Answer:

Explanation:
Given that :
mass of the SUV is = 2140 kg
moment of inertia about G , i.e
= 875 kg.m²
We know from the conservation of angular momentum that:

![mv_1 *0.765 = [I+m(0.765^2+0.895^2)] \omega_2](https://tex.z-dn.net/?f=mv_1%20%2A0.765%20%3D%20%5BI%2Bm%280.765%5E2%2B0.895%5E2%29%5D%20%5Comega_2)
![2140v_1*0.765 = [875+2140(0.765^2+0.895^2)] \omega_2](https://tex.z-dn.net/?f=2140v_1%2A0.765%20%3D%20%5B875%2B2140%280.765%5E2%2B0.895%5E2%29%5D%20%5Comega_2)



From the conservation of energy as well;we have :

^2 -2140(9.81)[\sqrt{0.76^2+0.895^2} -0.765]] =0](https://tex.z-dn.net/?f=%5B%5Cfrac%7B1%7D%7B2%7D%20%5B875%2B2140%280.765%5E2%2B0.895%5E2%29%5D%280.4262%20%5C%20v_1%29%5E2%20-2140%289.81%29%5B%5Csqrt%7B0.76%5E2%2B0.895%5E2%7D%20-0.765%5D%5D%20%3D0)






Answer:
2.36 x 10^6 J
Explanation:
Tc = 0°C = 273 K
TH = 22.5°C = 295.5 K
Qc = heat used to melt the ice
mass of ice, m = 85.7 Kg
Latent heat of fusion, L = 3.34 x 10^5 J/kg
Let Energy supplied is E which is equal to the work done
Qc = m x L = 85.7 x 3.34 x 10^5 = 286.24 x 10^5 J
Use the Carnot's equation


QH = 309.8 x 10^5 J
W = QH - Qc
W = (309.8 - 286.24) x 10^5
W = 23.56 x 10^5 J
W = 2.36 x 10^6 J
Thus, the energy supplied is 2.36 x 10^6 J.
It does take on new set of proerties
Answer: To determine acceleration ,Micah also needs the Time of the total trip in seconds.
Explanation:
Acceleration can be defined as rate of change of velocity.

for calculating acceleration, initial and final velocity are required in meter per second and the total time of the trip in seconds. Then acceleration is measured in meter per second square.
Thus, Micah knows that a car had a change in velocity of 15 m/s.To determine acceleration ,Micah also needs the <u>Time</u> of the total trip in seconds.
Answer:
Part 1) Voltage in secondary windings is 61.08 Volts
Part 2) Current in secondary windings is 0.53 Amperes
Explanation:
The potential developed in the primary and secondary winding of a transformer are related as

where
Np no of turns in primary coil
Ns no of turns in secondary coil
Vp Voltage of turns in primary coil
Vs Voltage of turns in secondary coil
Applying values in the formula we get

Part 2)
Using Ohm's law the current is given by
