Answer:
The velocity of the light will be 1.0c only
Explanation:
The velocity of the light measured in the case given in question will be 1.0c only.
This is due to the fact that the velocity of light is never relative. The velocity of the light is maximum
The velocity of the light cannot be scaled down in no case
Thus, the velocity of the light remains as constant.
Hence, the velocity of the light measured will be 1.0c although the ships have relative velocity.
Human hair
Yield strength (MPa)
140-160
Ultimate tensile strength (MPa)
200-250
Hope that helps.
Answer:
10 m/s
Explanation:
Momentum before collision = momentum after collision
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(8 kg)(8 m/s) + (6 kg)(6 m/s) = (8 kg)(5 m/s) + (6 kg) v
64 kg m/s + 36 kg m/s = 40 kg m/s + (6 kg) v
60 kg m/s = (6 kg) v
v = 10 m/s
Answer:
<em>The final speed of the vehicle is 36 m/s</em>
Explanation:
<u>Uniform Acceleration</u>
When an object changes its velocity at the same rate, the acceleration is constant.
The relation between the initial and final speeds is:

Where:
vf = Final speed
vo = Initial speed
a = Constant acceleration
t = Elapsed time
The vehicle starts from rest (vo=0) and accelerates at a=4.5 m/s2 for t=8 seconds. The final speed is:


The final speed of the vehicle is 36 m/s
Answer:
The height is 
A circular hoop of different diameter cannot be released from a height 30cm and match the sphere speed because from the conservation relation the speed of the hoop is independent of the radius (Hence also the diameter )
Explanation:
From the question we are told that
The height is 
The angle of the slope is 
According to the law of conservation of energy
The potential energy of the sphere at the top of the slope = Rotational kinetic energy + the linear kinetic energy

Where I is the moment of inertia which is mathematically represented as this for a sphere

The angular velocity
is mathematically represented as

So the equation for conservation of energy becomes
![mgh_s = \frac{1}{2} [\frac{2}{5} mr^2 ][\frac{v}{r} ]^2 + \frac{1}{2}mv^2](https://tex.z-dn.net/?f=mgh_s%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5B%5Cfrac%7B2%7D%7B5%7D%20mr%5E2%20%5D%5B%5Cfrac%7Bv%7D%7Br%7D%20%5D%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
![mgh_s = \frac{1}{2} mv^2 [\frac{2}{5} +1 ]](https://tex.z-dn.net/?f=mgh_s%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20mv%5E2%20%5B%5Cfrac%7B2%7D%7B5%7D%20%2B1%20%5D)
![mgh_s = \frac{1}{2} mv^2 [\frac{7}{5} ]](https://tex.z-dn.net/?f=mgh_s%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20mv%5E2%20%5B%5Cfrac%7B7%7D%7B5%7D%20%5D)
![gh_s =[\frac{7}{10} ] v^2](https://tex.z-dn.net/?f=gh_s%20%3D%5B%5Cfrac%7B7%7D%7B10%7D%20%5D%20v%5E2)

Considering a circular hoop
The moment of inertial is different for circle and it is mathematically represented as

Substituting this into the conservation equation above
![mgh_c = \frac{1}{2} (mr^2)[\frac{v}{r} ] ^2 + \frac{1}{2} mv^2](https://tex.z-dn.net/?f=mgh_c%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%28mr%5E2%29%5B%5Cfrac%7Bv%7D%7Br%7D%20%5D%20%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20mv%5E2)
Where
is the height where the circular hoop would be released to equal the speed of the sphere at the bottom



Recall that 


Substituting values
