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lidiya [134]
2 years ago
14

How much work do you do when you lift a 100 N child 0.5 m?

Physics
1 answer:
Papessa [141]2 years ago
8 0

Answer:

<h2>50 J</h2>

Explanation:

The work done by an object can be found by using the formula

workdone = force × distance

From the question we have

workdone = 100 × 0.5

We have the final answer as

<h3>50 J</h3>

Hope this helps you

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I need help with homework! what should i choose as the greatest british invention?
STatiana [176]
THE MINI Alec Issigonis, 1959
The Telephone Alexander Graham Bell, 1876

http://www.radiotimes.com/news/2013-01-08/the-50-greatest-british-inventions



4 0
3 years ago
Use the Pythagorean theorem to answer the following question. A ping-pong ball is shot straight north from a popgun at 4.0 m/s.
MaRussiya [10]
Resultant is 5 m/s using the Pythagorean theorem<span />
4 0
2 years ago
Read 2 more answers
Suppose that the sound level of a conversation is initially at an angry 70 db and then drops to a soothing 50 db. assuming that
stepan [7]
Angry sound level = 70 db
Soothing sound level = 50 db
Frequency, f = 500 Hz
Assuming speed of sound = 345 m/s
Density (assumed) = 1.21 kg/m^3
Reference sound intensity, Io = 1*10^-12 w/m^2

Part (a): Initial sound intensity (angry sound)
10log (I/Io) = Sound level
Therefore,
For Ia = 70 db
Ia/(1*10^-12) = 10^(70/10)
Ia = 10^(70/10)*10^-12 = 1*10^-5 W/m^2

Part (b): Final sound intensity (soothing sound)
Is = 50 db
Therefore,
Is = 10^(50/10)*10^-12 = 18*10^-7 W/m^2

Part (c): Initial sound wave amplitude
Now,
I (W/m^2) = 0.5*A^2*density*velocity*4*π^2*frequency^2

Making A the subject;
A = Sqrt [I/(0.5*density*velocity*4π^2*frequency^2)]

Substituting;
A_initial = Sqrt [(1*10^-5)/(0.5*1.21*345*4π^2*500^2)] = 6.97*10^-8 m = 69.7 nm

Part (d): Final sound wave amplitude
A_final = Sqrt [(1*10^-7)/(0.5*1.21*345*4π^2*500^2)] = 6.97*10^-9 m = 6.97 nm
4 0
3 years ago
Ceres is a dwarf planet located in the main asteroid belt.
Pie
Yes that is correct, it <span>lies between the orbits of Mars and Jupiter and is the largest in the belt.

</span>
5 0
3 years ago
As SCUBA divers go deeper underwater, the pressure from the weight of all the water above them increases tremendously which comp
slavikrds [6]

Answer: A.

As a diver rises, the pressure on their body decreases which allows the volume of the gas to decrease.

Explanation:

The problem is that a diver, experiences an increased pressure of water compresses nitrogen and more of it dissolves into the body. Just as there is a natural nitrogen saturation point at the surface, there are saturation points under water. Those depend on the depth, the type of body tissue involved, and also how long a diver is exposed to the extra pressure. The deeper a diver go, the more nitrogen the body absorbs.

The problem is getting rid of the nitrogen once you ascend again. As the pressure diminishes, nitrogen starts dissolving out of the tissues of the diver's body, a process called "off-gassing." That results in tiny nitrogen bubbles that then get carried to the lungs and breathed out. However, if there is too much nitrogen and/or it is released too quickly, small bubbles can combine to form larger bubbles, and those can do damage to the body, anything from minor discomforts all the way to major problems and even death.

4 0
2 years ago
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