Answer: Option (b) is the correct answer.
Explanation:
Since, there is a negative charge present on the ball and a positive charge present on the rod. So, when the negatively charged metal ball will come in contact with the rod then positive charges from rod get conducted towards the metal ball.
Hence, the rod gets neutralized. But towards the metal ball there is a continuous supply of negative charges. Therefore, after the neutralization of positive charge from the rod there will be flow of negative charges from the metal ball towards the rod.
Thus, we can conclude that negative charge spread evenly on both ends.
W = mg, Assuming g ≈ 9.8 m/s² on the earth surface.
735 N = m* 9.8
735/9.8 = m
75 = m
Mass , m = 75 kg. B.
Answer:
a. 79.1 N
b. 344 J
c. 344 J
d. 0 J
e. 0 J
Explanation:
a. Since the crate has a constant velocity, its net force must be 0 according to Newton's 1st law. The push force 
by the worker must be equal to the friction force 
on the crate, which is the product of friction coefficient μ and normal force N:
Let g = 9.81 m/s2
b. The work is done on the crate by this force is the product of its force and the distance traveled s = 4.35
c. The work is done on the crate by friction force is also the product of friction force and the distance traveled s = 4.35
This work is negative because the friction vector is in the opposite direction with the distance vector
d. As both the normal force and gravity are perpendicular to the distance vector, the work done by those forces is 0. In other words, these forces do not make any work.
e. The total work done on the crate would be sum of the work done by the pushing force and the work done by friction
Answer:
v = 120 m/s
Explanation:
We are given;
earth's radius; r = 6.37 × 10^(6) m
Angular speed; ω = 2π/(24 × 3600) = 7.27 × 10^(-5) rad/s
Now, we want to find the speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator.
The angle will be;
θ = ¾ × 90
θ = 67.5
¾ is multiplied by 90° because the angular distance from the pole is 90 degrees.
The speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator will be:
v = r(cos θ) × ω
v = 6.37 × 10^(6) × cos 67.5 × 7.27 × 10^(-5)
v = 117.22 m/s
Approximation to 2 sig. figures gives;
v = 120 m/s
Using V= vo +at with Vo = 0 and a= 4m/sec2.
V= 0+ 4x8= 32m/s
