|Momentum| = (mass) x (speed)
225 kg-m/s =(50kg) x (speed)
Divide each side by (50kg): Speed=(225 kg-m/s) / (50 kg) = 4.5 m/s .
Regarding the velocity, nothing can be said other than the speed, because
we have no information regarding the direction of the object's motion.
Answer:
7.22 × 10²⁹ kg
Explanation:
For the material to be in place, the gravitational force on the material must equal the centripetal force on the material.
So, F = gravitational force = GMm/R² where M = mass of neutron star, m = mass of object and R = radius of neutron star = 17 km
The centripetal force F' = mRω² where R = radius of neutron star and ω = angular speed of neutron star
So, since F = F'
GMm/R² = mRω²
GM = R³ω²
M = R³ω²/G
Since ω = 500 rev/s = 500 × 2π rad/s = 1000π rad/s = 3141.6 rad/s = 3.142 × 10³ rad/s and r = 17 km = 17 × 10³ m and G = universal gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg²
Substituting the values of the variables into M, we have
M = R³ω²/G
M = (17 × 10³ m)³(3.142 × 10³ rad/s)²/6.67 × 10⁻¹¹ Nm²/kg²
M = 4913 × 10⁹ m³ × 9.872 × 10⁶ rad²/s²/6.67 × 10⁻¹¹ Nm²/kg²
M = 48,501.942 × 10¹⁵ m³rad²/s² ÷ 6.67 × 10⁻¹¹ Nm²/kg²
M = 7217.66 × 10²⁶ kg
M = 7.21766 × 10²⁹ kg
M ≅ 7.22 × 10²⁹ kg
<h2><u>Q</u><u>u</u><u>e</u><u>s</u><u>t</u><u>i</u><u>o</u><u>n</u>:-</h2>
The speed of a wave is 40 m/s. If the wavelength is 80 centimeters, what is the frequency of the wave ?
<h2><u>A</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u>:-</h2>
<h3>Given:-</h3>
Velocity (V) = 40 m/s
Wavelength
= 80 cm = 0.8 m
<h3>To Find:-</h3>
The frequency (F) of the wave.
<h2>Solution:-</h2>
We know,

40 = F × 0.8
F = 
F = 50
<h3>The frequency of the wave is <u>5</u><u>0</u><u> </u><u>H</u><u>z</u>. [Answer]</h3>
Answer:
ω = 2.1 rad/sec
Explanation:
- As the rock is moving along with the merry-go-round, in a circular trajectory, there must be an external force, keeping it on track.
- This force, that changes the direction of the rock but not its speed, is the centripetal force, and aims always towards the center of the circle.
- Now, we need to ask ourselves: what supplies this force?
- In this case, the only force acting on the rock that could do it, is the friction force, more precisely, the static friction force.
- We know that this force can be expressed as follows:

where μs = coefficient of static friction between the rock and the merry-
go-round surface = 0.7, and Fn = normal force.
- In this case, as the surface is horizontal, and the rock is not accelerated in the vertical direction, this force in magnitude must be equal to the weight of the rock, as follows:
- Fn = m*g (2)
- This static friction force is just the same as the centripetal force.
- The centripetal force depends on the square of the angular velocity and the radius of the trajectory, as follows:

- Since (1) is equal to (3), replacing (2) in (1) and solving for ω, we get:

- This is the minimum angular velocity that would cause the rock to begin sliding off, due to that if it is larger than this value , the centripetal force will be larger that the static friction force, which will become a kinetic friction force, causing the rock to slide off.