A circle has a revolution of 360°. Since there are 12 hour markings, each hour interval has an angle of 30°. In radians, that would be equal to π/6 radians. So, in every 1 hour that passes, it covers π/6 of an angle. So, the angular velocity denoted as ω is π/6 ÷ 1 hour = π/6 rad/h. We can compute the average linear velocity, v, from the relationship:
v = rω, where r is the radius of the circle which is the length of the hour hand
v = (2.4 cm)(π/6 rad/h)
v = 1.257 cm/hour
Therefore, the average velocity is 1.257 cm per hour.
For the average acceleration, it is equal to zero. The hands of the clock move at a constant velocity. Since acceleration is the change of velocity per unit time, there is no change of velocity because it's constant. That's why it is zero.
Answer:
<h2>a) Time elapsed before the bullet hits the ground is 0.553 seconds.</h2><h2>b)
The bullet travels horizontally 110.6 m</h2>
Explanation:
a) Consider the vertical motion of bullet
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 0 m/s
Acceleration, a = 9.81 m/s²
Displacement, s = 1.5 m
Substituting
s = ut + 0.5 at²
1.5 = 0 x t + 0.5 x 9.81 xt²
t = 0.553 s
Time elapsed before the bullet hits the ground is 0.553 seconds.
b) Consider the horizontal motion of bullet
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 200 m/s
Acceleration, a = 0 m/s²
Time, t = 0.553 s
Substituting
s = ut + 0.5 at²
s = 200 x 0.553 + 0.5 x 0 x 0.553²
s = 110.6 m
The bullet travels horizontally 110.6 m
Answer:
Explanation:
Given the mass as M, the rotational inertia of the mower is;
-The roller doesn't slip while rolling;
Answer:
Explanation:
Block A sits on block B and force is applied on block A . Block A will experience two forces 1) force P and 2 ) friction force in opposite direction of motion . Block B will experience one force that is force of friction in the direction of motion .
Let force on block A be P . friction force on it will be equal to kinetic friction, that is μ mg , where μ is coefficient of friction and m is mass of block A
friction force = .4 x 2.5 x 9.8
= 9.8 N
net force on block A = P - 9.8
acceleration = ( P - 9.8 ) / 2.5
force on block B = 9.8
acceleration = force / mass
= 9.8 / 6
for common acceleration
( P - 9.8 ) / 2.5 = 9.8 / 6
( P - 9.8 ) / 2.5 = 1.63333
P = 13.88 N .
