The body system on the chart
Because the frictional force between the orange skin peel is great enough when you are walking for it to be carried on the tray, along with the gravitational force downwards onto the tray. When you stop, the force that you exerted moving forward it the same as on the tray and on the orange. So when you stop, the force is still on the orange as the same velocity as your we’re traveling, while the tray and you stop.
Answer:
Time taken to reach final velocity = 5.5 second
Explanation:
Given:
Initial velocity (Starting from rest)(u) = 0 m/s
Acceleration of ball (a) = 1 m/s²
Final velocity (v) = 5.5 m/s
Find:
Time taken to reach final velocity
Computation:
Using first equation of motion;
v = u + at
where,
v = final velocity
u = initial velocity
a = acceleration
t = time taken
5.5 = 0 + (1)(t)
5.5 = t
Time taken to reach final velocity = 5.5 second
Answer:
1.603 s
Explanation:
Given that
Initial mass, = 0.45 kg
Initial period, = 1.45 s
Initial radius, = 0.14 m
Final mass, = 0.55 kg
Final period, = ?
Final radios, = 0.14 m
Since we are finding the rotation period of two masses of same radius, we can assume that the outward force is the same in both cases. This means that
m₁r₁ω₁² = m₂r₂ω2²
Where, ω = 2π/T, on substituting, we have
0.45 * 0.14 * (2π / 1.45)² = 0.550 * 0.14 * (2π / T₂)²
0.45 / 1.45² = 0.550 / T₂²
T₂² = 0.550 * 1.45² / 0.45
T₂² = 2.56972
T₂ = √2.56972
T₂ = 1.603 sec
Answer:
(B) 1.23 x 10⁴ J
Explanation:
Given;
radius of the sphere, r = 7.0 m
diameter of the sphere, d = 2r = 14.0 m
mass of the person sitting on the sphere, m = 90.0 kg
The gravitational potential energy of the person is given by;
P.E = mgh
where;
g is acceleration due to gravity = 9.8 m/s²
h is the height above the ground level = d = 14.0 m
P.E = mgh
P.E = (90)(9.8)(14)
P.E = 12348 J
P.E = 1.2348 x 10⁴ J
Therefore, the gravitational potential energy of the person is 1.2348 x 10⁴ J
