Answer:
So, the force, F is the agent which provides the basic property of motion or rest to the body.While, the car is more obviously to have a mass, m and that the angle withe road or surface is also given which is normal to the road(i.e angle =90 degree). Then we say that lets say that the car is moving with the constant velocity of 20 m/sec and its kept unchanged by the car. So, we have the mass, m as 1 kg for the car and the value of the gravity we have the g=9.8 m/sec.
Now,
We have F=ma,
and a=v/t,
so we can have another equation for it as,
Now, providing the required data to, it; ∴t =2 sec,
F=(1)×(20/2),
- So, the car would be acting the force,F of about 10 N while the car is present on the lower region of the track.
Answer:
0.9904 m/s
Explanation:
To solve this problem we need to use the conservation of momentum:
m1v1 + m2v2 = m1'v1' + m2'v2'
Using this equation, we have:
m_bullet*v_bullet = m_bullet_after*v_bullet_after + m_block*v_block
0.00425 * 375 = 0.00425 * 114 + 1.12 * v_block
1.12 * v_block = 1.5938 - 0.4845
1.12 * v_block = 1.1093
v_block = 1.1093 / 1.12
v_block = 0.9904 m/s
Answer:
the speed of the car moving after accelerating for 56 m is 20 m/s
Explanation:
The computation is shown below:
As we know that
Vf^2 = Vi^2 + 2ad
= ( 8m/s )^2 + 2 ( 3m/s2 ) ( 56m)
= 400 m2/s2
Vf = 20 m/s
Hence, the speed of the car moving after accelerating for 56 m is 20 m/s
For circular motion.
Centripetal acceleration = mv²/r = mω²r
Where v = linear velocity, r = radius = diameter/2 = 1/2 = 0.5m
m = mass = 175g = 0.175kg.
Angular speed, ω = Angle covered / time
= 2 revolutions / 1 second
= 2 * 2π radians / 1 second
= 4π radians / second
Centripetal Acceleration = mω²r = 0.175*(4π)² * 0.5 Use a calculator
≈13.817 m/s²
The magnitude of acceleration ≈13.817 m/s² and it is directed towards the center of rotation.
Tension in the string = m*a
= 0.175*13.817
= 2.418 N