Answer:
44.6 N
Explanation:
Draw a free body diagram of the block. There are four forces on the block:
Weight force mg pulling down,
Normal force N pushing up,
Friction force Nμ pushing left,
and applied force F pulling right 30° above horizontal.
Sum of forces in the y direction:
∑F = ma
N + F sin 30° − mg = 0
N = mg − F sin 30°
Sum of forces in the x direction:
∑F = ma
F cos 30° − Nμ = 0
F cos 30° = Nμ
N = F cos 30° / μ
Substitute:
mg − F sin 30° = F cos 30° / μ
mg = F sin 30° + (F cos 30° / μ)
Plug in values:
mg = 20 N sin 30° + (20 N cos 30° / 0.5)
mg = 44.6 N
Answer:
Air resistance
Answer B is correct
Explanation:
The friction that occurs when air pushes against a moving object causing it to negatively accelerate is called as air resistance.
hope this helps
brainliest appreciated
good luck! have a nice day!
Answer:
<h2>C. <u>
0.55 m/s towards the right</u></h2>
Explanation:
Using the conservation of law of momentum which states that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision.
Momentum = Mass (M) * Velocity(V)
BEFORE COLLISION
Momentum of 0.25kg body moving at 1.0m/s = 0.25*1 = 0.25kgm/s
Momentum of 0.15kg body moving at 0.0m/s(body at rest) = 0kgm/s
AFTER COLLISION
Momentum of 0.25kg body moving at x m/s = 0.25* x= 0.25x kgm/s
<u>x is the final velocity of the 0.25kg ball</u>
Momentum of 0.15kg body moving at 0.75m/s(body at rest) =
0.15 * 0.75kgm/s = 0.1125 kgm/s
Using the law of conservation of momentum;
0.25+0 = 0.25x + 0.1125
0.25x = 0.25-0.1125
0.25x = 0.1375
x = 0.1375/0.25
x = 0.55m/s
Since the 0.15 kg ball moves off to the right after collision, the 0.25 kg ball will move at <u>0.55 m/s towards the right</u>
<u></u>
Answer:
The time elapsed at the spacecraft’s frame is less that the time elapsed at earth's frame
Explanation:
From the question we are told that
The distance between earth and Retah is 
Here c is the peed of light with value 
The time taken to reach Retah from earth is 
The velocity of the spacecraft is mathematically evaluated as
substituting values
The time elapsed in the spacecraft’s frame is mathematically evaluated as
substituting value
![T = 90000 * \sqrt{ 1 - \frac{[2.4*10^{8}]^2}{[3.0*10^{8}]^2} }](https://tex.z-dn.net/?f=T%20%20%3D%20%2090000%20%2A%20%20%5Csqrt%7B%201%20-%20%20%5Cfrac%7B%5B2.4%2A10%5E%7B8%7D%5D%5E2%7D%7B%5B3.0%2A10%5E%7B8%7D%5D%5E2%7D%20%7D)
=> 
So The time elapsed at the spacecraft’s frame is less that the time elapsed at earth's frame
