Answer:
3×10^7 m/s or 0.10c (e)
Explanation: If the actual value of the speed of light were to be put into consideration.
Given that the speed of light is c = 3.0×10^8m/s
The alien spaceship is approaching at the rate of 10% of the speed of light.
10% of 3.0×10^8m/s
10/100 × 3.0×10^8m/s
0.1 ×3.0×10^8m/s
3×10^7 m/s. Which is the same thing as 0.1 of c = 0.1×c
E = (1/2)CV²
1 = (1/2)*(2*10⁻⁶)V²
10⁶ = V²
1000 = V
You should charge it to 1000 volts to store 1.0 J of energy.
Answer:
B
Explanation:
Work done can be said to be positive if the applied force has a component to be in the direction of the displacement and when the angle between the applied force and displacement is positive.
In 1 and 2 work done is positive
Answer:
temperature on left side is 1.48 times the temperature on right
Explanation:
GIVEN DATA:

T1 = 525 K
T2 = 275 K
We know that


n and v remain same at both side. so we have

..............1
let final pressure is P and temp 

..................2
similarly
.............3
divide 2 equation by 3rd equation
![\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}](https://tex.z-dn.net/?f=%5Cfrac%7B21%7D%7B11%7D%5E%7B-2%2F3%7D%20%5Cfrac%7B21%7D%7B11%7D%5E%7B5%2F3%7D%20%3D%20%5B%5Cfrac%7BT_1%20%7Bf%7D%7D%7BT_2%20%7Bf%7D%7D%5D%5E%7B5%2F3%7D)

thus, temperature on left side is 1.48 times the temperature on right
Answer:
, 
Explanation:
The jet is flying at constant velocity: this means that its acceleration is zero, so the net force acting on the jet is also zero.
Therefore, we can write:

where
is the thrust force generated by each engine of the jet
is the drag force
Solving for Fd,

The velocity of the jet is

So, the rate at which the drag force does work (which is the power) is

and substituting

we find

Converting into horsepower,
