Explanation:
Given:
v₀ = 250 mph
v = 0 mph
t = 25 s
Find: a
v = at + v₀
(0 mph) = a (25 s) + (250 mph)
a = -10 mph/s
The mass of the hoop is the only force which is computed by:F net = 2.8kg*9.81m/s^2 = 27.468 N
the slow masses that must be quicker are the pulley, ring, and the rolling sphere.
The mass correspondent of M the pulley is computed by torque τ = F*R = I*α = I*a/R F = M*a = I*a/R^2 --> M = I/R^2 = 21/2*m*R^2/R^2 = 1/2*m
The mass equal of the rolling sphere is computed by: the sphere revolves around the contact point with the table. So using the proposition of parallel axes, the moment of inertia of the sphere is I = 2/5*mR^2 for spin about the midpoint of mass + mR^2 for the distance of the axis of rotation from the center of mass of the sphere. I = 7/5*mR^2 M = 7/5*m
the acceleration is then a = F/m = 27.468/(2.8 + 1/2*2 + 7/5*4) = 27.468/9.4 = 2.922 m/s^2
Work = Force x Distance
Assuming that this work is being done parallel to the displacement that is, but under that assumption:
W = (50)(10)
W = 500 J
Answer:
Explanation:
We can use Newton's Universal Law of Gravitation to solve this problem:
., where 
is acceleration due to gravity at the planet's surface, 
is gravitational constant 
, 
is the mass of the planet, and 
is the radius of the planet.
Since acceleration due to gravity is given as 
, our radius should be meters. Therefore, convert 
kilometers to meters:
.
Now plugging in our values, we get:
,
Solving for :
.
