The red box must way more. Gravitational potential energy is the product of a an objects mass times the acceleration due to gravity (which is constant on earth) times its height. Since the objects are on the same shelf they are at the same height, and since gravitational acceleration is constant as long as we stay on planet earth, then the mass is the only possible thing that could have changed. This means that the red box must weigh more than the blue box.
Answer:
a) 567J
b) 283.5J
c)850.5J
Explanation:
The expression for the translational kinetic energy is,

Substitute,
14kg for m
9m/s for v

The translational kinetic energy of the center of mass is 567J
(B)
The expression for the rotational kinetic energy is,

The expression for the moment of inertia of the cylinder is,

The expression for angular velocity is,

substitute
1/2mr² for I
and vr for w
in equation for rotational kinetic energy as follows:



The rotational kinetic energy of the center of mass is 283.5J
(c)
The expression for the total energy is,

substitute 567J for E(r) and 283.5J for E(R)

The total energy of the cylinder is 850.5J
Answer:
2874.33 m/s²
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
g = Acceleration due to gravity = 9.81 m/s²

Now H-h = 0.588 - 0.002 = 0.586 m
The final velocity will be the initial velocity

Acceleration of the frog is 2874.33 m/s²
A. 10 rations = 1 deca-ration.
b. 2000 mockingbirds = 2 x 10³ = 2 kilo-mockingbirds.
c. 10⁻⁵ phones = 1 micro-phones.
d. 10⁻⁹ goats = 1 nano-goats.
e. 1018 miners = 1.018 x 10³ = 1.018 kilo-miners.
By using third law of equation of motion, the final velocity V of the rubber puck is 8.5 m/s
Given that a hockey player hits a rubber puck from one side of the rink to the other. The parameters given are:
mass m = 0.170 kg
initial speed u = 6 m/s.
Distance covered s = 61 m
To calculate how fast the puck is moving when it hits the far wall means we are to calculate final speed V
To do this, let us first calculate the kinetic energy at which the ball move.
K.E = 1/2m
K.E = 1/2 x 0.17 x 
K.E = 3.06 J
The work done on the ball is equal to the kinetic energy. That is,
W = K.E
But work done = Force x distance
F x S = K.E
F x 61 = 3.06
F = 3.06/61
F = 0.05 N
From here, we can calculate the acceleration of the ball from Newton second law
F = ma
0.05 = 0.17a
a = 0.05/0.17
a = 0.3 m/
To calculate the final velocity, let us use third equation of motion.
=
+ 2as
=
+ 2 x 0.3 x 61
= 36 + 36
= 72
V = 
V = 8.485 m/s
Therefore, the puck is moving at the rate of 8.5 m/s (approximately) when it hits the far wall.
Learn more about dynamics here: brainly.com/question/402617