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Vanyuwa [196]
3 years ago
14

Newton's first law equations like velocity and stuff like that

Physics
1 answer:
Klio2033 [76]3 years ago
4 0

Answer:

Newton's first law: An object at rest remains at rest, or if in motion, remains in motion at a constant velocity unless acted on by a net external force. ... An object sliding across a table or floor slows down due to the net force of friction acting on the object.

Explanation:

please give me a heart

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How many different parts parts of the solar system can you see?
photoshop1234 [79]

Out of the solar system parts, we can see the sun, a few planets, some meteorites and comets.

Rest all is not visible through naked eyes

6 0
3 years ago
If a ball has the weight of 3.76, how much work would it take to lift it 2 meters above the ground? (sorry if this question does
Aloiza [94]
We know, work done = Weight * Displacement 
w = 3.76 * 2
w = 7.52 J

In short, Your Answer would be: 7.52 Joules

Hope this helps!
7 0
2 years ago
A car accelerates at 3 m/s*2. Assuming the car starts from rest, how much time does it need to
uranmaximum [27]
<h3><u>Given</u><u>:</u><u>-</u></h3>

Acceleration,a = 3 m/s²

Initial velocity,u = 0 m/s

Final velocity,v = 12 m/s

<h3><u>To</u><u> </u><u>be</u><u> </u><u>calculated:-</u><u> </u></h3>

Calculate the time take by a car.

<h3><u>Solution:-</u><u> </u></h3>

According to the first equation of motion:

v = u + at

★ Substituting the values in the above formula,we get:

⇒ 12 = 0 + 3 × t

⇒ 12 = 3t

⇒ 3t = 12

⇒ t = 12/3

⇒ t = 4 sec

5 0
3 years ago
Read 2 more answers
The cylinder with piston locked in place is immersed in a mixture of ice and water and allowed to come to thermal equilibrium wi
lukranit [14]

Answer:

a. volume of gas:  (decreases)

b. temperature of gas:  (same)

c. internal energy of gas: (same)

d. pressure of gas: (increases)

Explanation:

We have a gas (let's suppose that is ideal) in a piston with a fixed volume V.

Then we put in a reservoir at 0°C (the mixture of water and ice)

remember that the state equation for an ideal gas is:

P*V = n*R*T

and:

U = c*n*R*T

where:

P = pressure

V = volume

n = number of mols

R = constant

c = constant

T = temperature.

Now, we have equilibrium at T = 0°C, then we can assume that T is also a constant.

Then in the equation:

P*V = n*R*T

all the terms in the left side are constants.

P*V = constant

And knowing that:

U = c*n*R*T

then:

n*R*T = U/c

We can replace it in the other equation to get:

P*V = U/c = constant.

Now, the piston is (slowly) moving inwards, then:

a) Volume of the gas: as the piston moves inwards, the volume where the gas can be is smaller, then the volume of the gas decreases.

b) temperature of the gas: we know that the gas is a thermal equilibrium with the mixture (this happens because we are in a slow process) then the temperature of the gas does not change.

c) Internal energy of the gas:

we have:

P*V = n*R*T = constant

and:

P*V = U/c = constant.

Then:

U = c*Constant

This means that the internal energy does not change.

d) Pressure of the gas:

Here we can use the relation:

P*V = constant

then:

P = (constant)/V

Now, if V decreases, the denominator in that equation will be smaller. We know that if we decrease the value of the denominator, the value of the quotient increases.

And the quotient is equal to P.

Then if the volume decreases, we will see that the pressure increases.

4 0
2 years ago
You are trying to overhear a juicy conversation, but from your distance of 20.0 m , it sounds like only an average whisper of 30
12345 [234]

Answer:

r₂ = 0.2 m

Explanation:

given,

distance = 20 m

sound of average whisper = 30 dB

distance moved closer = ?

new frequency = 80 dB

using formula

\beta = 10 log(\dfrac{I_1}{I_0})

   I₀ = 10⁻¹² W/m²

now,

30 = 10 log(\dfrac{I_1}{10^{-12}})

\dfrac{I_1}{10^{-12}}= 10^3

I_1= 10^{-8}\ W/m^2

to hear the whisper sound = 80 dB

80 = 10 log(\dfrac{I_2}{10^{-12}})

\dfrac{I_2}{10^{-12}}= 10^8

I_2= 10^{-4}\ W/m^2

we know intensity of sound is inversely proportional to square of distances

\dfrac{I_1}{I_2}=\dfrac{r_2^2}{r_1^2}

\dfrac{10^{-8}}{10^{-4}}=\dfrac{r_2^2}{20^2}

10^{-4}=\dfrac{r_2^2}{20^2}

  r₂ = 0.2 m

6 0
3 years ago
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