This phenomenon is called Nuclear Magnetic Resonance.
When I met my future wife, she was working in the medical research building next door to the communications building where I worked. (We shared a parking lot.) MRI was not a thing yet, and she was doing research in Nuclear Magnetic Resonance. I learned a lot about it when I walked next door to visit her in her lab. Strange as it may seem, several years earlier, her older brother was involved in the invention of the CAT scan. When we got married, I figured that our kids had at least a 50% chance of inheriting some brains. So we had some, and they've done OK.
Answer:
yes it is essential
Explanation:
a switch is an electrical component that can disconnect or connect the conducting path in an electrical circuit; controls current flow into a circuit
Answer:
2 m/s and -2 m/s
Explanation:
The object travels with an angle of
60.0°
with the positive direction of the y-axis: this means that it lies either in the 1st quadrant (positive x) or in the 2nd quadrant (negative x).
If it lies in the 1st quadrant, the value of vx (component of v along x direction) is:
If it lies in the 2nd quadrant, the value of vx (component of v along x direction) is:
Choices A, B, and D are false statements.
I think choice-C is trying to say the right thing, but it
might have gotten copied incorrectly.
Electric fields and electric forces both increase as the distance
decreases, and decrease as the distance increases.
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.