How much work is done ON THE BOOK witih mass of 10 kg that is being pushed across a 2m table at 5m/s?

Which of the following are examples of both projectile motion and 2-dimensional motion?

Elanso [62]

Answer:

A

is answer

because of questions

5 0

3 years ago

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A container with volume 1.64 L is initially evacuated. Then it is filled with 0.226 g of N2N

vaieri [72.5K]

Answer:

0.015 atm

Explanation:

The pressure of the gas can be calculated using Ideal Gas Law:

$p = \frac{nRT}{V}$

Where:

n: is the number of moles of the gas

R: is the gas constant = 0.082 L*atm/(K*mol)

V: is the volume of the container = 1.64 L

T: is the temperature

We need to find the number of moles and the temperature. The number of moles is:

$n = \frac{m}{M}$

Where:

M: is the molar mass of the N₂ = 14.007 g/mol*2 = 28.014 g/mol

m: is the mass of the gas = 0.226 g

$n = \frac{0.226 g}{28.014 g/mol} = 8.07 \cdot 10^{-3} moles$

Now, the temperature can be found using the following equation:

$v_{rms} = \sqrt{\frac{3RT}{M}}$

Where:

R: is the gas constant = 0.082 L*atm/K*mol = 8.314 J/K*mol

$v_{rms}$ : is the root-mean-square speed of the gas = 182 m/s

By solving the above equation for T, we have:

$T = \frac{v_{rms}^{2}*M}{3R} = \frac{(182 m/s)^{2}*28.014 \cdot 10^{-3} Kg/mol}{3*8.314 J K^{-1}mol^{-1}} = 37.20 K$

Finally, we can find the pressure of the gas:

$p = \frac{nRT}{V} = \frac{8.07 \cdot 10^{-3} mol*0.082 L*atm* K^{-1}*mol^{-1}*37.20 K}{1.64 L} = 0.015 atm$

Therefore, the pressure of the gas is 0.015 atm.

I hope it helps you!

8 0

3 years ago

7. A toy car of mass 1.2 kg is driving vertical circles inside a hollow cylinder of radius 2.0m. It is moving at a constant spee

wlad13 [49]

Answer:

a)

$N_{top}=9.8N\\N_{bottom}=33.4N$

b) $v_{min}=4.4m/s$

Explanation:

The net force on the car must produce the centripetal acceleration necessary to make this circle, which is $a_{cp}=\frac{v^2}{R}$ . At the top of the circle, the normal force and the weight point downwards (like the centripetal force should), while at the bottom the normal force points upwards (like the centripetal force should) and the weight downwards, so we have (taking the upwards direction as positive):

$-m\frac{v^2}{R}=-N_{top}-mg\\m\frac{v^2}{R}=N_{bottom}-mg$

Which means:

$N_{top}=m\frac{v^2}{R}-mg=(1.2kg)\frac{(6m/s)^2}{2m}-(1.2kg)(9.8m/s^2)=9.8N\\N_{bottom}=m\frac{v^2}{R}+mg=(1.2kg)\frac{(6m/s)^2}{2m}+(1.2kg)(9.8m/s^2)=33.4N$