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Oksi-84 [34.3K]
2 years ago
12

Ok ok ik ik this is not the answer but I think you need to hear this

Physics
1 answer:
leonid [27]2 years ago
4 0

Answer

oh thanksssss i hope u have a great day

Explanation:

ur a really awesome person and i thank u for that

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As a woman walks, her entire weight is momentarily placed on one heel of her high-heeled shoes. Calculate the pressure exerted o
Soloha48 [4]

Answer:

4.15 x 10^6 N

Explanation:

Area, A = 1.43 cm^2 = 1.43 x 10^-4 m^2

mass, m = 60.5 kg

Weight, F = m g = 60.5 x 9.8 = 592.9 N

Pressure = Force / Area

P = Weight / Area

P = 592.9 / (1.43 x 10^-4)

P = 4.15 x 10^6 N

3 0
3 years ago
When you push a child on a swing, you are doing work on the child because _____.
max2010maxim [7]

ur answer is A or also known as

When you push a child on a swing, you are doing work on the child because  you are pushing against the force of gravity

hope this helps :)

8 0
3 years ago
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In a belly-flop diving contest, the winner is the diver who makes the biggest splash upon hitting the water. the size
ad-work [718]

The second diver have to leap to make a competitive splash by 4.08 m high.

<h3>What is potential energy?</h3>

The energy by virtue of its position is called the potential energy.

PE = mgh

where, g = 9.81 m/s²

Given is the diver jumps from a 3.00-m platform. one diver has a mass of 136 kg and simply steps off the platform. another diver has a mass of 100 kg and leaps upward from the platform.

The potential energy of the first diver must be equal to the second diver.

P.E₁ = P.E₂

m₁gh₁ = m₂gh₂

Substitute the vales, we have

136 x 3  = 100 x h₂

h₂ = ₂4.08 m

Thus, the second diver need to leap by 4.08 m high.

Learn more about potential energy.

brainly.com/question/24284560

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7 0
2 years ago
What can fall but never get hurt
Vladimir [108]

It’s either snow or rain if it’s a riddle sort of.

4 0
3 years ago
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Work of 5 Joules is done in stretching a spring from its natural length to 19 cm beyond its natural length. What is the force (i
densk [106]

Answer:

Explanation:

Given that,

5J work is done by stretching a spring

e = 19cm = 0.19m

Assuming the spring is ideal, then we can apply Hooke's law

F = kx

To calculate k, we can apply the Workdone by a spring formula

W=∫F.dx

Since F=kx

W = ∫kx dx from x = 0 to x = 0.19

W = ½kx² from x = 0 to x = 0.19

W = ½k (0.19²-0²)

5 = ½k(0.0361-0)

5×2 = 0.0361k

Then, k = 10/0.0361

k = 277.008 N/m

The spring constant is 277.008N/m

Then, applying Hooke's law to find the applied force

F = kx

F = 277.008 × 0.19

F = 52.63 N

The applied force is 52.63N

6 0
3 years ago
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