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3 years ago

Two men pushing a stalled car generate a net force of +690Nfor

Physics

1 answer:

Karolina [17]3 years ago

7 0

Answer:

The final momentum of the car is 4554 kg m/s.

Explanation:

Given that,

F = 690 N

Time = 6.6 sec

Momentum :

The momentum is the product of the force and change of time.

We need to calculate the final momentum

Using formula of momentum

$\Delta p=F\times \Delta t$

Where, F = force

t = time

Put the value into the formula

$\Delta p=690\times6.6$

$\Delta p=4554\ kg-m/s$

Hence, The final momentum of the car is 4554 kg m/s.

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An air mass is a large pocket of air with _____.

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An air mass is a large pocket of air with a uniform temperature and humidity

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3 years ago

A 40kg rock falls off a cliff that is 50 meters high. How fast is the speed of the rock when it hits the ground

Oksi-84 [34.3K]

31.3m/s

Explanation:

Given parameters:

Mass of rock = 40kg

Height of cliff = 50m

Unknown:

Speed of rock when it hits ground = ?

Solution:

We are going to use the appropriate motion equation to solve this problem

The rock is falling with the aid of gravitational force. The force is causing it to accelerate with an amount of velocity.

Using;

V² = U² + 2gH

V = unknown velocity

U = initial velocity = O

g = acceleration due to gravity = 9.8m/s²

H = height of fall

since the initial velocity of the bodyg is 0

V² = 2gH

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Velocity brainly.com/question/4460262

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7 0

3 years ago

Calculate the equivalent of 20 degrees Celsius in degrees Fahrenheit and Kelvin.

alekssr [168]

Answer:

68 °F, 293.15 K

Explanation:

Fahrenheit, Kelvin and Celsius are the different scales of temperature in which temperature is measured.

Given : T = 20°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

The conversion of T( °C) to T(F) is shown below:

T (°F) = (T (°C) × 9/5) + 32

So,

3 0

3 years ago

Two cylinders each contain 0.30 mol of a diatomic gas at 320 K and a pressure of 3.0 atm. Cylinder A expands isothermally and cy

Svetllana [295]

Answer :

(a). The final temperature of the gas in the cylinder A is 320 K.

(b). The final temperature of the gas in the cylinder B is 233.7 K.

(c). The final volume of the gas in the cylinder A is $7.86\times10^{-3}\ m^3$

(d). The final volume of the gas in the cylinder B is $5.7\times10^{-3}\ m^3$

Explanation :

Given that,

Number of mole n = 0.30 mol

Initial temperature = 320 K

Pressure = 3.0 atm

Final pressure = 1.0 atm

We need to calculate the initial volume

Using formula of ideal gas

$P_{1}V_{1}=nRT$

$V_{1}=\dfrac{nRT}{P_{1}}$

Put the value into the formula

$V_{1}=\dfrac{0.30\times8.314\times320}{3.039\times10^{5}}$

$V_{1}=2.62\times10^{-3}\ m^3$

(a). We need to calculate the final temperature of the gas in the cylinder A

Using formula of ideal gas

In isothermally, the temperature is not change.

So, the final temperature of the gas in the cylinder A is 320 K.

(b). We need to calculate the final temperature of the gas in the cylinder B

Using formula of ideal gas

$T_{2}=T_{1}\times(\dfrac{P_{1}}{P_{2}})^{\frac{1}{\gamma}-1}$

Put the value into the formula

$T_{2}=320\times(\dfrac{3}{1})^{\frac{1}{1.4}-1}$

$T_{2}=233.7\ K$

(c). We need to calculate the final volume of the gas in the cylinder A

Using formula of volume of the gas

$P_{1}V_{1}=P_{2}V_{2}$

$V_{2}=\dfrac{P_{1}V_{1}}{P_{2}}$

Put the value into the formula

$V_{2}=\dfrac{3\times2.62\times10^{-3}}{1}$

$V_{2}=0.00786\ m^3$

$V_{2}=7.86\times10^{-3}\ m^3$

(d). We need to calculate the final volume of the gas in the cylinder B

Using formula of volume of the gas

$V_{2}=V_{1}(\dfrac{P_{1}}{P_{2}})^{\frac{1}{\gamma}}$

$V_{2}=2.62\times10^{-3}\times(\dfrac{3}{1})^{\frac{1}{1.4}}$

$V_{2}=0.0057\ m^3$

$V_{2}=5.7\times10^{-3}\ m^3$

Hence, (a). The final temperature of the gas in the cylinder A is 320 K.

(b). The final temperature of the gas in the cylinder B is 233.7 K.

(c). The final volume of the gas in the cylinder A is $7.86\times10^{-3}\ m^3$

(d). The final volume of the gas in the cylinder B is $5.7\times10^{-3}\ m^3$

6 0

3 years ago