Answer:
(a) W=1.20×10⁴J
(b) U= -5.46×10⁴J
(c) Q= -4.26×10⁴J
Explanation:
Given that student does 1.20×10⁴J work
(a) W=1.20×10⁴J
Work done by student,so positive sign
During the process, his internal energy decreases by 5.46×10⁴J.
(b) U= -5.46×10⁴J.
As the Energy decreases therefore negative sign
For (c) Q
We know the formula

Answer:
E. The wheel with spokes has about twice the KE.
See explanation in: https://quizlet.com/100717504/physics-8-mc-flash-cards/
Answer:
The time is 
The speed is 
Explanation:
From the question we are told that
The height of the cliff is 
Generally from kinematic equation we have that

before the jump the persons initial velocity is u = 0 m/s
So

=> 
Generally from kinematic equation

=> 
=> 
Answer:
(a) T= 38.4 N
(b) m= 26.67 kg
Explanation:
We apply Newton's second law:
∑F = m*a (Formula 1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Kinematics
d= v₀t+ (1/2)*a*t² (Formula 2)
d:displacement in meters (m)
t : time in seconds (s)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s²
v₀=0, d=18 m , t=5 s
We apply the formula 2 to calculate the accelerations of the blocks:
d= v₀t+ (1/2)*a*t²
18= 0+ (1/2)*a*(5)²
a= (2*18) / ( 25) = 1.44 m/s²
to the right
We apply Newton's second law to the block A
∑Fx = m*ax
60-T = 15*1.44
60 - 15*1.44 = T
T = 38.4 N
We apply Newton's second law to the block B
∑Fx = m*ax
T = m*ax
38.4 = m*1.44
m= (38.4) / (1.44)
m = 26.67 kg