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1 year ago

15

A car accelerates uniformly from rest tona speed of 30.0 mi/h in 12.0s. Find the distance the car travels during this time? Find

the constant acceleration of the car?

1 answer:

meriva1 year ago

5 0

Answer:

2.5 mi/s^2

Explanation:

please see paper for work!

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Coders play an important role in

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3 years ago

umka21 [38]

Answer:

Part a)

$a = 3.68 m/s^2$

Part b)

$a = 11.8 m/s^2$

Explanation:

Part a)

For force conditions of two blocks we will have

$m_1g - T = m_1 a$

$T - m_2g = m_2 a$

now from above equations we have

$(m_1 - m_2) g = (m_1 + m_2) a$

$a = \frac{m_1 - m_2}{m_1 + m_2} g$

now we know that

$m_1 = \frac{908}{9.8} = 92.65 kg$

$m_2 = \frac{412}{9.8} = 42 kg$

now from above equation we have

$a = \frac{92.65 - 42}{92.65 + 42}(9.8)$

$a = 3.68 m/s^2$

Part b)

When heavier block is removed and F = 908 N is applied at the end of the string then we have

$F - mg = ma$

$908 - 412 = 42 a$

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3 years ago

Which of these galaxies would you most likely find at the center of a large cluster of galaxies?

Arada [10]

Answer:

b. a large elliptical galaxy

Explanation:

In elliptical galaxies the stars are grouped in an elliptical shape, it has a low quantity of gas and dust in comparison to spiral galaxies, and its stars belong to an old population, there is not new stellar formation in it.

The stars orbit in a messy way which made to believe that they form from the merger of galaxies.

They are also really massive (around $10^{4}$ solar masses).

The most massive and luminous can be found in the center of cluster of galaxies.

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3 years ago

A litre of a gas weigh 2 gram at 300 kelvin temperature and 1 atm pressure,if the pressure is made 75 atm then at which temperat

liberstina [14]

Answer:

45000 K .

Explanation:

Given :

A liter of a gas weigh 2 gram at 300 kelvin temperature and 1 atm pressure

We need to find the temperature in which 1 litre of the same gas weigh 1 gram

in pressure 75 atm.

We know, by ideal gas equation :

$PV=nRT$

Here , n is no of moles , $n=\dfrac{Given \ Weight }{Molecular\ Mass}=\dfrac{w}{M}$

Putting initial and final values and dividing them :

$\dfrac{P_1V_1}{P_2V_2}=\dfrac{\dfrac{w_1}{M}T_1}{\dfrac{w_2}{M}T_2}$

$\dfrac{1\times 1}{75\times 1}=\dfrac{\dfrac{2}{M}\times 300}{\dfrac{1}{M}\times T_2}\\ \\T_2=45000\ K.$

Hence , this is the required solution.

7 0

3 years ago