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Crazy boy [7]
3 years ago
10

arlene is to walk across a high wire strung horizontally between two buildings 10.0m apart. The sag in the rope when she is at t

he midpoint is 10.0 degrees, if her mass is 50.0 kg, what is the tension in the rope at this point?...can you help me...thanks!
Physics
2 answers:
Ipatiy [6.2K]3 years ago
4 0
We are given with a force in the middle of a rope measuring 10 m and an angle of sagging of 10 degrees from the horizontal. The tension of the rope is equal to the hypotenuse of the right triangle. 

sin 10 = 50 kg * 9.8 m/s2 / T
Tension = 2821. 80 N 
Greeley [361]3 years ago
3 0

Answer:

T = 1412.3 N

Explanation:

As we know that weight of the Arlene is counterbalanced by the tension in the string

So here it is given that string makes 10 degree angle at mid point

So we will have

T sin\theta + T sin\theta = mg

so we have

T = \frac{mg}{2sin\theta}

here we know that

m = 50 kg

\theta = 10 degree

now we have

T = \frac{50 \times 9.81}{2sin10}

T = 1412.3 N

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In the united states a standard letter sized piece of paper is 8.5 inches wide by 11 inches long. The international standard for
Andre45 [30]

This question is incomplete

Complete Question

In the United States, a standard letter-sized piece of paper is 8.5 inches wide by 11 inches long. The international standard for a letter-sized piece of paper is different. The international standard is based on SI units: 21.0 cm wide by 29.7 cm long.

a. Convert 21.0 cm to inches. Show your dimensional analysis setup.

b. Convert 29.7 cm to inches. Show your dimensional analysis setup.

c. State the dimensions, in inches, of the international standard for a letter-sized piece of paper.

d. Which piece of paper is longer: a U.S. letter-sized piece of paper, or an international letter-sized piece of paper?

Answer:

a) 8.267721 inches ≈ 8.3 inches

b) 11.6929197 inches ≈ 11.7 inches

c) It's dimensions in inches for the international standard for letter - sized for paper = 8.3 wide inches by 11.7 inches long

d) The International standard letter - sized paper is longer.

Explanation:

a. Convert 21.0 cm to inches. Show your dimensional analysis setup.

1 cm = 0.393701inch

21 cm =

Cross Multiply

21 cm × 0.393701inch/ 1 cm

= 8.267721 inches

Approximately 8.3 inches

b. Convert 29.7 cm to inches. Show your dimensional analysis setup.

1 cm = 0.393701inch

29.7 cm =

Cross Multiply

29.7 × 0.393701 inch/ 1 cm

= 11.6929197 inches

Approximately 11.7 inches

c. State the dimensions, in inches, of the international standard for a letter-sized piece of paper.

The international standard for a letter-sized has dimensions 21.0 cm wide by 29.7 cm long.

Where

21.0cm = 8.267721 inches

≈ 8.3 inches

29.7cm = 11.6929197 inches

≈ 11.7 inches

Hence, it's dimensions in inches = 8.3 inches by 11.7 inches.

d. Which piece of paper is longer: a U.S. letter-sized piece of paper, or an international letter-sized piece of paper?

U.S letter - sized paper = 8.5 inches wide by 11 inches long

International standard letter- sized paper = 8.3 wide inches by 11.7 inches long.

Hence, the International standard letter - sized paper is longer.

5 0
3 years ago
What is a free electron? Can someone help?
Aleksandr [31]
A free electron is one which has become detached from a covalent bond between two atoms and is able to move around from atom to atom and possibly take part in electric current flow.
3 0
3 years ago
Which best describes a difference between laser light and regular light?
Mashcka [7]

Answer:

B

Explanation:

6 0
2 years ago
If a force of 12 N is applied to an object
Varvara68 [4.7K]

Answer:

<h3>The answer is 3 kg</h3>

Explanation:

The mass of the object can be found by using the formula

m =  \frac{f}{a}  \\

f is the force

a is the acceleration

From the question we have

m =  \frac{12}{4}  \\

We have the final answer as

<h3>3 kg</h3>

Hope this helps you

6 0
3 years ago
A mass of 5kg accelerates at 3m/s/s, how much force was put on it?
goldfiish [28.3K]

Answer:

15N

Explanation:

According to Newton's Second Law of Motion

F = m*a

mass = m = 5Kg

acceleration = a = 3m/s^2

=> F = 5kg * 3m/s^2

=> F = 15 N

5 0
3 years ago
Read 2 more answers
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