Complete Question
The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s.
(a) What is its angular acceleration in revolutions per minute-squared
(b) How many revolutions does the engine make during this 20 s interval?
rev
Answer:
a

b

Explanation:
From the question we are told that
The initial angular speed is 
The angular speed after
is 
The time for revolution considered is
Generally the angular acceleration is mathematically represented as

=>
=> 
Generally the number of revolution made is
is mathematically represented as

=> 
=> 
Answer:
T = 0.003 s
(Period is written as T)
Explanation:
Period = time it takes for one wave to pass (measured in seconds)
frequency = number of cycles that occur in 1 second
(measured in Hz / hertz / 1 second)
Period : T
frequency : f
So, if we know that the frequency of a wave is 300 Hz, we can find the period of the wave from the relation between frequency and period
T =
f = 
to find the period (T) of this wave, we need to plug in the frequency (f) of 300
T = 
T = 0.00333333333
So, the period of a wave that has a frequency of 300 Hz is 0.003 s
[the period/T of this wave is 0.003 s]
Answer:

I'm just in jss2 but I read physics. this is what I think
Answer:
Solving for time :
(There are 4 formulas from linear motion. These formulas are very helpful as it allows us to prevent complicated calculations. Choose among the four that has : 1. The most constants known
2. The unknown constant that we want to solve)
s = (1/2)(u+v)t <--- one of the formulas
from linear motion
s (distance) = 0.05m
u (initial velocity) = 100m/s
v (final velocity) = 0 m/s (it stops)
t (time taken for change in velocity) = to be found
0.05 = (1/2)(100+0)t
t = 0.001 seconds
Solving for the resistant force :
Since the bullet hits the bag with an impulsive force and stops, the force that stops the bullet is the resistant force.
When the bullet stops :
F net = 0
F r = F imp
F r = (mu -mv)/t
F r = (0.01x100-0.01x0)/0.001
F r = 1/0.001
F r = 1000N
Answer:
The width of the central bright fringe on the screen is observed to be unchanged is 
Explanation:
To solve the problem it is necessary to apply the concepts related to interference from two sources. Destructive interference produces the dark fringes. Dark fringes in the diffraction pattern of a single slit are found at angles θ for which

Where,
w = width
wavelength
m is an integer, m = 1, 2, 3...
We here know that as
as w are constant, then

We need to find
, then

Replacing with our values:


Therefore the width of the central bright fringe on the screen is observed to be unchanged is 