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3 years ago

What is the work energy transfer equation?

Physics

2 answers:

rosijanka [135]3 years ago

8 0

Answer:

The equation used to calculate the work done is: work done = force × distance. W = F × d. This is when: work done (W) is measured in joules (J)

Andru [333]3 years ago

5 0

Answer:

The equation used to calculate the work done is: work done = force × distance. W = F × d. This is when: work done (W) is measured in joules (J)

Explanation:

The net work done on a particle equals the change in the particle's kinetic energy:

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Why does a rock weigh less in water than it does on land?

Cloud [144]

Water is more dense then air so it sorta holds the rock as it sinks

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3 years ago

What are the three parts of Kinetic Theory

anyanavicka [17]

1. No energy is gained or lost when molecules collide.
2. The molecules in a gas take up a negligible (able to be ignored) amount of space in relation to the container they occupy.
3. The molecules are in constant, linear motion.

8 0

3 years ago

Read 2 more answers

A rock is suspended by a light string. When the rock is in air, the tension in the string is 43.8 N . When the rock is totally i

sergeinik [125]

Answer:

Density of unknown liquid is $2047 kg/m^3$

Explanation:

When rock is suspended in air then the weight of the rock is counter balanced by the tension force in the string

So here we have

$T = mg = 43.8 N$

now when the rock is immersed in water then the tension in the string is and buoyancy force due to water is counter balanced by the weight of the object

so here we have

$T_1 + F_b = mg$

$31.3 + F_b = 43.8$

$F_b = 43.8 - 31.3 = 12.5 N$

now we have

$(1000)V(9.81) = 12.5$

$V = 1.27 \times 10^{-3} m^3$

now when the rock is immersed into other liquid then we have

$T_2 + F_b' = mg$

$18.3 + F_b' = 43.8$

$F_b' = 25.5 N$

now we have

$\rho(1.27 \times 10^{-3})(9.81) = 25.5$

$\rho = 2047 kg/m^3$

4 0

3 years ago

We are sending a 30 Mbit file from source host A to destination host B. All links in the path between source and destination hav

Dmitrij [34]

Answer:

$t=1.5\times 10^{-4}\ s$

Explanation:

Given:

file size to be transmitted, $D=30\ Mb$

transmission rate of data, $\dot D=10\ Mb.s^{-1}$

propagation speed, $v=2\times 10^8\ m.s^{-1}$

distance of data transfer, $s=10000\ km=10^4\ m$

Now the delay in data transfer from source to destination for each 10 Mb:

$t'=\frac{s}{v}$

$t'=\frac{10^4}{2\times 10^8}$

$t'=5\times 10^{-5}\ s$

Now this time is taken for each 10 Mb of data transfer and we have 30 Mb to transfer:

So,

$t=3\times t'$

$t=3\times 5\times 10^{-5}$

$t=1.5\times 10^{-4}\ s$

3 0

3 years ago

The sound level at a distance of 1.48 m from a source is 120 dB. At what distance will the sound level be 70.7 dB?

Tju [1.3M]

Answer:

The second distance of the sound from the source is 431.78 m..

Explanation:

Given;

first distance of the sound from the source, r₁ = 1.48 m

first sound intensity level, I₁ = 120 dB

second sound intensity level, I₂ = 70.7 dB

second distance of the sound from the source, r₂ = ?

The intensity of sound in W/m² is given as;

$dB = 10 Log[\frac{I}{I_o} ]\\\\For \ 120 dB\\\\120 = 10Log[\frac{I}{1*10^{-12}}]\\\\12 = Log[\frac{I}{1*10^{-12}}]\\\\10^{12} = \frac{I}{1*10^{-12}}\\\\I = 10^{12} \ \times \ 10^{-12}\\\\I = 1 \ W/m^2$

$For \ 70.7 dB\\\\70.7 = 10Log[\frac{I}{1*10^{-12}}]\\\\7.07 = Log[\frac{I}{1*10^{-12}}]\\\\10^{7.07} = \frac{I}{1*10^{-12}}\\\\I = 10^{7.07} \ \times \ 10^{-12}\\\\I = 1 \times \ 10^{-4.93} \ W/m^2$

The second distance, r₂, can be determined from sound intensity formula given as;

$I = \frac{P}{A}\\\\I = \frac{P}{\pi r^2}\\\\Ir^2 = \frac{P}{\pi }\\\\I_1r_1^2 = I_2r_2^2\\\\r_2^2 = \frac{I_1r_1^2}{I_2} \\\\r_2 = \sqrt{\frac{I_1r_1^2}{I_2}} \\\\r_2 = \sqrt{\frac{(1)(1.48^2)}{(1 \times \ 10^{-4.93})}}\\\\r_2 = 431.78 \ m$

Therefore, the second distance of the sound from the source is 431.78 m.

7 0

3 years ago