A girl is sledding down a slope that is inclined at 30º with respect to the horizontal. The wind is aiding the motion by providing a steady force of 161 N that is parallel to the motion of the sled. The combined mass of the girl and the sled is 87.7 kg, and the coefficient of kinetic friction between the snow and the runners of the sled is 0.151. How much time is required for the sled to travel down a 271-m slope, starting from rest?
Answers
Answer:
The sled required 9.96 s to travel down the slope.
Explanation:
Please, see the figure for a description of the problem. In red are the x and y-components of the gravity force (Fg). Since the y-component of Fg (Fgy) is of equal magnitude as Fn but in the opposite direction, both forces get canceled.
Then, the forces that cause the acceleration of the sled are the force of the wind (Fw), the friction force (Ff) and the x-component of the gravity force (Fgx).
The sum of all these forces make the sled move. Finding the resulting force will allow us to find the acceleration of the sled and, with it, we can find the time the sled travel.
The magnitude of the friction force is calculated as follows:
Ff = μ · Fn
where :
μ = coefficient of kinetic friction
Fn = normal force
The normal force has the same magnitude as the y-component of the gravity force:
Fgy = Fg · cos 30º = m · g · cos 30º
Where
m = mass
g = acceleration due to gravity
Then:
Fgy = m · g · cos 30º = 87.7 kg · 9.8 m/s² · cos 30º
Fgy = 744 N
Then, the magnitude of Fn is also 744 N and the friction force will be:
Ff = μ · Fn = 0.151 · 744 N = 112 N
The x-component of Fg, Fgx, is calculated as follows:
Fgx = Fg · sin 30º = m·g · sin 30º = 87.7 kg · 9.8 m/s² · sin 30º = 430 N
The resulting force, Fr, will be the sum of all these forces:
Fw + Fgx - Ff = Fr
(Notice that forces are vectors and the direction of the friction force is opposite to the other forces, then, it has to be of opposite sign).
Fr = 161 N + 430 N - 112 N = 479 N
With this resulting force, we can calculate the acceleration of the sled:
F = m·a
where:
F = force
m = mass of the object
a = acceleration
Then:
F/m = a
a = 479N/87.7 kg = 5.46 m/s²
The equation for the position of an accelerated object moving in a straight line is as follows:
x = x0 + v0 · t + 1/2 · a · t²
where:
x = position at time t
x0 = initial position
v0 = initial velocity
t = time
a = acceleration
Since the sled starts from rest and the origin of the reference system is located where the sled starts sliding, x0 and v0 = 0.
x = 1/2· a ·t²
Let´s find the time at which the position of the sled is 271 m:
271 m = 1/2 · 5.46 m/s² · t²
2 · 271 m / 5.46 m/s² = t²
t = 9.96 s
The sled required almost 10 s to travel down the slope.
Related Questions
A bullet in a gun is accelerated from the firing chamber to the end of the barrel at an average rate of 5.70 x 10^5 m/s^2 for 9.60 x 10^−4 s. What is its muzzle velocity (in m/s) (that is, its final velocity)? (Enter the magnitude.)
Answers
Final answer:
To calculate the muzzle velocity, we use the equation for constant acceleration (v = u + at) with an initial velocity (u) of 0, the given acceleration (a) of 5.70 x 10^5 m/s^2, and the time (t) of 9.60 x 10^-4 s to find the final velocity (v), which is approximately 547 m/s.
Explanation:
The question concerns calculating the muzzle velocity of a bullet as it's accelerated from the firing chamber to the end of the barrel of a gun.
Using the formula v = u + at, where v is the final velocity, u is the initial velocity (which is zero before the gun is fired), a is the acceleration, and t is the time for which the acceleration is applied, we can find the muzzle velocity. The bullet experiences an acceleration (a) of 5.70 x 105 m/s2 for a time (t) of 9.60 x 10−4 seconds.
Plugging the values into the formula, we get:
v = 0 + (5.70 x 105 m/s2)(9.60 x 10−4 s)
v = 5.47 x 102 m/s
Therefore, the final muzzle velocity of the bullet as it leaves the barrel is approximately 547 m/s.
Find the frequency (in hertz) of the first overtone for a 1.75 m pipe that is closed at only one end. Use 350 m/s for the speed of sound in air. Also, would increasing the air temperature raise, lower, or have no effect upon this frequency?
Answers
Answer:
The first overtone frequency is 100 Hz.
Solution:
According to the question:
Length of pipe, l = 1.75 m
Speed of sound in air, v_{sa} = 350 m/s
Frequency of first overtone, [tex]f_{1}[/tex] is given by:
[tex]f_{1} = \frac{v_{sa}}{2l}[/tex]
[tex]f_{1} = \frac{350}{2\times 1.75} = 100 Hz[/tex]
Since, the frequency, as clear from the formula depends only on the speed
and the length. It is independent of the air temperature.
Thus there will be no effect of air temperature on the frequency.
(a) With what speed must a ball be thrown vertically from ground level to rise to a maximum height of 52 m? (b) How long will it be in the air?
Answers
Explanation:
Maximum height reached by the ball, s = 52 m
Let u is the initial speed of the ball and v is the final speed of the ball, v = 0 because at maximum height the final speed goes to 0. We need to find u.
(a) The third equation of motion as :
[tex]v^2-u^2=2as[/tex]
Here, a = -g
[tex]0-u^2=-2gs[/tex]
[tex]u^2=2\times 9.8\times 52[/tex]
u = 31.92 m/s
(b) Let t is the time when the ball is in air. It is given by :
[tex]v=u+at[/tex]
[tex]u=gt[/tex]
[tex]t=\dfrac{31.92\ m/s}{9.8\ ms/^2}[/tex]
t = 3.25 seconds
Hence, this is the required solution.
Discuss whether the kinetic energy or the momentum of the ""bullet"" is more important in causing damage to the bulletproof vest.
Answers
Answer:
Answered
Explanation:
The Kinetic energy of a bullet does not remain constant. It dissipates as the bullet travel through the air. It is partly converted into heat, sound, air resistance etc.
Whereas the momentum of a travelling body remains conserved and hence constant. therefore a bullet with more momentum will cause more damage. that is momentum is more important than kinetic energy.
A bicyclist pedals at a speed of 5.0 km/h. How far does he travel in 80 minutes? A. 0.08 km В. 300 m C. 400 m D. 6.7 km
Answers
Answer:d-6.7 km
Explanation:
Given
Bicyclist pedals at a speed of 5 km/h
so his speed in meter per second
[tex]5\times \frac{5}{18}=\frac{25}{18} m/s[/tex]
In 80 minutes he would travel
Distance traveled[tex]=\frac{25}{18}\times 80\times 60=6666.667 m\approx 6.67 km[/tex]
A monument has a height of 348 ft, 8 in. Express this height in meters. Answer in units of m.
Answers
Answer:
The height of mountain in meter will be 106.2732 m
Explanation:
We have given height of mountain = 348 ft,8 in
We know that 1 feet = 0.3048 meter
So 348 feet [tex]=348\times 0.3048=106.07meter[/tex]
And we know that 1 inch = 0.0254 meter
So 8 inch [tex]8\times 0.0254=0.2032m[/tex]
So the total height of mountain in meter = 106.07+0.2032 = 106.2732 m
The height of mountain in meter will be 106.2732 m
Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a large toy rocket to the back of a sled and take the modified sled to a large, flat, snowy field. You ignite the rocket and observe that the sled accelerates from rest in the forward direction at a rate of 13.513.5 m/s2 for a time period of 3.503.50 s. After this time period, the rocket engine abruptly shuts off, and the sled subsequently undergoes a constant backward acceleration due to friction of 5.155.15 m/s2. After the rocket turns off, how much time does it take for the sled to come to a stop?
By the time the sled finally comes to a rest, how far has it traveled from its starting point?
Answers
1) 9.18 s
In the first part of the motion, the rocket accelerates at a rate of
[tex]a_1=13.5 m/s^2[/tex]
For a time period of
[tex]t_1=3.50 s[/tex]
So we can calculate the velocity of the rocket after this time period by using the SUVAT equation:
[tex]v_1=u+a_1t_1[/tex]
where u = 0 is the initial velocity of the rocket. Substituting a1 and t1,
[tex]v_1=(13.5)(3.50)=47.3 m/s[/tex]
In the second part of the motion, the rocket decelerates with a constant acceleration of
[tex]a_2 = -5.15 m/s^2[/tex]
Until it comes to a stop, to reach a final velocity of
[tex]v_2 = 0[/tex]
So we can use again the same equation
[tex]v_2 = v_1 + a_2 t_2[/tex]
where [tex]v_1 = 47.3 m/s[/tex]. Solving for t2, we find after how much time the rocket comes to a stop:
[tex]t_2 = -\frac{v_1}{a_2}=-\frac{47.3}{5.15}=9.18 s[/tex]
2) 299.9 m
We have to calculate the distance travelled by the rocket in each part of the motion.
The distance travelled in the first part is given by:
[tex]d_1 = ut_1 + \frac{1}{2}a_1 t_1^2[/tex]
Using the numbers found in part a),
[tex]d_1 = 0 + \frac{1}{2}(13.5) (3.50)^2=82.7 m[/tex]
The distance travelled in the second part of the motion is
[tex]d_2= v_1 t_2 + \frac{1}{2}a_2 t_2^2[/tex]
Using the numbers found in part a),
[tex]d_2 = (47.3)(9.18) + \frac{1}{2}(-5.15) (9.18)^2=217.2 m[/tex]
So, the total distance travelled by the rocket is
d = 82.7 m + 217.2 m = 299.9 m
You toss a ball straight up at 6.8 m/s ; it leaves your hand at 2.0 m above the floor. Suppose you had tossed a second ball straight down at 6.8 m/s (from the same place 2.0 m above the floor). When would the second ball hit the floor?
Answers
Answer:0.249 s
Explanation:
Given
Ball is tossed down with a velocity of 6.8 m/s downward
height from ground=2 m
therefore time to reach ground is
[tex]s=ut+\frac{gt^2}{2}[/tex]
[tex]2=6.8\times t+\frac{9.81\times t^2}{2}[/tex]
[tex]9.81t^2+13.6t-4=0[/tex]
[tex]t=\frac{-13.6\pm \sqrt{13.6^2+4\times 4\times 9.81}}{2\times 9.81}[/tex]
[tex]t=\frac{-13.6+18.49}{19.62}=0.249 s[/tex]
Does percent error give indication of accuracy or precision? Discuss.
Answers
Answer:
Accuracy
Explanation:
Percent error is the ratio of the difference of the measured and actual value to the actual value multiplied by 100.
It gives the percent deviation of the value obtained from the actual value.
Accuracy is the measure of how close the readings are to the actual value or set standard and can be improved by increase the no. of readings in an experiment.
Precision is the measure of the closeness of the obtained values to one another.
Thus accuracy of the reading can be sensed by the percent error.
A heat pump has a coefficient of performance that is 60% of the Carnot heat pump coefficient of performance. The heat pump is used to heat a home to 24.0°C during the winter with the low temperature reservoir at the outdoor temperature. At which outdoor temperature would it be more efficient to add the energy directly to the interior of the home than use it to run the heat pump? -154°C -40.0°C -4.00°C -83.4°C -25.2°C
Answers
Answer:[tex]T_L=-154.2^{\circ}[/tex]
Explanation:
Given
COP= 60 % of carnot heat pump
[tex]COP=\frac{60}{100}\times \frac{T_H}{T_H-T_L}[/tex]
For heat added directly to be as efficient as via heat pump
[tex]Q_s=W[/tex]
[tex]COP=\frac{Q_s}{W}=\frac{60}{100}\times \frac{T_H}{T_H-T_L}[/tex]
[tex]1=\frac{60}{100}\times \frac{T_H}{T_H-T_L}[/tex]
[tex]1=\frac{60}{100}\times \frac{24+273}{24+273-T_L}[/tex]
[tex]T_L=118.8 K[/tex]
[tex]T_L=-154.2^{\circ}[/tex]
One end of a horizontal spring with force constant 130.0 N/m is attached to a vertical wall. A 3.00 kg block sitting on the floor is placed against the spring. The coefficient of kinetic friction between the block and the floor is μk = 0.400. You apply a constant force F⃗ to the block. F⃗ has magnitude 88.0 N and is directed 3 toward the wall. The spring is compressed 80.0 cm. (a) What is the speed of the block? (b) What is the magnitude of the block’s acceleration? (c) What is the direction of the block’s acceleration?
Answers
Final answer:
The speed of the block is 4.08 m/s, the magnitude of the block’s acceleration is 25.41 m/s^2, and the direction of the block’s acceleration is toward the wall.
Explanation:
(a) To find the speed of the block, we can use the principle of conservation of mechanical energy. The potential energy stored in the spring when it is compressed is converted into the kinetic energy of the block when it is released. The potential energy stored in the spring is given by:
PE = 0.5 * k * x^2
where k is the force constant of the spring and x is the compression of the spring. Plugging in the values, we get:
PE = 0.5 * 130.0 N/m * 0.80 m * 0.80 m = 41.60 J
The kinetic energy of the block when it is released is given by:
KE = 0.5 * m * v^2
where m is the mass of the block and v is its speed. Equating the potential and kinetic energies, we have:
PE = KE
41.60 J = 0.5 * 3.00 kg * v^2
Solving for v, we get:
v = √(41.60 J / (0.5 * 3.00 kg)) = 4.08 m/s
(b) The magnitude of the block's acceleration can be calculated using Newton's second law, which states that the net force acting on an object is equal to its mass multiplied by its acceleration. In this case, the net force is the force applied to the block minus the force of friction. The force applied to the block is given by F = 88.0 N. The force of friction can be calculated using the equation:
f = μk * m * g
where μk is the coefficient of kinetic friction, m is the mass of the block, and g is the acceleration due to gravity. Plugging in the values, we get:
f = 0.400 * 3.00 kg * 9.8 m/s^2 = 11.76 N
The net force is therefore:
net force = F - f = 88.0 N - 11.76 N = 76.24 N
Using Newton's second law, we have:
76.24 N = 3.00 kg * a
Solving for a, we get:
a = 76.24 N / 3.00 kg = 25.41 m/s^2
(c) The direction of the block's acceleration can be determined by considering the net force acting on the block. In this case, the applied force and the force of friction are in opposite directions, resulting in a net force in the direction of the applied force. Therefore, the direction of the block's acceleration is toward the wall.
Two 3.5-cm-diameter disks face each other, 2.0 mm apart. They are charged to ± 11 nC . a) What is the electric field strength between the disks?
Express your answer in newtons per coulomb.
b) A proton is shot from the negative disk toward the positive disk. What launch speed must the proton have to just barely reach the positive disk?
Express your answer in meters per second.
Answers
Answer:
a) 1.29*10^6 N/C
b) 0.703 *10^6 m/s
Explanation:
This is a parallel plates capacitor. In a parallel plates capacitor the electric field depends on the charge of the disks, its area and the vacuum permisivity (Assuming there is no dielectric) and can be found using the expression:
[tex]E = \frac{Q}{A*e_0} =\frac{11*10^{-9}C}{(\frac{1}{4}\pi*(0.035m)^2)*8.85*10^{-12}C^2/Nm^2} = 1.29 *10^6 N/C[/tex]
For the second part, we use conservation of energy. The change in kinetic energy must be equal to the change in potential energy. The potential energy is given by:
[tex]PE = V*q[/tex]
V is the electric potential or voltage, q is the charge of the proton. The electric potential is equal to:
[tex]V = -E*d[/tex]
Where d is the distance to the positive disk. Then:
[tex]\frac{1}{2}mv_1^2 +V_1q = \frac{1}{2}mv_2^2 +V_2q\\\frac{1}{2}m(v_1^2 - v_2^2)=(V_2-V_1)q = (r_1-r_2)Eq|r_2 = 0m, v_2=0m/s\\v_1 = \sqrt{2\frac{(0.002m)*1.29*10^6 N/C*1.6*10^{-19}C}{1.67*10^{-27}kg}}= 0.703 *10^6 m/s[/tex]
A skiy diver, with parachute unopened, falls 625 m in 15.0s.
Then she opens her parachute and falls another 356 m in142 s. What
is her average velocity (both magnitude anddirection) for the
entire fall?
Answers
Answer:
average velocity = 6.25m/sec
Explanation:
given data:
for unopened
height = 625 m
time = 15 sec
for opened
height = 356 m
time = 142 sec
Unopened:
[tex]V1 = \frac{625\ m}{15\ sec} = 41.67m/sec[/tex]
Opened:
[tex]V2 = \frac{356\ m}{142\ sec} = 2.51m/sec[/tex]
we know that
Total Average Velocity[tex] = \frac{Total\ distance}{Total\ time}[/tex]
average velocity[tex] = \frac{981\ m}{157\ sec}[/tex]
average velocity = 6.25m/sec
downward direction.
Calculate the speed of a proton that has moved in a uniform electric field of 180.0 N/C from rest over a distance of 12.5-cm. Assume it began at rest.
Answers
Answer:
Given:
Electric field = 180 N/C
[tex]Force\ on\ proton = 1.6\times10^{-19} C[/tex]
[tex]Force\ on\ proton = 180\times1.6\times10^{-19} =288\times10^{-19} N[/tex]
[tex]Mass\ of\ proton = 1.673\times10^{-27} kg[/tex]
[tex]Acceleration of proton = \frac{force}{mass}[/tex]
[tex]Acceleration\ of\ proton = \frac{288\times10^{-19}}{1.673*10^{-27}} =172\times108 m/s^{2}[/tex]
Let the speed of proton be "x"
x = [tex]\sqrt{Acceleration}[/tex]
[tex]x = \sqrt{(2\times172\times108\times0.125)}=65602.2 m/s[/tex]
Answer:
the velocity of the proton is 65574.38 m/s
Explanation:
given,
uniform electric field = 180 N/C
Distance = 12.5 cm = 0.125 m
charge of proton = 1.6 × 10⁻¹⁹ C
force = E × q
=180 × 1.6 × 10⁻¹⁹
F= 2.88 × 10⁻¹⁷ N
mass of proton = 1.673 × 10⁻²⁷ kg
acceleration =[tex]\dfrac{force}{mass}[/tex]
=[tex]\dfrac{2.88 \times 10^{-17}}{1.673\times 10^{-27}}[/tex]
=1.72 × 10¹⁰ m/s²
velocity = [tex]\sqrt{2\times 0.125 \times 1.72 \times 10^{10}}[/tex]
=65574.38 m/s
hence , the velocity of the proton is 65574.38 m/s
Light is incident normally on two narrow parallel slits a distance of 1.00 mm apart. A screen is placed a distance of 1.2 m from the slits. The distance on the screen between the central maximum and the centre of the n=4 bright spot is measured to be 3.1 mm. a Determine the wavelength of light. b This experiment is repeated in water (of refractive index 1.33). Suggest how the distance of 3.1 mm would change, if at all.
Answers
Answer:
Explanation:
distance between slits d = 1 x 10⁻³ m
Screen distance D = 1.2 m
Wave length of light = λ
Distance of n th bright fringe fro centre
= n λ D / d where n is order of bright fringe . Here n = 4
Given
3.1 x 10⁻³ = (4 x λ x 1.2) / 1 x 10⁻³
λ = 3.1 x 10⁻⁶ / 4.8
= .6458 x 10⁻⁶
6458 x 10⁻¹⁰m
λ= 6458 A.
The distance will reduce 1.33 times
New distance = 3.1 /1.33
= 2.33 mm.
Xylene (a common solvent in the petroleum industry) boils at 281.3°F at one atmosphere pressure. At what temperature does Xylene boil in °C, R and K?
Answers
Answer:
Xylene boils at 138.5 °C, 740.97 R and 411.65 K
Explanation:
To convert the temperature in Fahrenheit to Celsius you need to use this formula[tex]T_{\°C}=(T_{\°F}-32)\cdot \frac{5}{9}[/tex]
We know that temperature is 281.3 °F so in °C is:
[tex]\°C=(281.3-32)\cdot \frac{5}{9}= 138.5 \°C[/tex]
To convert Fahrenheit to Rankine you need to use this formula[tex]T_{R}=T_{\°F}+459.67\\T_{R}=281.3\°F+459.67=740.97 R[/tex]
To convert Fahrenheit to Kelvin you need to use this formula[tex]T_{K}=(T_{\°F}+459.67)\cdot \frac{5}{9} \\T_{K}=(281.3 \°F+459.67)\cdot \frac{5}{9} \\T_{K}=411.65K[/tex]
A piano wire of length 2.5 m vibrates so that one-half wavelength is contained on the string. If the frequency of vibration is 35 Hz, the amplitude of vibration is 3.0 mm, and the density is 20 g/m, how much energy is transmitted per second down the wire?
Answers
Answer:
The energy transmitted per second down the wire is 0.761 watt.
Explanation:
Given that,
Length = 2.5 m
Amplitude = 3.0 mm
Density = 20 g/m
Frequency = 35 Hz
We need to calculate the wavelength
Using formula of wavelength
[tex]L = \dfrac{\lambda}{2}[/tex]
[tex]\lambda=2L[/tex]
Put the value into the formula
[tex]\lambda=2\times2.5[/tex]
[tex]\lambda=5\ m[/tex]
We need to calculate the speed
Using formula of speed
[tex]v = f\lambda[/tex]
Put the value into the formula
[tex]v =35\times5[/tex]
[tex]v =175\ m/s[/tex]
We need to calculate the energy is transmitted per second down the wire
Using formula of the energy is transmitted per second
[tex]P=\dfrac{1}{2}\mu A^2\omega^2\times v[/tex]
[tex]P=\dfrac{1}{2}\mu\times A^2\times(2\pi f)^2\times v[/tex]
Put the value into the formula
[tex]P=\dfrac{1}{2}\times20\times10^{-3}\times(3.0\times10^{-3})^2\times4\times\pi^2\times(35)^2\times175[/tex]
[tex]P=0.761\ watt[/tex]
Hence, The energy transmitted per second down the wire is 0.761 watt.
Leady oxide is a material that?s usually composed of A. 25% free lead and 75% lead oxide. B. 10% free lead and 90% lead oxide. C. 60% free lead and 40% lead oxide. D. 50% free lead and 50% lead oxide.
Answers
Answer:
option A is correct
25% free lead and 75% lead oxide
Explanation:
we have given free lead and lead oxide %
so here we know lead oxide material usually composed in the range of for free lead and lead oxide as
lead oxide material contain free lead range is 22 % to 38 %
so here we have only option 1 which contain in free lead in the range of 22 % to 38 % i.e 25 % free lead
so
option A is correct
25% free lead and 75% lead oxide
A 5.0 A current is set up in a circuit for 6.0 min by a rechargeable battery with a 6.0 V emf. By how much is the chemical energy of the battery reduced?
Answers
Answer:
The chemical energy of the battery was reduced in 10800J
Explanation:
The first thing to take into account is that the stored energy in a battery is in Watts per second or Joules ([tex]W\cdot s=J[/tex]). It means that the battery provides a power for a certain time.
The idea is to know how much [tex]W\cdot s[/tex] has been consumed by the circuit.
The first step is to know the power that is consumed by the circuit. It is [tex]P=V\cdot I[/tex]. The problem says that the circuit consumes a current of 5.0A with a voltage of 6.0V. It means that the power consumed is:
[tex]P=V\cdot I=(6.0V)\cdot (5.0A)=30W[/tex]
The previous value (30W) is the power that the circuit consumes.
Now, you must find the total amount of power that is consumed by the circuit in 6.0 minutes. You just have to multiply the power that the circuit consumed by the time it worked, it means, 6.0 minutes.
[tex]energy=P\cdot t=(30W)\cdot (6.0min)=180W\cdot min[/tex]
You must convert the minutes unit to seconds. Remember that 1 minute has 60 seconds.
[tex]energy=P\cdot t=(30W)\cdot (6.0min) \cdot \frac{60s}{1min}=10800W\cdot s=10800J[/tex]
Thus, the chemical energy of the battery was reduced in 10800J
A large asteroid of mass 33900 kg is at rest far away from any planets or stars. A much smaller asteroid, of mass 610 kg, is in a circular orbit about the first at a distance of 146 meters as a result of their mutual gravitational attraction. What is the speed of the second asteroid? Now suppose that the first and second asteroids carry charges of 1.18 C and -1.18 C, respectively. How fast would the second asteroid have to be moving in order to occupy the same circular orbit as before?
Answers
Answer:
a) 1.2*10^-4 m/s
b) 375 m/s
Explanation:
I assume the large asteroid doesn't move.
The smaller asteroid is affected by an acceleration determined by the universal gravitation law:
a = G * M / d^2
Where
G: universal gravitation constant (6.67*10^-11 m^3/(kg*s^2))
M: mass of the large asteroid (33900 kg)
d: distance between them (146 m)
Then:
a = 6.67*10^-11 * 33900 / 146^2 = 10^-10 m/s^2
I assume the asteroid in a circular orbit, in this case the centripetal acceleration is:
a = v^2/r
Rearranging:
v^2 = a * r
[tex]v = \sqrt{a * r}[/tex]
v = \sqrt{10^-10 * 146} = 1.2*10^-4 m/s
If the asteroids have electric charges of 1.18 C and -1.18 C there will be an electric force of:
F = 1/(4π*e0)*(q1*q2)/d^2
Where e0 is the electrical constant (8.85*10^-12 F/m)
F = 1/(4π*8.85*10^-12) (-1.18*1.18)/ 146^2 = -587 kN
On an asteroid witha mass of 610 kg this force causes an acceleration of:
F = m * a
a = F / m
a = 587000 / 610 = 962 m/s^2
With the electric acceleration, the gravitational one is negligible.
The speed is then:
v = \sqrt{962 * 146} = 375 m/s
A muon has a kinetic energy equal to 4 times its rest energy of 105 MeV. (a) What is its velocity, in units of c?
(b) What is its momentum in energy units (i.e., units of MeV/c)?
Answers
Answer:
v = 0.9798*c
Explanation:
E0 = 105 MeV
The mass of a muon is
m = 1.78 * 10^-30 kg
The kinetic energy is:
[tex]Ek = \frac{E0}{\sqrt{1 - \frac{v^2}{c^2}}}-E0[/tex]
The kinetic energy is 4 times the rest energy.
[tex]4*E0 = \frac{E0}{\sqrt{1 - \frac{v^2}{c^2}}}-E0[/tex]
[tex]4 = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}-1[/tex]
[tex]5 = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}[/tex]
[tex]\sqrt{1 - \frac{v^2}{c^2}} = \frac{1}{5}[/tex]
[tex]1 - \frac{v^2}{c^2} = \frac{1}{25}[/tex]
v^2 / c^2 = 1 - 1/25
v^2 / c^2 = 24/25
v^2 = 24/25 * c^2
v = 0.9798*c
Assuming that the limits of the visible spectrum are approximately 380 and 700 nm, find the angular range of the first-order visible spectrum produced by a plane grating that has 900 grooves/mm with the light incident normally on the grating.
Answers
Answer:
angular range is ( 0.681 rad , 0.35 rad )
Explanation:
given data
wavelength λ = 380 nm = 380 × [tex]10^{-9}[/tex] m
wavelength λ = 700 nm = 700 × [tex]10^{-9}[/tex] m
to find out
angular range of the first-order
solution
we will apply here slit experiment equation that is
d sinθ = m λ ...........1
here m is 1 for single slit and d is = [tex]\frac{1}{900*10^3 m}[/tex]
so put here value in equation 1 for 380 nm
we get
d sinθ = m λ
[tex]\frac{1}{900*10^3}[/tex] sinθ = 1 × 380 × [tex]10^{-9}[/tex]
θ = 0.35 rad
and for 700 nm
we get
d sinθ = m λ
[tex]\frac{1}{900*10^3}[/tex] sinθ = 1 × 700 × [tex]10^{-9}[/tex]
θ = 0.681 rad
so angular range is ( 0.681 rad , 0.35 rad )
Charge g is distributed in a spherically symmetric ball of radius a. (a) Evaluate the average volume charge density p. (b) Now assume p(r) is directly proportional to r. (i) Derive the formula for p(r) in terms of r, Q, and a. At what value of r does p(r)= ? [Ans: 340] (ii) Find q(r), and graph it.
Answers
Answer:
Explanation:
The volume of a sphere is:
V = 4/3 * π * a^3
The volume charge density would then be:
p = Q/V
p = 3*Q/(4 * π * a^3)
If the charge density depends on the radius:
p = f(r) = k * r
I integrate the charge density in spherical coordinates. The charge density integrated in the whole volume is equal to total charge.
[tex]Q = \int\limits^{2*\pi}_0\int\limits^\pi_0 \int\limits^r_0 {k * r} \, dr * r*d\theta* r*d\phi[/tex]
[tex]Q = k *\int\limits^{2*\pi}_0\int\limits^\pi_0 \int\limits^r_0 {r^3} \, dr * d\theta* d\phi[/tex]
[tex]Q = k *\int\limits^{2*\pi}_0\int\limits^\pi_0 {\frac{r^4}{4}} \, d\theta* d\phi[/tex]
[tex]Q = k *\int\limits^{2*\pi}_0 {\frac{\pi r^4}{4}} \, d\phi[/tex]
[tex]Q = \frac{\pi^2 r^4}{2}}[/tex]
Since p = k*r
Q = p*π^2*r^3 / 2
Then:
p(r) = 2*Q / (π^2*r^3)
A perfectly flexible cable has length L, and initially it is at rest with a length Xo of it hanging over the table edge. Neglecting friction, compute the length hanging over the table edge after an elapsed time t, assuming cable sections remain straight during the subsequent motion.
Answers
Answer:
[tex]X=X_o+\dfrac{1}{2}gt^2[/tex]
Explanation:
Given that
Length = L
At initial over hanging length = Xo
Lets take the length =X after time t
The velocity of length will become V
Now by energy conservation
[tex]\dfrac{1}{2}mV^2=mg(X-X_o)[/tex]
So
[tex]V=\sqrt{2g(X-X_o)}[/tex]
We know that
[tex]\dfrac{dX}{dt}=V[/tex]
[tex]\dfrac{dX}{dt}=\sqrt{2g(X-X_o)}[/tex]
[tex]\sqrt{2g}\ dt=(X-X_o)^{-\frac{1}{2}}dX[/tex]
At t= 0 ,X=Xo
So we can say that
[tex]X=X_o+\dfrac{1}{2}gt^2[/tex]
So the length of cable after time t
[tex]X=X_o+\dfrac{1}{2}gt^2[/tex]
A 100 meter rope is 20 kg and is stretched with a tension of 20 newtons. If one end of the rope is vibrated with small amplitude at 10Hz, what would the velocity of waves traveling down it be? What would the velocity be if it rained and the rope soaked up 5 kg of water?
Answers
Answer:
The velocity waves before rain is 10 m/s
The velocity of wave after the rope soaked up 5 kg more is 8.944 m/s
Solution:
As per the question:
Length of the rope, l = 100 m
Mass of the rope, m = 20 kg
Force due to tension in the rope, [tex]T_{r} = 20 N[/tex]
Frequency of vibration in the rope, f = 10 Hz
Extra mass of the rope after being soaked in rain water, m' = 5 kg
Now,
In a rope, the wave velocity is given by:
[tex]v_{w} = \sqrt{\frac{T_{r}}{M_{d}}}[/tex] (1)
where
[tex]M_{d}[/tex] = mass density
Mass density before soaking, [tex]M_{d} = \frac{m}{l} = \frac{20}{100} = 0.20[/tex]
Mass density after being soaked, [tex]M_{d} = \frac{m + m'}{l} = \frac{25}{100} = 0.25[/tex]
Initially, the velocity is given by using eqn (1):
[tex]v_{w} = \sqrt{\frac{20}{0.20}} = 10 m/s[/tex]
The velocity after being soaked in rain:
[tex]v_{w} = \sqrt{\frac{20}{0.25}} = 8.944 m/s[/tex]
With each beat of your heart the aortic valve opens and closes. The valve opens and closes very rapidly, with a peak velocity as high as 4 m/s. If we image it with 7 MHz sound and the speed of sound is approximately 1500 m/s in human tissue, what is the frequency shift between the opening and closing of the valve?
Answers
Answer:
|Δf| = 37.3 kHz
Explanation:
given,
peak velocity = 4 m/s
speed of the sound = 1500 m/s
frequency = 7 MHz
[tex]v = C\dfrac{\pm \dlta f}{2 f_0}[/tex]
[tex]\delta f = \pm 2 f_0 (\dfrac{V}{C})[/tex]
[tex]\delta f = \pm 2\times 7 (\dfrac{4}{1500})[/tex]
[tex]=\pm 0.0373 MHz[/tex]
= 37.3 kHz
|Δf| = 37.3 kHz
hence, frequency shift between the opening and closing valve is 37.3 kHz
Hot air enters a rectangular duct (20cm wide, 25cm high, and 5m long) at 100 kPa and 60 degrees C at an average velocity of 5 m/s. While air flows the duct, it gets cool down (loses energy) so that air leave the duct at 54 degrees C. Determine the rate of heat loss from the air under steady condition
Answers
Answer:
1.57 kW
Explanation:
The rate of heat loss is given by:
q = Gm * Cp * (tfin - ti)
Where
q: rate of heat loss
Gm: mass flow
Cp: specific heat at constant pressure
The Cp of air is:
Cp = 1 kJ/(kg*K)
The mass flow is the volumetric flow divided by the specific volume
Gm = Gv / v
The volumetric flow is the air speed multiplied by the cruss section of the duct.
Gv = s * h * w (I name speed s because I have already used v)
The specific volume is obtained from the gas state equation:
p * v = R * T
60 C is 333 K
The gas constant for air is 287 J/(kg*K)
Then:
v = (R * T)/p
v = (287 * 333) / 100000 = 0.955 m^3/kg
Then, the mass flow is
Gm = s * h * w / v
And rthe heat loss is of:
q = s * h * w * Cp * (tfin - ti) / v
q = 5 * 0.25 * 0.2 * 1 * (54 - 60) / 0.955 = -1.57 kW (negative because it is a loss)
When a honeybee flies through the air, it develops a charge of +20 pC . Part A How many electrons did it lose in the process of acquiring this charge? Express your answer as a number of electrons.
Answers
The number of electrons lost by the by the honeybee in acquiring the charge of +20 pC is;
n = 1.25 × 10^(8) electrons
We are given;
Charge of honeybee; Q = 20 pC = 20 × 10^(-12) C
Now, formula for number of electrons is;
n = Q/e
Where;
e is charge on electron = 1.6 × 10^(-19) C
Thus;
n = (20 × 10^(-12))/(1.6 × 10^(-19))
n = 1.25 × 10^(8) electrons
Read more at; https://brainly.com/question/14653647
A rigid adiabatic container is divided into two parts containing n1 and n2 mole of ideal gases respectively, by a movable and thermally conducting wall. Their pressure and volume are P1, V1 for part 1 and P2, V2 for part 2 respectively. Find the final pressure P and temperature T after the two gas reaches equilibrium. Assume the constant volume specific heats of the two gas are the same.
Answers
Answer:
Explanation:
Given
Pressure, Temperature, Volume of gases is
[tex]P_1, V_1, T_1 & P_2, V_2, T_2 [/tex]
Let P & T be the final Pressure and Temperature
as it is rigid adiabatic container therefore Q=0 as heat loss by one gas is equal to heat gain by another gas
[tex]-Q=W+U_1----1[/tex]
[tex]Q=-W+U_2-----2[/tex]
where Q=heat loss or gain (- heat loss,+heat gain)
W=work done by gas
[tex]U_1 & U_2[/tex] change in internal Energy of gas
Thus from 1 & 2 we can say that
[tex]U_1+U_2=0[/tex]
[tex]n_1c_v(T-T_1)+n_2c_v(T-T_2)=0[/tex]
[tex]T(n_1+n_2)=n_1T_1+n_2T_2[/tex]
[tex]T=\frac{n_1+T_1+n_2T_2}{n_1+n_2}[/tex]
where [tex]n_1=\frac{P_1V_1}{RT_1}[/tex]
[tex]n_2=\frac{P_2V_2}{RT_2}[/tex]
[tex]T=\frac{\frac{P_1V_1}{RT_1}\times T_1+\frac{P_2V_2}{RT_2}\times T_2}{\frac{P_1V_1}{RT_1}+\frac{P_2V_2}{RT_2}}[/tex]
[tex]T=\frac{P_1V_1+P_2V_2}{\frac{P_1V_1}{T_1}+\frac{P_2V_2}{T_2}}[/tex]
and [tex]P=\frac{P_1V_1+P_2V_2}{V_1+V_2}[/tex]
A 1100 kg car is traveling around a flat 82.3 m radius curve. The coefficient of static friction between the car tires and the road is .521. What is the maximum speed in m/s at which the car can take the curve?
Answers
Answer:
The maximum speed of car will be 20.5m/sec
Explanation:
We have given mass of car = 1100 kg
Radius of curve = 82.3 m
Static friction [tex]\mu _s=0.521[/tex]
We have to find the maximum speed of car
We know that at maximum speed centripetal force will be equal to frictional force [tex]m\frac{v^2}{r}=\mu _srg[/tex]
[tex]v=\sqrt{\mu _srg}=\sqrt{0.521\times 82.3\times 9.8}=20.5m/sec[/tex]
So the maximum speed of car will be 20.5m/sec
Answer:20.51 m/s
Explanation:
Given
Mass of car(m)=1100 kg
radius of curve =82.3 m
coefficient of static friction([tex]\mu [/tex])=0.521
here centripetal force is provided by Friction Force
[tex]F_c(centripetal\ force)=\frac{mv^2}{r}[/tex]
Friction Force[tex]=\mu N[/tex]
where N=Normal reaction
[tex]\frac{mv^2}{r}=\mu N[/tex]
[tex]\frac{1100\times v^2}{82.3}=0.521\times 1100\times 9.81[/tex]
[tex]v^2=0.521\times 9.81\times 82.3[/tex]
[tex]v=\sqrt{420.63}=20.51 m/s [/tex]