Given :
An electron moving in the positive x direction experiences a magnetic force in the positive z direction.
To Find :
The direction of the magnetic field.
Solution :
We know, force is given by :

Here, q = -e.

Now, for above condition to satisfy :

So, 
Therefore, direction of magnetic field is negative y direction.
Hence, this is the required solution.
The answer is C. You must divide your wavelength and your frequency to get your answer.
Answer:
F = 7,916,955.0N
Explanation:
According to newtons second law
Force = mass * acceleration
Given
mass = 52.0kg
distance S = 22.0m
time t = 17.0 ms = 0.017s
We need to get the acceleration first using the formula;
S = ut+ 1/2at²
22 = 0 + 1/2 a(0.017²)
22 = 0.0001445a
a = 22/0.0001445
a = 152,249.13m/s²
The magnitude of the average force exerted will be;
F = ma
F = 52 * 152,249.13
F = 7,916,955.0N
Answer:
39.81 N
Explanation:
I attached an image of the free body diagrams I drew of crate #1 and #2.
Using these diagram, we can set up a system of equations for the sum of forces in the x and y direction.
∑Fₓ = maₓ
∑Fᵧ = maᵧ
Let's start with the free body diagram for crate #2. Let's set the positive direction on top and the negative direction on the bottom. We can see that the forces acting on crate #2 are in the y-direction, so let's use Newton's 2nd Law to write this equation:
- ∑Fᵧ = maᵧ
- T₁ - m₂g = m₂aᵧ
Note that the tension and acceleration are constant throughout the system since the string has a negligible mass. Therefore, we don't really need to write the subscripts under T and a, but I am doing so just so there is no confusion.
Let's solve for T in the equation...
- T₁ = m₂aᵧ + m₂g
- T₁ = m₂(a + g)
We'll come back to this equation later. Now let's go to the free body diagram for crate #1.
We want to solve for the forces in the x-direction now. Let's set the leftwards direction to be positive and the rightwards direction to be negative.
The normal force is equal to the x-component of the force of gravity.
- (F_n · μ_k) - m₁g sinΘ = m₁aₓ
- (F_g cosΘ · μ_k) - m₁g sinΘ = m₁aₓ
- [m₁g cos(30) · 0.28] - [m₁g sin(30)] = m₁aₓ
- [(6)(9.8)cos(30) · 0.28] - [(6)(9.8)sin(30)] = (6)aₓ
- [2.539595871] - [-58.0962595] = 6aₓ
- 60.63585537 = 6aₓ
- aₓ = 10.1059759 m/s²
Now let's go back to this equation:
We have 3 known variables and we can solve for the tension force.
- T = 2(10.1059759 + 9.8)
- T = 2(19.9059759)
- T = 39.8119518 N
The tension force is the same throughout the string, therefore, the tension in the string connecting M2 and M3 is 39.81 N.
A. scatter plot (100% positive, btw quizzes is a good tool for answers)