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3 years ago

Find the force of gravity of 1000 kg car.

Chemistry

1 answer:

ANEK [815]3 years ago

6 0

Answer:

The force of gravity acting on the car is <u>9800 N vertically downward.</u>

Explanation:

Given:

Mass of the car given is 1000 kg.

We know that the force of gravity is the force applied by the center of Earth on any body. The force of gravity is also called the weight of the body and always act towards the center of the Earth.

From Newton's second law, we know that the force acting on a body is equal to its mass and acceleration.

Here, the acceleration acting on the car is due to gravity and thus has a constant value of 9.8 m/s² on the surface of Earth.

Therefore, the force of gravity acting on the car is given using the Newton's second law as:

Force of gravity = Mass of car $\times$ Acceleration due to gravity.

Force of gravity = (1000 kg) $\times$ (9.8 m/s²)

Force of gravity = 9800 N [1 kg.m/s² = 1 N]

Therefore, the force of gravity acting on the car is 9800 N vertically downward.

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Answer:

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3 years ago

How much percent of shops are in the city? How about in the country?

kolezko [41]

Answer:

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1 year ago

Will mark the brainiest later for correct answers! Please show work.

Elodia [21]

Answer:

According to avogadro's law, 1 mole of every substance contains avogadro's number $6.023\times 10^{23}$ of particles and weighs equal to its molecular mass.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

$\text{Number of moles}=\frac{\text{Given molecules}}{\text {Avogadros number}}$

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b. moles in 17.75 g of NaCl = $\frac{17.75g}{58.5g/mol}=0.3034moles$

molecules in 17.75 g of $NaCl$ = $0.3034\times 6.023\times 10^{23}=1.827\times 10^{23}$

formula units 17.75 g of $NaCl$ = $0.3034\times 6.023\times 10^{23}=1.827\times 10^{23}$

c. moles in 20.06 g of $CuSO_4.5H_2O$ = $\frac{20.06g}{249.68g/mol}=0.08034moles$

formula units in 20.06 g of $CuSO_4.5H_2O$ = $0.08034\times 6.023\times 10^{23}=0.4839\times 10^{23}$

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3 years ago