Answer: 0.78m/s^2
Explanation:
Force is the product of mass of an object and its acceleration.
i.e Force = Mass x Acceleration
In this case,
Force applied on cart = 35N
Mass of cart and contents = 45Kg
Acceleration = ?
Then, Force = Mass x Acceleration
35N = 45Kg x Acceleration
Acceleration = (35N/45kg)
Acceleration = 0.78 m/s^2 (The unit of acceleration is metre per second square)
Thus, the shopping cart moved with an accelerations of 0.78m/s^2
<h2>
Option A is the correct answer.</h2>
Explanation:
For a simple pendulum we have at any position
Total energy = Constant
Kinetic energy + Potential energy = Constant
At maximum displacement of pendulum, velocity is zero, hence kinetic energy is zero.
At maximum displacement the pendulum only have potential energy.
Given that maximum potential energy is 10 J.
That is at any position
Kinetic energy + Potential energy = 10 J
Now we need to find kinetic energy when potential energy is 5 J.
Kinetic energy + 5 J = 10 J
Kinetic energy = 5 J
Option A is the correct answer.
Answer:
The magnitude is 
The direction is the positive x axis
Explanation:
From the question we are told that
The electric field is E = 450 V/m in the negative y ‑direction
The speed of the proton is
in the positive z direction
Generally the overall force acting on the proton is mathematical represented as

Now for the beam of protons to continue un-deflected along its straight-line trajectory then 
So

=> 
Generally from unit vector cross product vector multiplication

So the direction of B (magnetic field must be in the positive x -axis )
So

=> 
=>
=>
=> 
=>
Answer:
0.6 A
Explanation:
As the motor gears towards full speed, the voltage of the circuit tends to become the difference that exists between the line voltage and the that of the back emf. Remembering Ohm's law, we then apply it to get the final, lower current that is based on the reduced voltage Ef. We use the provided resistance in the question, that is 3 ohms.
Ef = (120 V) - (117 V) = 3 V
If = Ef/R = (3 V) / (5.0 Ω) = 0.6 A
Complete Question
A thin, horizontal, 12-cm-diameter copper plate is charged to 4.4 nC . Assume that the electrons are uniformly distributed on the surface. What is the strength of the electric field 0.1 mm above the center of the top surface of the plate?
Answer:
The values is 
Explanation:
From the question we are told that
The diameter is 
The charge is 
The distance from the center is 
Generally the radius is mathematically represented as

=> 
=> 
Generally electric field is mathematically represented as
![E = \frac{Q}{ 2\epsilon_o } [1 - \frac{k}{\sqrt{r^2 + k^2 } } ]](https://tex.z-dn.net/?f=E%20%3D%20%20%5Cfrac%7BQ%7D%7B%202%5Cepsilon_o%20%7D%20%5B1%20-%20%5Cfrac%7Bk%7D%7B%5Csqrt%7Br%5E2%20%2B%20%20k%5E2%20%7D%20%7D%20%5D)
substituting values
![E = \frac{4.4 *10^{-9}}{ 2* (8.85*10^{-12}) } [1 - \frac{(1.00 *10^{-4})}{\sqrt{(0.06)^2 + (1.0*10^{-4})^2 } } ]](https://tex.z-dn.net/?f=E%20%3D%20%20%5Cfrac%7B4.4%20%2A10%5E%7B-9%7D%7D%7B%202%2A%20%288.85%2A10%5E%7B-12%7D%29%20%7D%20%5B1%20-%20%5Cfrac%7B%281.00%20%2A10%5E%7B-4%7D%29%7D%7B%5Csqrt%7B%280.06%29%5E2%20%2B%20%20%281.0%2A10%5E%7B-4%7D%29%5E2%20%7D%20%7D%20%5D)
