Answer:
h = 104.13 m
Explanation:
Given that,
Horizontal velocity of the car, v = 13 m/s
It hits the ground 60 m from the shoreline
We need to find the height of the cliff. Let t be the time of fall. We can find it as follows :
![t=\dfrac{d}{v}\\\\t=\dfrac{60\ m}{13\ m/s}\\\\t=4.61\ s](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7Bd%7D%7Bv%7D%5C%5C%5C%5Ct%3D%5Cdfrac%7B60%5C%20m%7D%7B13%5C%20m%2Fs%7D%5C%5C%5C%5Ct%3D4.61%5C%20s)
So, the time of fall is 4.61 seconds.
Let h is the height of the cilff. Using second equation of motion to find it as follows :
![h=\dfrac{1}{2}gt^2\\\\h=\dfrac{1}{2}\times 9.8\times 4.61^2\\\\=104.13\ m](https://tex.z-dn.net/?f=h%3D%5Cdfrac%7B1%7D%7B2%7Dgt%5E2%5C%5C%5C%5Ch%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%209.8%5Ctimes%204.61%5E2%5C%5C%5C%5C%3D104.13%5C%20m)
So, the height of the cliff is 104.13 m.
Answer:
Initial velocity, U = 28.73m/s
Explanation:
Given the following data;
Final velocity, V = 35m/s
Acceleration, a = 5m/s²
Distance, S = 40m
To find the initial velocity (U), we would use the third equation of motion.
V² = U² + 2aS
Where;
V represents the final velocity measured in meter per seconds.
U represents the initial velocity measured in meter per seconds.
a represents acceleration measured in meters per seconds square.
S represents the displacement measured in meters.
Substituting into the equation, we have;
35² = U + 2*5*40
1225 = U² + 400
U² = 1225 - 400
U² = 825
Taking the square root of both sides, we have;
Initial velocity, U = 28.73m/s
Acceleration = r w² radius r = 0.82 meter angular velocity w
4.7 = 0.82 w²
So w = 2.394 radians / sec
Time period T = time duration for completing one revolution = 2 π / w
= 2π / 2.394 = 2.624 seconds
Answer
given,
force per unit length = 350 µN/m
current, I = 22.5 A
y = y = 0.420 m
![\dfrac{F}{L}= \dfrac{KI_1I_2}{d}](https://tex.z-dn.net/?f=%5Cdfrac%7BF%7D%7BL%7D%3D%20%5Cdfrac%7BKI_1I_2%7D%7Bd%7D)
![I_2 = \dfrac{F}{L}\dfrac{d}{KI_1}](https://tex.z-dn.net/?f=I_2%20%3D%20%5Cdfrac%7BF%7D%7BL%7D%5Cdfrac%7Bd%7D%7BKI_1%7D)
![I_2 = 350\times 10^{-6}\times \dfrac{0.42}{2 \times 10^{-7}\times 22.5}](https://tex.z-dn.net/?f=I_2%20%3D%20350%5Ctimes%2010%5E%7B-6%7D%5Ctimes%20%5Cdfrac%7B0.42%7D%7B2%20%5Ctimes%2010%5E%7B-7%7D%5Ctimes%2022.5%7D)
I₂ = 32.67 A
distance where the magnetic field is zero
![\dfrac{4\pi \times 10^{-7}\times 32.67}{2\pi y_1}=\dfrac{4\pi \times 10^{-7}\times 22.5}{2\pi (0.42-y_1)}](https://tex.z-dn.net/?f=%5Cdfrac%7B4%5Cpi%20%5Ctimes%2010%5E%7B-7%7D%5Ctimes%2032.67%7D%7B2%5Cpi%20y_1%7D%3D%5Cdfrac%7B4%5Cpi%20%5Ctimes%2010%5E%7B-7%7D%5Ctimes%2022.5%7D%7B2%5Cpi%20%280.42-y_1%29%7D)
![y_1 = 0.248\ m](https://tex.z-dn.net/?f=y_1%20%3D%200.248%5C%20m)
there the distance at which the magnetic field is zero in the two wire is at 0.248 m.