Answer:
F = 263.51 N
Explanation:
given,
diameter of wheel = 78 cm
diameter of axle = 14.8 cm
Force exerted on the rim of wheel = 150 N
Force applied outside the axle = ?
To prevent rotation wheel from rotating the Force 'F' should be applied outside of the axle.
Net momentum about the center of mass should be zero
now,
Moment of about center due to 150 N = moment about center due to F on axle
7.4 F = 1950
F = 263.51 N
Hence, Force exerted outside of the axle in order to prevent the wheel from rotating is equal to 263.51 N.
When you are in the front passenger seat of a car turning to the left, you may find yourself pressed against the right-side door. because of <span> Newton's first and third law</span>
M = mass of the bowling ball = 4 kg
V = speed of bowling ball = 3.93 m/s
P = magnitude of momentum of bowling ball = ?
magnitude of momentum of bowling ball is given as
P = MV
inserting the values
P = 4 x 3.93
P = 15.72 kgm/s
m = mass of ping-pong ball = 2.293 g = 2.293 x 10⁻³ kg
v = speed of the ping-pong ball = ?
p = magnitude of momentum of ping-pong ball
Given that :
magnitude of momentum of ping-pong ball = magnitude of momentum of bowling ball
p = P
m v = 15.72
(2.293 x 10⁻³) v = 15.72
v = 6.86 x 10³ m/s
<span>How much kinetic energy does a 7.2-kg dog need to make a vertical jump of 1.2 m?
</span>Answer:
85 Joule (approx)
Explanation:
Potential energy at highest point
<span><span>P.E = mgh = 7.2 kg × 9.8 m/s2</span>×1.2 m≈85 Joule</span>
This is the kinetic energy required for dog to jump a height of 1.2 m.
hope this helps!
