Answer:
It's your first question!!!
Explanation:
Enjoy it!!!
By definition, speed is the integral of acceleration with respect to time.
We have then:

As the acceleration is constant, then integrating we have:

Where,
vo: constant of integration that corresponds to the initial velocity
We observe then that the speed varies linearly when the acceleration is constant
.
Therefore, for constant acceleration, the velocity is changing.
Answer:
an object with a constant acceleration always have:
A. changing velocity
Answer:
The voltage drop across the bulb is 115 V
Explanation:
The voltage drop equation is given by:

Where:
ΔW is the total work done (4.6kJ)
Δq is the total charge
We need to use the definition of electric current to find Δq

Where:
I is the current (2 A)
Δt is the time (20 s)


Then, we can put this value of charge in the voltage equation.

Therefore, the voltage drop across the bulb is 115 V.
I hope it helps you!
Answer:
See below
Explanation:
Vertical position = 45 + 20 sin (30) t - 4.9 t^2
when it hits ground this = 0
0 = -4.9t^2 + 20 sin (30 ) t + 45
0 = -4.9t^2 + 10 t +45 = 0 solve for t =4.22 sec
max height is at t= - b/2a = 10/9.8 =1.02
use this value of 't' in the equation to calculate max height = 50.1 m
it has 4.22 - 1.02 to free fall = 3.2 seconds free fall
v = at = 9.81 * 3.2 = 31.39 m/s VERTICAL
it will <u>also</u> still have horizontal velocity = 20 cos 30 = 17.32 m/s
total velocity will be sqrt ( 31.39^2 + 17.32^2) = 35.85 m/s
Horizontal range = 20 cos 30 * t = 20 * cos 30 * 4.22 = 73.1 m
Answer:
The right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²
Explanation:
Thickness of the wall is L= 20cm = 0.2m
Thermal conductivity of the wall is K = 2.79 W/m·K
Temperature at the left side surface is T₁ = 50°C
Temperature of the air is T = 22°C
Convection heat transfer coefficient is h = 15 W/m2·K
Heat conduction process through wall is equal to the heat convection process so

Expression for the heat conduction process is

Expression for the heat convection process is

Substitute the expressions of conduction and convection in equation above


Substitute the values in above equation

Now heat flux through the wall can be calculated as

Thus, the right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²