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4 years ago

What experiment should I make using Gravitational Force? PLEASE HELP ME :)

1 answer:

8 0

You could try the "Spinning Bucket" or the "Center Of Gravity" experiment. There are plenty more that you could research! Hope this helped :)

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A mass of 4 kg is traveling over a quarter circular ramp with a radius of 10 meters. At the bottom of the incline the mass is mo

amid [387]

Answer:

499.7 J

Explanation:

Since total mechanical energy is conserved,

U₁ + K₁ + W₁ = U₂ + K₂ + W₂ where U₁ = potential energy at bottom of incline = mgh₁, K₁ = kinetic energy at bottom of incline = 1/2mv₁² and W₁ = work done by friction at bottom of incline, and U₂ = potential energy at top of incline = mgh₂, K₁ = kinetic energy at top of incline = 1/2mv₂² and W₂ = work done by friction at top of incline. m = mass = 4 kg, h₁ = 0 m, v₁ = 21.3 m/s, W₁ = 0 J, h₂ = radius of circular ramp = 10 m, v₂ = 2.8 m/s, W₂ = unknown.

So, U₁ + K₁ + W₁ = U₂ + K₂ + W₂

mgh₁ + 1/2mv₁² + W₁ = mgh₂ + 1/2mv₂² + W₂

Substituting the values of the variables into the equation, we have

mgh₁ + 1/2mv₁² + W₁ = mgh₂ + 1/2mv₂² + W₂

4 kg × 9.8 m/s²(0) + 1/2 × 4 kg × (21.3 m/s)² + 0 = 4 kg × 9.8 m/s² × 10 m + 1/2 × 4 kg × (2.8 m/s)² + W₂

0 + 2 kg × 453.69 m²/s² = 392 kgm²/s² + 2 kg × 7.84 m²/s² + W₂

907.38 kgm²/s² = 392 kgm²/s² + 15.68 kgm²/s² + W₂

907.38 kgm²/s² = 407.68 kgm²/s² + W₂

W₂ = 907.38 kgm²/s² - 407.68 kgm²/s²

W₂ = 499.7 kgm²/s²

W₂ = 499.7 J

Since friction is a non-conservative force, the work done by all the non-conservative forces is thus W₂ = 499.7 J

4 0

3 years ago

Not sure what im doing wrong on this, I've been using the equation <img src="https://tex.z-dn.net/?f=y_%7Bf%7D%20%3Dy_%7Bi%7D%20

denpristay [2]

Answer:

for text

Explanation:

I think it's correct

6 0

3 years ago

A tuning fork of frequency 254 Hz and an open orang pipe of slightly lower frequency are at 15oC. When

katovenus [111]

The temperature of the air in the open orang pipe has been altered by 18.73° C

The frequency of an open orang pipe is estimated by using the formula:

$\mathbf{f = \dfrac{v}{2L}}$

Then, the combination of the frequency of the tuning fork and the open orang pipe is:

$\mathbf{254 - \dfrac{v}{2L} }$

These combinations of frequency produce 4 beats per sound.

i.e.

$\mathbf{254 - \dfrac{v}{2L} =4}$

$\mathbf{ \dfrac{v}{2L} = 254-4 }$

$\mathbf{ \dfrac{v}{2L} = 250 ----(1)}$

When it is altered, the beats first diminish and increase again by 4.

i.e.

$\mathbf{ \dfrac{v'}{2L} = 254+4 }$

$\mathbf{ \dfrac{v'}{2L} = 258 --- (2) }$

If we equate both equations (1) and (2) together, we have:

$\mathbf{\dfrac{v'}{v}= \dfrac{258}{250}}$

However, from our previous knowledge, we understand that the velocity of an object varies directly proportional to the square root of its temperature.

Hence;

when the temperature of the pipe = unknown ???
the temperature of the open orang pipe = 15

∴

$\implies \mathbf{\sqrt{\Big(\dfrac{273 + T}{273 + 15}\Big)}= \dfrac{258}{250}}$

By squaring both sides, we have:

$\implies \mathbf{\Big(\dfrac{273 + T}{273 + 15}\Big)}= \Big (\dfrac{258}{250}\Big )^2}$

$\implies \mathbf{\Big(\dfrac{273 + T}{273 + 15}\Big)= \Big (\dfrac{66564}{62500}\Big )}$

$\implies \mathbf{\Big(\dfrac{273 + T}{288}\Big)= \Big (1.065024\Big )}$

$\implies \mathbf{273 +T =306.726912 }$

T = 306.726912 - 273

T ≅ 33.73 ° C

∴

The change in temperature ΔT = 33.73° C - 15° C

The change in temperature ΔT = 18.73° C

Learn more about wave frequency here:

brainly.com/question/14316711?referrer=searchResults

4 0

3 years ago

Children are ready for more complex storybooks:

FromTheMoon [43]

Answer:

Explanation:

Because that is when they're mind learns how to read

5 0

3 years ago

If an object is travelling faster than its own sound waves, which of the following is created?

Andrej [43]

Answer:

Explanation:

When an object travels faster than the speed of sound, a sonic boom is created. A ring of pressure waves (probably why wave destruction and pressure cooker are options lol) are created, which is why when you see a jet make a sonic boom there looks like a ring shaped cloud behind it. Hope this helps! For further confirmation/information, just look up sonic boom science (or else you will get sonic boom the video game)

7 0

3 years ago

Read 2 more answers