Answer:
d = 6.43 cm
Explanation:
Given:
- Speed resistance coefficient in silicon n = 3.50
- Memory takes processing time t_p = 0.50 ns
- Information is to be obtained within T = 2.0 ns
Find:
- What is the maximum distance the memory unit can be from the central processing unit?
Solution:
- The amount of time taken for information pulse to travel to memory unit:
t_m = T - t_p
t_m = 2.0 - 0.5 = 1.5 ns
- We will use a basic relationship for distance traveled with respect to speed of light and time:
d = V*t_m
- Where speed of light in silicon medium is given by:
V = c / n
- Hence, d = c*t_m / n
-Evaluate: d = 3*10^8*1.5*10^-9 / 3.50
d = 0.129 m 12.9 cm
- The above is the distance for pulse going to and fro the memory and central unit. So the distance between the two is actually d / 2 = 6.43 cm
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.
This is an insidious question. Quite frankly, I would not have
expected to see it here on Brainly. But I'm ready to play the
cards that you have dealt me.
None of the choices offered is a correct solution.
If the output of the AC generator is nice and sinusoidal, and
its maximum (peak) emf is 150 volts, then its RMS emf is
(1/2) (150) (√2) = 106.07 volts.
The resistor's dissipation is
Power = (current) x (voltage) .
If the resistor is dissipating its full rated 35W, then
35W = (current) x (106.07 V)
Divide each side by 106.07 V:
RMS Current = (35W) / (106.07 V) = 0.33 Ampere .
_________________________________________
Looking over the choices offered . . .
The largest choice ... 3.1 A ... is the current in a resistor
that is dissipating 35W if the voltage is
(35W / 3.1A) = 11.29 volts .
The smallest choice ... 1.2 A ... is the current in a resistor
that is dissipating 35W if the voltage is
(35W / 1.2A) = 29.17 volts .
Whatever you meant the so-called "150 V" of the generator
to represent ... whether the RMS sinusoidal, peak sinusoidal,
peak square-wave, RMS square-wave, DC, average, etc. ...
none of the choices for current, in combination with any of these
generators, would dissipate 35W.
That will be A melting a substance that's because the rest are examples of chemical changes and also its the same thing but in a different form.
