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3 years ago

The moon doesn't fall to earth because of it's friction?

Physics

2 answers:

otez555 [7]3 years ago

6 0

Answer:

I think No since the gravitational force of earth is not eneogh to pull moon to it

TiliK225 [7]3 years ago

4 0

Answer:

The Moon does not fall towards Earth right now because Earth rotates itself. The energy from the Earth's own rotation around its axis is gradually tranferred into energy of the Moon's orbital motion. That's why the Earth's rotating speed decreases but the distance to the Moon increases.

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A roller coaster travels 41.1 m at an angle of 40.0°

Iteru [2.4K]

Answer:

41.1 ÷ 40.0

Explanation:

Did you learn about Newton

3 0

2 years ago

A gas expands against a constant external pressure of 2.00 atm until its volume has increased from 6.00 to 10.00 L. During this

mars1129 [50]

Answer:

ΔU = - 310.6 J (negative sign indicates decrease in internal energy)

W = 810.6 J

Explanation:

Using first law of thermodynamics:

Q = ΔU + W

where,

Q = Heat Absorbed = 500 J

ΔU = Change in Internal Energy of Gas = ?

W = Work Done = PΔV =

P = Pressure = 2 atm = 202650 Pa

ΔV = Change in Volume = 10 L - 6 L = 4 L = 0.004 m³

Therefore,

Q = ΔU + PΔV

500 J = ΔU + (202650 Pa)(0.004 m³)

ΔU = 500 J - 810.6 J

ΔU = - 310.6 J (negative sign indicates decrease in internal energy)

The work done can be simply calculated as:

W = PΔV

W = (202650 Pa)(0.004 m³)

W = 810.6 J

7 0

3 years ago

In a photoelectric experiment, you shine light onto an electrode and record a current of 25 μA. When you apply +500 mV to the el

kkurt [141]

Answer:

2.083 V.

Explanation:

Stopping potential is the potential that is required to stop the current to zero . This potential is applied externally to oppose the potential created by the photoelectric effect . It gives the measure the photoelectric potential being generated .

Here current drops to 25 μA to 19 μA by a potential of 500mV

Change in current

= 25 - 19 = 6 μA

Voltage requirement for unit reduction in current

= 500 / 6 μA

To reduce current 0f 25 μA

requirement of V = (500 / 6 ) x 25 = 2083.33 mV = 2.083 V.

7 0

3 years ago

The rate at which heat enters an air conditioned building is often roughly proportional to the difference in temperature between

erma4kov [3.2K]

Answer:

Considering first question

Generally the coefficient of performance of the air condition is mathematically represented as

$COP = \frac{T_i}{T_o - T_i}$

Here $T_i$ is the inside temperature

while $T_o$ is the outside temperature

What this coefficient of performance represent is the amount of heat the air condition can remove with 1 unit of electricity

So it implies that the air condition removes $\frac{T_i}{T_o - T_i}$ heat with 1 unit of electricity

Now from the question we are told that the rate at which heat enters an air conditioned building is often roughly proportional to the difference in temperature between inside and outside. This can be mathematically represented as

$Q \ \alpha \ (T_o - T_i)$

=> $Q= k (T_o - T_i)$

Here k is the constant of proportionality

since 1 unit of electricity removes $\frac{T_i}{T_o - T_i}$ amount of heat

E unit of electricity will remove $Q= k (T_o - T_i)$

$E = \frac{k(T_o - T_i)}{\frac{T_i}{ T_h - T_i} }$

=> $E = \frac{k}{T_i} (T_o - T_i)^2$

given that $\frac{k}{T_i}$ is constant

=> $E \ \alpha \ (T_o - T_i)^2$

From this above equation we see that the electricity required(cost of powering and operating the air conditioner) is approximately proportional to the square of the temperature difference.

Considering the second question

Assuming that $T_i = 30 ^oC$

and $T_o = 40 ^oC$

Hence

$E = K (T_o - T_i)^2$

Here K stand for a constant

$E = K (40 - 30)^2$

=> $E = 100K$

Now if the $T_i = 20 ^oC$

Then

$E = K (40 - 20)^2$

=> $E = 400 \ K$

So from this see that the electricity require (cost of powering and operating the air conditioner)when the inside temperature is low is much higher than the electricity required when the inside temperature is higher

Considering the third question

Now in the case where the heat that enters the building is at a rate proportional to the square-root of the temperature difference between inside and outside

We have that

$Q = k (T_o - T_i )^{\frac{1}{2} }$

$E = \frac{k (T_o - T_i )^{\frac{1}{2} }}{\frac{T_i}{T_o - T_i} }$

=> $E = \frac{k}{T_i} * (T_o - T_i) ^{\frac{3}{2} }$

Assuming $\frac{k}{T_i}$ is a constant

Then

$E \ \alpha \ (T_o - T_i)^{\frac{3}{2} }$

From this above equation we see that the electricity required(cost of powering and operating the air conditioner) is approximately proportional to the square root of the cube of the temperature difference.

4 0

3 years ago

A dog is running at an initial speed of 10 m/s. He covers 50 m in 4 seconds. What is the acceleration of the dog?

nikdorinn [45]

D i hope this helps
:))

8 0

2 years ago

The moon doesn't fall to earth because of it's friction?​

The moon doesn't fall to earth because of it's friction?