43. A mother pushes her 120-newton child, who is
1 answer:
5.0 N.s ( 5.0 newton second) is the magnitude of the impulse imparted to the child by the mother
Option A
<u>Explanation:</u>
An impulse is a variation in the object's momentum when the object is subjected to force for a time period. Therefore, you can use the pulse to calculate the change in pulse or use the pulse to calculate the average impact force. The equation for impulse is given by,
Given data:
F = 10 N
t = 0.50 s
We need to find the magnitude of the impulse. By applying the given values to the above equation, we get
You might be interested in
Answer:
19.21ms-¹
Explanation:
that is the solution above
The question is incomplete. Here is the complete question.
A uniform electric field of 2kN/C points in the +x-direction.
(a) What is the change in potential energy of a +2.00nC test charge, as it is moved from point a at x = -30.0 cm to point b at x = +50.0 cm?
(b) The same test charge is released from rest at point a. What is the kinetic energy when it passes through point b?
(c) If a negative charge instead of a positive charge were used in this problem, qualitatively, how would your answers change?
Answer: (a) ΔU = 3.2× J
(b) KE = 2× J
Explanation: <u>Potential</u> <u>Energy</u> (U) is the amount of work done due to its position or condition and its unit is Joule (J). <u>Kinetic</u> <u>Energy</u> (KE) is the ability to do work by virtue of velocity and the unit is also (J). <u>Mechanical</u> <u>Energy</u> is the sum of Potential and Kinetic Energies of a system.
(a) Related to electricity, Potential Energy can be calculated as:
ΔU = Eqd
where E is the electric field (in N/C);
q is the charge (in C);
d is the distance between plaques (in m);
For a at x = - 30cm and b at x = 50 cm:
E = 2× N/C
q = 2× C
d = 50 - (-30) = 80× = 8×
m
ΔU = = Eqd
= 2×
. 2×
. 8×
ΔU = 3.2× J
(b) Mechanical Energy is constant, so:
Since the initial position is zero and there is no initial kinetic energy:
- (2×
. 2×
. 5×
)
J
(c) If the charge is negative, electric field does positive work, which diminishes the potential energy. The charge flows from the negative side towards the positive side and stays, not doing anything.