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the resistance of the cable is 582.9 ohms
we are given the length of the cable which is 3 km, of 1.5 mm in, the diameter and resistivity of copper which is 1.72 m
The formula we are referring to for calculating the resistance of the cable is
R = ρl/A.
As there are 19 strands of copper conductors, so the resistance will be
R = 19( ρl/A)
Here ρ is the resisitivity = 1.72 , l is the length = 3(1+0.05)*10³3= 3150 m
A=pie/4(1.5 x 10⁻³)^2 =1.766 x 10⁻⁶ =1.766 x 10^-6
Substituting the values in the formula we get
R = 19 ( 1.72*3150 )/1.766 x 10⁻⁶
= 582.9 ohm
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Answer:
(a) θ = 33.86°
(b) Ay = 49.92 N
Explanation:
You have that the magnitude of a vector is A = 89.6 N
The x component of such a vector is Ax = 74.4 N
(a) To find the angle between the vector and the x axis you use the following formula for the calculation of the x component of a vector:
(1)
Ax: x component of vector A
A: magnitude of vector A
θ: angle between vector A and the x axis
You solve the equation (1) for θ, by using the inverse of cosine function:

the angle between the A vector and the x axis is 33.86°
(b) The y component of the vector is given by:

the y comonent of the vecor is Ay = 49.92 N
Answer: In this lab we wanted to know how motion can be described. So the hypothesis is if the starting height of a sloped racetrack is increased, then the speed at which a toy car travels along the track will increase because the toy car will have a greater acceleration. My prediction is that cars travel faster on higher tracts. So the heighten the track was intentionally manipulated. So it is the independent variable the speed of the car is the dependent variable. The speed at the first quarter checkpoint is 1.09 m/s. The speed at the second quarter checkpoint is 1.95 m/s. The speed at the third quarter checkpoint is 2.373.36 m/s. The speed at the finish line is 2.803.00 m/s. The average speed increases as the height increases.
The cars on the higher track travel farther than the cars on the lower track, in the same time.
This means that the cars on the higher track have a greater average speed than those on the lower track. This is demonstrated by the
slope of the higher track line being greater than the slope of the lower track line.
Explanation: put it in notes then send it to files to compress it to submit it.
Answer:
The velocity of the arrow after 3 seconds is 30.02 m/s.
Explanation:
It is given that,
An arrow is shot upward on the moon with velocity of 35 m/s, its height after t seconds is given by the equation:

We know that the rate of change of displacement is equal to the velocity of an object.

Velocity of the arrow after 3 seconds will be :

So, the velocity of the arrow after 3 seconds is 30.02 m/s. Hence, this is the required solution.