Complete Question
An airplane takes off a runway at a constant speed of 49 m/s at constant angle 30 to the horizontal.How high (in meters ) is the airplane above the ground 13 seconds after takeoff?
Answer:
The height is 
Explanation:
From the question we are told that
The speed at which the plane takes off is 
The angle at which it takes off is 
The time taken is 
The vertical distance traveled is mathematically represented as

Substituting values


Answer:
The specific heat capacity is the heat or energy required to change one unit mass of a substance of a constant volume by 1 °C. The formula is Cv = Q / (ΔT ⨉ m)
Answer:
1200 W
Explanation:
Power is given by the ratio between work done and time taken:

where W is the work done and t the time taken.
In this problem, W = 3600 J and t = 3.0 s. Therefore, the power in this exercise is

"The equation can be used to calculate the power absorbed by any surface" statement concerning the Stefan-Boltzmann equation is correct.
Answer: Option A
<u>Explanation:</u>
According to Stefan Boltzmann equation, the power radiated by black body radiation source is directly proportionate to the fourth power of temperature of the source. So the radiation transferred is absorbed by another surface and that absorbed power will also be equal to the fourth power of the temperature. So the equation describes the relation of net radiation loss with the change in temperature from hotter temperature to cooler temperature surface.

So this law is application for calculating power absorbed by any surface.
Answer:
Explanation:
There is electric field between the plates whose value is given by the following expression
electric field E = V /d where V is potential between the plates and d is distance between them
E = 300 / 5 x 10⁻³
= 60 x 10³ N/c
Force on electron = q E where q is charge on the electron
F = 1.6 X 10⁻¹⁹ X 60 X 10³ = 96 X 10⁻¹⁶ N.
Acceleration a = force / mass
a = 96 x 10⁻¹⁶/ mass = 96 x 10⁻¹⁶ / 9.1 x 10⁻³¹
= 10.55 x 10¹⁵ m / s²
For midway , distance travelled
s = 2.5 x 10⁻³ m
s = 1\2 a t²
t = 
= 
t = .474 x 10⁻¹⁸ s
For striking the plate time is calculated as follows
t =
[/tex]
t = 0.67 x 10⁻¹⁸ s