Answer:82. Since you have a distance and a force, then the easiest principle to use is energy, i.e. work.
The work done by friction is F * d. This work cancels out the kinetic energy of the stone (1/2)mv^2
Fd = (1/2)mv^2
F = (1/2)mv^2/d.
Plug in m = 20 kg, v = 3 m/sec, d = 40 m.
83. With more mass, the kinetic energy is higher now. The work needed is higher. W = F * d and F is the same.
Explanation:Hope I helped :)
How much work would be done on a particle with 5.0 C of charge on it if it moved from an equipotential line at 5.5 volts to another equipotential line at 3.5 volts?
Answer:
10J
Explanation:
In this question we have the following information
The charge of the particle is q = 5 C
The equipotenetial level is V1 = 5.5 v
and also the
equipotenetial level is V2 = 3.5 v
So we calculate the
work done W=q x (v1-v2)
workdone = 5 x (5.5-3.5)
= 5x2
=10 J
Workdone = 10 J
So we conclude that the workdone on a particle with these information is 10j
A baseball is thrown across the field. The ____________is measured from where the ball is thrown to where landed was 75 feet.
motion
direction
distance
reference point
Answer:
distance i think
Explanation:
A car moves forward up a hill at 12 m/s with a uniform backward acceleration of 1.6 m/s2. What is its displacement after 6 s?
Answer:
The displacement of the car after 6s is 43.2 m
Explanation:
Given;
velocity of the car, v = 12 m/s
acceleration of the car, a = -1.6 m/s² (backward acceleration)
time of motion, t = 6 s
The displacement of the car after 6s is given by the following kinematic equation;
d = ut + ¹/₂at²
d = (12 x 6) + ¹/₂(-1.6)(6)²
d = 72 - 28.8
d = 43.2 m
Therefore, the displacement of the car after 6s is 43.2 m
A man walks south at a speed of 2.00 m/s for 60.0 minutes. He then turns around and walks north a distance 3000 m in 25.0 minutes. What is the average velocity of the man during his entire motion?
Answer:
v = 0.823 m/s
Explanation:
A man walks south at a speed of 2.00 m/s for 60.0 minutes.
The distance covered in South = 60 × 60 × 2 = 7200 m
He then turns around and walks north a distance 3000 m in 25.0 minutes.
As they moved in opposite direction, net displacement will be : 7200 - 3000 = 4200 m
Average velocity of the man = net displacement/time
[tex]v=\dfrac{4200\ m}{(60+25)\times 60}\\\\=0.823\ m/s[/tex]
So, the average velocity of the man is 0.823 m/s.