Answer:
Explanation:
Given:
length of ladder 
weight of ladder 
position of firefighter 
weight of firefighter 
angle of ladder 
Unknown:
force of the wall on the ladder 
force of friction on base of ladder 
normal force on base of ladder 
From the free body diagram of the sketch you get 3 equations:

Solving the equations gives:

a)

b)

c) Using the result from b and solving for 

Answer:
A is the answer .(A) is correct
Answer:
The mass of the ice block is equal to 70.15 kg
Explanation:
The data for this exercise are as follows:
F=90 N
insignificant friction force
x=13 m
t=4.5 s
m=?
applying the equation of rectilinear motion we have:
x = xo + vot + at^2/2
where xo = initial distance =0
vo=initial velocity = 0
a is the acceleration
therefore the equation is:
x = at^2/2
Clearing a:
a=2x/t^2=(2x13)/(4.5^2)=1.283 m/s^2
we use Newton's second law to calculate the mass of the ice block:
F=ma
m=F/a = 90/1.283=70.15 kg
By working with percentages, we want to see how many inches is the center of gravity out of the limits. We will find that the CG is 1.45 inches out of limits.
<h3>What are the limits?</h3>
First, we need to find the limits.
We know that the MAC is 58 inches, and the limits are from 26% to 43% MAC.
So if 58 in is the 100%, the 26% and 43% of that are:
- 26% → (26%/100%)*58in = 0.26*58 in = 15.08 in
- 43% → (43%/100%)*58in = 0.43*58 in = 24.94 in.
But we know that the CG is found to be 45.5% MAC, then it measures:
(45.5%/100%)*58in = 0.455*58in = 26.39 in
We need to compare it with the largest limit, so we get:
26.39 in - 24.94 in = 1.45 in
This means that the CG is 1.45 inches out of limits.
If you want to learn more about percentages, you can read:
brainly.com/question/14345924
Answer:
6.9066 × 10⁻⁵ m
Explanation:
For constructive interference, the expression is:
Where, m = 1, 2, .....
d is the distance between the slits.
The formula can be written as:
....1
The location of the bright fringe is determined by :
Where, L is the distance between the slit and the screen.
For small angle ,
So,
Formula becomes:
Using 1, we get:

Thus, the distance between the central maximum is 3.00 cm
First bright fringe , m = 1 occur at 3.00 / 2 = 1.50 cm
Since,
1 cm = 0.01 m
y = 0.0150 m
Given L = 2.00 m
λ = 518 nm
Since, 1 nm = 10⁻⁹ m
So,
λ = 518 × 10⁻⁹ m
Applying the formula as:

<u>⇒ d, distance between the slits = 6.9066 × 10⁻⁵ m</u>