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In-s [12.5K]
4 years ago
5

which scenario involves the most work? a.pulling a box with a force of 127 newtons across the floor for 8 meters b.holding a mas

s with a force of 509 newtons in your arms for 150 seconds c.pushing against a brick wall with a force of 20 newtons for 42 seconds d.lifting a rock by exerting an upward force of 500 newtons for 1 meter
Physics
1 answer:
abruzzese [7]4 years ago
6 0
<span>Work can only be said to be done when an applied force results in displacement of the body in the direction of the applied force.

Work can only be said to be done in options a and d

a.pulling a box with a force of 127 newtons across the floor for 8 meters

work = 127 * 8 = 1016J

</span>
<span>d.lifting a rock by exerting an upward force of 500 newtons for 1 meter

Work = 500 N * 1 m = 500 J

Option D has more work.

D</span>
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Why does the area around the equator stay about the same temperature year round?
Verdich [7]
Axial Tilt and Sun Energy

This axial tilt means that during the Earth's journey around the sun the poles receive varying amounts of sunlight. The equator, however, receives relatively consistent sunlight all year. The consistency of energy means the equator's temperature stays relatively constant all year.
3 0
3 years ago
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}

<u>Where:</u>

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}

<u>Where:</u>

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}}    

<u>Where:</u>

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
What is the difference between static friction and kinetic friction
sergejj [24]
The Force of Static Friction<span> keeps a stationary object at rest! Once the Force of</span>Static Friction<span> is overcome, the Force of </span>Kinetic Friction<span> is what slows down a moving object.</span>
3 0
3 years ago
Read 2 more answers
What force is required to accelerate a 1,100 kg car to 0.5 meters squared
Neporo4naja [7]

Answer:550N

Explanation:

mass=1100kg

Acceleration=0.5m/s^2

Force=mass x acceleration

Force=1100 x 0.5

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8 0
4 years ago
A motor has an armature resistance of 3.75 Ω . Part A If it draws 9.10 A when running at full speed and connected to a 120-V lin
bulgar [2K]

Back emf is 85.9 V.

<u>Explanation:</u>

Given-

Resistance, R = 3.75Ω

Current, I = 9.1 A

Supply Voltage, V = 120 V

Back emf = ?

Assumption - There is no effects of inductance.

A motor will have a back emf that opposes the supply voltage, as the motor speeds up the back emf increases and has the effect that the difference between the supply voltage and the back emf is what causes the current to flow through the armature resistance.

So if 9.1 A flows through the resistance of 3.75Ω then by Ohms law,

The voltage across the resistance would be

v = I x R

  = 9.1 x 3.75

  = 34.125 volts

We know,

supply voltage = back emf + voltage across the resistance

By plugging in the values,

120 V = back emf + 34.125 V

Back emf = 120 - 34.125

                = 85.9 Volts

Therefore, back emf is 85.9 V.

4 0
3 years ago
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