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trapecia [35]
4 years ago
13

A chair is pushed forward with a force of 185 N. The gravitational force of earth on the chair is 155 N downward. The force of f

riction acting on the chair is 55 N. Draw a free-body diagram showing all forces acting on the chair. Determine the magnitude of any missing force values AND determine the net force acting on the chair.

Physics
2 answers:
zhenek [66]4 years ago
7 0

The net force on the chair is 100N

<u>Explanation:</u>

The free body diagram is attached below.

According to the condition,

when a downward gravitational force of 155N of earth acts then an equal opposite upward force of same magnitude also acts on the chair.

Therefore, the normal force acting on the chair is 155N

When a chair is pushed forward on a surface, then a frictional force also acts on it in opposite direction.

The net force acting on the chair is = F - f

Where,

F is the applied force

f is the force of friction

Net force = 155N - 55N

Fnet = 100N

Therefore, net force on the chair is 100N

andreyandreev [35.5K]4 years ago
5 0

Answer:

100 N

Explanation:

Me big brain

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Answer:

<em>The car will be moving at 5.48 m/s at the bottom of the hill</em>

Explanation:

<u>Principle of Conservation of Mechanical Energy</u>

In the absence of friction, the total mechanical energy is conserved. That means that

E_m=U+K is constant, being U the potential energy and K the kinetic energy

U=mgh

\displaystyle K=\frac{mv^2}{2}

When the car is at the top of the hill, its speed is 0, but its height h should be enough to produce the needed speed v down the hill.

The Kinetic energy is then, zero. When the car gets enough speed we assume it is achieved at ground level, so the potential energy runs out to zero but the Kinetic is at max. So the initial potential energy is transformed into kinetic energy.

We are given the initial potential energy U=45 J. It all is transformed to kinetic energy at the bottom of the hill, thus:

\displaystyle \frac{mv^2}{2}=45

Multiplying by 2:

\displaystyle mv^2=90

Dividing by m:

\displaystyle v^2=\frac{90}{m}

Taking square roots:

\displaystyle v=\sqrt{\frac{90}{m}}

\displaystyle v=\sqrt{\frac{90}{3}}

v=\sqrt{30}

v = 5.48 m/s

The car will be moving at 5.48 m/s at the bottom of the hill

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5000 j of heat are added to a system. it does 3500 j of work on the surroundings. what is δe in j?
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<span>The Answer is 5200 Joules</span>
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Bob weighs 80 pounds. He ran up the steps which rose 14 feet. Fred weighs 110 pounds. He ran up the same steps as Bob. If both b
pav-90 [236]
Bob gained (80lbs x 14ft) = 1120 ft-lbs of energy.

Fred gained (110lbs x 14ft) = 1540 ft-lbs of energy

Since they both took the same amount of time, Fred's power (rate
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If a girl running along a straight road with a uniform velocity 1.5m/s,find her acceleration
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Earth has a mass of 5.97 × 1024 kg and a radius of 6.38 × 106 m, while Saturn has a mass of 5.68 × 1026 kg and a radius of 6.03
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Answer: mass on earth = 625.792 N, mass on saturn = 666.75 N

Explanation: weight of an object = mg

Where g = acceleration due to gravity =GM/r²

Where G = gravitational constant, M = mass of planet and r = radius of planet.

Let us start with earth.

G =gravitational constant of the earth =6.67×10^-11 m³ /kgs²

Me = mass of earth = 5.97×10^24 kg

re = radius of earth = 6.38×10^6 m

Acceleration due to gravity on earth = (6.67×10^-11×5.97×10^24) /(6.38×10^6)²

Acceleration due to gravity on earth = 3.98×10^14 /4.07×10^13

Acceleration due gravity on earth = 9.778 m/s²

The mass of the person is 64 kg, hence his weight on earth is given as

W = mass × Acceleration due gravity on earth = 64 ×9.778 = 625.792 N

For saturn.

G =gravitational constant of the earth =6.67×10^-11 m³ /kgs²

Ms = mass of saturn = 5.68×10^26kg

rs= radius of saturn= 6.03×10^7 m

Acceleration due to gravity on saturn= (6.67×10^-11×5.68×10^26 /(6.03×10^7)²

Acceleration due to gravity on saturn= 3.788×10^16/3.636×10^15

Acceleration due gravity on saturn= 10.417 m/s²

The mass of the person is 64 kg, hence his weight on saturn is given as

W = mass × Acceleration due gravity on saturn = 64 ×10.417 = 666.75 N

3 0
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