Answer:
0.0170m or 1.70cm
Explanation:
L=(μ₀*N^2*A)/(l)
N=x/2*pi*r so r=x/2*pi*N
A=pi*r^2=(pi*x^2)/(4*pi^2*N^2)
L=(μ₀*N^2*A)/(l)=(μ₀*N^2*pi*x^2)/(4*pi^2*N^2*l)=(μ₀*x^2)/(4*pi*l)
l=(μ₀*x^2)/4*pi*L)=[(4*10^-7)*(1.15m)/4*pi*2.47*10^-6H)=0.0170m or 1.70cm
Hope this helps. Any questions please feel free to ask. Thanks!
The cart with speed increasing to the right ... cart-b ...
has a net force to the right acting on it.
The others ...
... Cart-a has no net force at all acting on it, so its speed isn't changing.
... Cart-c has a net force toward the left acting on it, so its speed
toward the left is INcreasing. (That's the same thing as its speed
toward the right DEcreasing.
5. The jogger's velocity is a constant 3.55 m/s between t = 4 s and t = 8 s.
6. Given a linear plot of velocity, the acceleration is determined by the slope of the line. Take any two points on the part of the plot after t = 8 s - for instance, we see it passes through (8 s, 3.5 m/s) and (10 s, 4 m/s) - and compute the slope:
(4 m/s - 3.5 m/s)/(10 s - 8 s) = (0.5 m/s)/(2 s) = 0.25 m/s^2
7. This amounts to finding the area between the velocity function and the time axis and between t = 4 s and t = 8 s. During this time, the velocity is 3.5 m/s. The time interval lasts 4 s. So the distance covered is
(3.5 m/s)*(4 s) = 14 m
8. After 4 seconds, Jimmy's speed decreases from 30.0 m/s to 27.2 m/s, so his acceleration (assuming it was constant) was
a = (27.2 m/s - 30.0 m/s)/(4 s) = -0.200 m/s^2
It's unclear what is meant by "rate of acceleration", since the acceleration is itself a rate. But maybe they just mean to ask for the acceleration, or possibly the magnitude?
The text looks incomplete. Here the complete question found on google:
<em>"The acceleration of a bus is given by ax(t)=αt, where α = 1.28m/s3 is a constant. Part A If the bus's velocity at time t1 = 1.13s is 5.09m/s , what is its velocity at time t2 = 2.02s ? If the bus's position at time t1 = 1.13s is 5.92m , what is its position at time t2 = 2.02s ?"</em>
A) 6.88 m/s
The velocity of the bus can be found by integrating the acceleration. Therefore:
(1)
where
C is a constant term
We know that at 
, the velocity is 
. Substituting these values into (1), we can find the exact value of C:
So now we can find the velocity at time 
:
B) 11.2 m
To find the position, we need to integrate the velocity:
(2)
where D is another constant term.
We know that at , the position is 
. Substituting these values into (2), we can find the exact value of D:
And so now we can find the position at time using eq.(2):
Answer:
The larger the area, the lower the pressure. The force does not depend on the area
Explanation:
- The force exerted by the object on your hand does not depend on the area of contact between the object and the hand, but only on the weight of the object:
where mg is the weight of the object.
- The pressure exerted by the object on your hand is given by:
where F is the force and A is the area of contact between the object and the hand. From this equation, we see that when the area of contact A is larger, the pressure p exerted on your hand is lower, and vice-versa.
