the electric field strength.
A woman is standing in the ocean, and she notices that after a wave crest passes by, five more crests pass in a time of 39.5 s. The distance between two successive crests is 37.9 m. What is the wave's (a) period, (b) frequency, (c) wavelength, and (d) speed?
(a) 7.9 s
The period of a wave is time that passes between two consecutive crests (or two consecutive troughs).
In this case, we are told that five crests pass in a time of 39.5 s. Therefore we can find the period by using the proportion:
[tex]\frac{5}{39.5 s}=\frac{1}{T}[/tex]
Where T is the period. Re-arranging the equation, we find
[tex]T=\frac{(39.5)(1)}{5}=7.9 s[/tex]
(b) 0.127 Hz
The frequency of a wave is equal to the reciprocal of the period:
[tex]f=\frac{1}{T}[/tex]
where
f is the frequency
T is the period
For this wave, we have T = 7.9 s, so its frequency is
[tex]f=\frac{1}{7.9 s}=0.127 Hz[/tex]
(c) 37.9 m
The wavelength of a wave is the distance between two consecutive crests (or two consecutive troughs). For this wave, the distance between two successive crests is 37.9 m, so the wavelength of the wave is
[tex]\lambda=37.9 m[/tex]
(d) 4.81 m/s
The speed of a wave is given by
[tex]v=\lambda f[/tex]
where
[tex]\lambda[/tex] is the wavelength
f is the frequency
For the wave in the problem, we have
[tex]\lambda=37.9 m\\f=0.127 Hz[/tex]
Therefore, the speed of the wave is
[tex]v=(37.9)(0.127)=4.81 m/s[/tex]
A 230 g air-track glider is attached to a spring. The glider is pushed in 8.2 cm and released. A student with a stopwatch finds that 13 oscillations take 19.0 s . What is the spring constant? Express your answer using two significant figures.
Answer:
4.3 N/m
Explanation:
m = 230 g = 0.230 kg, x = 8.2 cm
in 13 oscillations, time taken = 19 s
In 1 oscillation, time taken = 19 / 13 = 1.46 s
By the use of formula of time period , Let k be the spring constant.
[tex]T = 2\pi \sqrt{\frac{m}{k}}[/tex]
[tex]1.46 = 2\pi \sqrt{\frac{0.230}{k}}[/tex]
0.054 = 0.230 / k
k = 4.26 N/m
k = 4.3 N/m
0. A 3.00-kg block is dropped from rest on a vertical spring whose spring constant is 750 N/m. The block hits the spring, compresses it, and then comes to rest, having fallen through a total vertical distance of 0.45 m. How far is the spring compressed? (Hint: Use conservation of energy.)
Answer:
the spring compressed is 0.1878 m
Explanation:
Given data
mass = 3 kg
spring constant k = 750 N/m
vertical distance h = 0.45
to find out
How far is the spring compressed
solution
we will apply here law of mass of conservation
i.e
gravitational potential energy loss = gain of eastic potential energy of spring
so we say m×g×h = 1/2× k × e²
so e² = 2×m×g×h / k
so
we put all value here
e² = 2×m×g×h / k
e² = 2×3×9.81×0.45 / 750
e² = 0.0353
e = 0.1878 m
so the spring compressed is 0.1878 m
At what rate is soda being sucked out of a cylindrical glass that is 6 in tall and has radius of 2 in? The depth of the soda decreases at a constant rate of 0.25 in/sec.
Answer:
The soda is being sucket out at a rate of 3.14 cubic inches/second.
Explanation:
R= 2in
S= π*R²= 12.56 inch²
rate= 0.25 in/sec
rate of soda sucked out= rate* S
rate of soda sucked out= 3.14 inch³/sec
Final answer:
Soda is being pulled out of a cylindrical glass at the rate of -3.14 cubic inches per second.
Explanation:
The question asks at what rate soda is being absorbed from a cylindrical glass. Given the cylindrical glass's dimensions and the rate at which the depth of soda decreases, we need to find the volume rate of change.
He formula for the volume of a cylinder is V = πr2h, where r is the radius and h is the height of the cylinder. To find the rate at which volume changes, we differentiate with respect to time to obtain dV/dt = πr2(dh/dt). Plugging in the values, we have r = 2 inches and dh/dt = -0.25 inches/second (negative because the depth is decreasing).
Using the given values, the rate at which soda is being pulled out of the glass is
dV/dt = π(2 in)2(-0.25 in/sec) = π(4 in2)(-0.25 in/sec) = -3.14 in3/sec.
Therefore, the rate at which the soda is being pulled out of the glass is -3.14 in3/sec.
At one instant, a 19.0-kg sled is moving over a horizontal surface of snow at 4.50 m/s. After 7.80 s has elapsed, the sled stops. Use a momentum approach to find the magnitude of the average friction force acting on the sled while it was moving.
Answer:
Frictional force, F = -10.96 N
Explanation:
It is given that,
Mass of the sled, m = 19 kg
Initial velocity of the sled, u = 4.5 m/s
After 7.80 s has elapsed, the sled stops. We need to find the magnitude of the average friction force acting on the sled while it was moving. The change in momentum is equal to the impulse imparted i.e.
[tex]F.\Delta t=m(v-u)[/tex]
[tex]F=\dfrac{m(v-u)}{\Delta t}[/tex]
[tex]F=\dfrac{19\ kg(0-4.5\ m/s)}{7.8\ s}[/tex]
F = -10.96 N
So, the average frictional force acting on the sled is 10.96 N. Hence, this is the required solution.
A powerful grasshopper launches itself at an angle of 45° above the horizontal and rises to a maximum height of 1.01 m during the leap. With what speed v did it leave the ground? Neglect air resistance. Hint: Use conservation of energy. A. 6.29 m/s B. 7.15 m/s C. 5.98 m/s D. 6.72 m/s E. 5.37 m/s
Answer:
6.29 m/s option (A)
Explanation:
theta = 45 degree, H = 1.01 m
let v be the launch speed
Use the formula for the maximum height for the projectile
H = v^2 Sin^θ / 2g
1.01 = v^2 x Sin^2(45) / (2 x 9.8)
1.01 = 0.0255 v^2
v^2 = 39.59
v = 6.29 m/s
The initial velocity of the grasshopper is 6.29 m/s.
Initial velocity of the grasshopper
The Initial velocity of the grasshopper is calculated from the following kinematic equation.
[tex]H = \frac{v_0^2 sin^2 \theta}{2g}[/tex]
where;
H is the maximum heightv is the initial velocity[tex]v_0^2 = \frac{2gH}{sin^2\theta} \\\\v_0^2 = \frac{2 \times 9.8 \times 1.01 }{(sin45)^2} \\\\v_0^2 = 39.6\\\\v_0 = 6.29 \ m/s[/tex]
Thus, the initial velocity of the grasshopper is 6.29 m/s.
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The thickness of a $1 bill is 0.11 mm. If you have a stack of $1 bills 450 m tall, how much money do you have?
Answer:
You will have 4.5 million dollar
Explanation:
The thickness of a $1 bill is 0.11 mm
So we have
1 $ = 0.1 mm
0.1 mm = 1 $
0.0001 m = 1 $
1 m = 10000 $
450 m = 450 x 10000 = 4500000 $
So you will have 4.5 million dollar
Answer:
4090909
Explanation:
Thickness of one bill = 0.11 mm
Total thickness = 450 m
No of $1 bills = total thickness / thickness of one bill
No of $1 bills = 450 / 0.11 × 10^-3
= 4090909
A spring-loaded gun, fired vertically, shoots a marble 9.0 m straight up in the air. What is the marble's range if it is fired horizontally from 1.8 m above the ground?
The range of the marble when fired horizontally from 1.8m above the ground can be calculated using the equations of motion in physics. First, the time of flight is found using the vertical motion and then the range is calculated using the time of flight and the initial velocity determined from the vertical launch. The marble's range is approximately 8.4m.
Explanation:To solve this problem, we need to make use of the concept of projectile motion in physics. The most crucial part in solving this type of problem is to break the motion into its horizontal and vertical components.
First, we find the time the projectile is in the air using the vertical motion. Ignoring air resistance, the time a projectile is in the air is determined by the initial vertical velocity and the height from which it drops. Here, the height is given as 1.8m and we can use the equation h = 0.5gt^2, where h is the height, g is the acceleration due to gravity (9.8 m/s^2), and t is the time. After calculating, we find that the time the marble is in the air is about 0.6 seconds.
Now, we can use the time to find the horizontal distance traveled by the marble, a.k.a the range. The range is given by R = vt, where v is the horizontal velocity, which is the same as the initial vertical velocity. From the problem, we know the marble reached a height of 9.0m when shot vertically, which we can use to find the initial velocity using the equation v = sqrt(2gh), where g is the acceleration due to gravity (9.8 m/s^2) and h is the height. We find that the initial velocity is about 14 m/s.
So, the range R = vt = 14m/s * 0.6s = 8.4m. Therefore, the marble's range when fired horizontally from 1.8m above the ground is approximately 8.4m.
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The marble's range if it is fired horizontally from 1.8 m above the ground is 7.968 m.
At the highest point, the velocity of the gun will be zero. The initial velocity of the shot can be determined using the kinematic equation,
[tex]v^2 = u^2 + 2gs[/tex]
here, v = 0 m/s
g = 9.8 m/s² (downward, hence taken negative)
s = 9.0 m (upwards, hence taken positive.)
Using proper sign convention, we get:
[tex](0 \hspace{0.8 mm} m/s)^2 = u^2 - 2 \times (9.8 \hspace{0.8 mm} m/s^2) \times (9.0 \hspace{0.8 mm} m )[/tex]
u = 13.28 m/s
Now, when the marble is fired at a height of 1.8 m above the ground, the time taken by the marble to reach the ground can be determined by the kinematic equation:
[tex]h = ut + \frac{1}{2}gt^2[/tex],
using u = 0 m/s (as we will consider the time of fall of the marble from highest point), g = 9.8 m/s² (downward, hence taken to be negative), h = 1.8 m (downwards, hence taken positive), we get:
[tex]- 1.8 \hspace{0.8 mm} m = (0 \hspace{0.8 mm} m/s)t - \frac{1}{2}(9.8 \hspace{0.8 mm} m/s^2)t^2[/tex]
or, [tex]1.8 \hspace{0.8 mm} m = (4.9 \hspace{0.8 mm} m/s^2)t^2[/tex]
or, t = 0.60 s
Range of the marble will be determined by the horizontal velocity of the marble. It will be the maximum horizontal distnace covered by the marble as it falls down in time t = 0.60 s.
R = [tex]v_x[/tex]t
Since, the horizontal velocity is not influenced by any acceleration and will remain constant.
∴ R = [tex]v_x[/tex]t = 13.28 m/s × 0.60 s = 7.968 m