Explanation:
Given that,
Charge acting on the object, 
Force acting on the object, 
(in downward direction)
(a) The electric force acting in the electric field is given by :
E is the electric field
E = 4.75 N/C
The direction of electric field is as same as electric force. But it is negative charge. So, the direction of electric field is in upward direction.
(b) The charge on the proton is, 
The force acting on the proton is :
If the charge on the proton is positive, the force on the proton is in upward direction.
Hence, this is the required solution.
Answer:
The work done on the hose by the time the hose reaches its relaxed length is 776.16 Joules
Explanation:
The given spring constant of the of the spring, k = 88.0 N/m
The length by which the hose is stretched, x = 4.20 m
For the hose that obeys Hooke's law, and the principle of conservation of energy, the work done by the force from the hose is equal to the potential energy given to the hose
The elastic potential energy, P.E., of a compressed spring is given as follows;
P.E. = 1/2·k·x²
∴ The potential energy given to hose, P.E. = 1/2 × 88.0 N/m × (4.20 m)²
1/2 × 88.0 N/m × (4.20 m)² = 776.16 J
The work done on the hose = The potential energy given to hose, P.E. = 776.16 J
Chemical to thermal to electrical current: Burning of coal or natural gases. Gravitational potential to kinetic to electrical current.
I would answer but you have no choices lol
Answer:
<h2>a) Time elapsed before the bullet hits the ground is 0.553 seconds.</h2><h2>b)
The bullet travels horizontally 110.6 m</h2>
Explanation:
a) Consider the vertical motion of bullet
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 0 m/s
Acceleration, a = 9.81 m/s²
Displacement, s = 1.5 m
Substituting
s = ut + 0.5 at²
1.5 = 0 x t + 0.5 x 9.81 xt²
t = 0.553 s
Time elapsed before the bullet hits the ground is 0.553 seconds.
b) Consider the horizontal motion of bullet
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 200 m/s
Acceleration, a = 0 m/s²
Time, t = 0.553 s
Substituting
s = ut + 0.5 at²
s = 200 x 0.553 + 0.5 x 0 x 0.553²
s = 110.6 m
The bullet travels horizontally 110.6 m
