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Helga [31]
4 years ago
12

Consider the free body diagram. If the sum of the tension forces is equal to the force of gravity, which description BEST appli

es?
A) A book is at rest on a tabletop.
B) A physics student rests a backpack upon one shoulder.
C) A girl hangs by both hands, motionless, from a trapeze.
D) A girl falls slowly to Earth while strapped to a parachute.

Physics
2 answers:
Naily [24]4 years ago
8 0

C.)    A girl hangs by both hands, motionless, from a trapeze.

mestny [16]4 years ago
4 0

Answer:

C) A girl hangs by both hands, motionless, from a trapeze.

Explanation:

The free body diagram in the figure gives us several pieces of information:

- There are two forces upward and one force downward. --> these could correspond to the forces of the two harms of the girl (upward) and the weight of the girl (downward)

- The sum of the tension forcs is equal to the force gravity --> this tells us that the net force is zero, therefore this could corresponds to the girl hanging motionless from the trapeze.

Let's see why the other options are not acceptable:

A) A book is at rest on a tabletop.   --> In this case we would only one have one force downward (the force of gravity) and one force upward (the normal reaction of the tabletop)

B) A physics student rests a backpack upon one shoulder.  --> this does not corresponds to two forces upward and one force downward

D) A girl falls slowly to Earth while strapped to a parachute. --> Again, in this case we would have one force downward (force of gravity) and one force upward (the air resistance on the parachute)

So, the only possible option is C.

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Explanation:

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And the ball will keep on moving until unless the net of a goal post exerts the external force to stop the ball.

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An air conditioner has a power rating of 3000 W and is used daily. Its electrical
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5.714 hours / day

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2 years ago
Read 2 more answers
NEED HELP ASAP
Dafna11 [192]

Answers:

a) -2.54 m/s

b) -2351.25 J

Explanation:

This problem can be solved by the <u>Conservation of Momentum principle</u>, which establishes that the initial momentum p_{o} must be equal to the final momentum p_{f}:  

p_{o}=p_{f} (1)  

Where:  

p_{o}=m_{1} V_{o} + m_{2} U_{o} (2)  

p_{f}=(m_{1} + m_{2}) V_{f} (3)

m_{1}=110 kg is the mass of the first football player

V{o}=-7 m/s is the velocity of the first football player (to the south)

m_{2}=75 kg  is the mass of the second football player

U_{o}=4 m/s is the velocity of the second football player (to the north)

V_{f} is the final velocity of both football players

With this in mind, let's begin with the answers:

a) Velocity of the players just after the tackle

Substituting (2) and (3) in (1):

m_{1} V_{o} + m_{2} U_{o}=(m_{1} + m_{2}) V_{f} (4)  

Isolating V_{f}:

V_{f}=\frac{m_{1} V_{o} + m_{2} U_{o}}{m_{1} + m_{2}} (5)

V_{f}=\frac{(110 kg)(-7 m/s) + (75 kg) (4 m/s)}{110 kg + 75 kg} (6)

V_{f}=-2.54 m/s (7) The negative sign indicates the direction of the final velocity, to the south

b) Decrease in kinetic energy of the 110kg player

The change in Kinetic energy \Delta K is defined as:

\Delta K=\frac{1}{2} m_{1}V_{f}^{2} - \frac{1}{2} m_{1}V_{o}^{2} (8)

Simplifying:

\Delta K=\frac{1}{2} m_{1}(V_{f}^{2} - V_{o}^{2}) (9)

\Delta K=\frac{1}{2} 110 kg((-2.5 m/s)^{2} - (-7 m/s)^{2}) (10)

Finally:

\Delta K=-2351.25 J (10) Where the minus sign indicates the player's kinetic energy has decreased due to the perfectly inelastic collision

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