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Agata [3.3K]
3 years ago
7

How much work is done when a 25-N horizontal force is applied to a 10-kg box, causing it to move 5 meters?

Physics
1 answer:
Katarina [22]3 years ago
8 0

Answer:

W=50*sin(25)

Explanation:

W=F*D*sin(25)

W=10*5*sin(25)

W=50*sin(25)

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The MAC is 58 inches, The CG limits are from 26% to 43% MAC. If the CG is found to be
eimsori [14]

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

6 0
2 years ago
Is anyone good with 9-10th grade physics?? Please pm me I have a few questions.
Tcecarenko [31]
Let's give it a whirl ..... 
3 0
3 years ago
Please help i need it to be answered by tomorrow may 1st i’m very frustrated
butalik [34]

Answer: At least one of his parents had brown eyes

Explanation: whoever had the brown eyes he got more genetics from them when it came to his eyes  

8 0
3 years ago
An athlete prepares to throw a 2.0-kilogram discus. His arm is 0.75 meters long. He spins around several times with the discus a
wlad13 [49]
Centripetal Force (Fcp) = ?

His arm length = Radius (R) = 0.75 m

Discus velocity = Linear Velocity (V) = 5 m/s

Discus mass (m) = 2 kg

Centripetal Acceleration (Acp) = V^2/R or W^2 x R
In this case i will use the V^2/R formula, because it uses the discus velocity (V).

fcp = m \times acp \\ fcp = m \times {v}^{2} \div r \\ fcp = 2 \times {5}^{2} \div 0.75 \\ fcp = 2 \times 25 \div 0.75 \\ fcp = 50 \div 0.75
fcp = 66.666... = 66 \: newtons

Answer: Last option, 66 N.
7 0
3 years ago
Read 2 more answers
Help me please....................
maksim [4K]
You need to write the question for the solution.
Add the question then only can get the answer
8 0
4 years ago
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