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shutvik [7]
2 years ago
6

A 3.0 kg mass is released from rest at point A. The mass slides along the curved surface to point B in 6.0 seconds. Point B is 2

.0 m lower than A and the mass is moving at 2.0 m/s at point B. Find the average power exerted by kinetic friction on the mass during this segment. a) 0 W b) -3 W c) -6 W d) -9 W e) -18 W
Physics
1 answer:
timurjin [86]2 years ago
5 0

Answer:

option d) -9 J

Explanation:

Given:

Mass, m = 3.0 kg

time, t = 6.0 seconds

Velocity of mass, v = 2.0 m/s

height, h = 2 m

Now, using the concept of work-Energy theorem

we have

Net work done = change in kinetic energy

or

Work done by gravity + work done by the friction = Final kinetic energy - Initial kinetic energy

mgh +W_f = \frac{1}{2}mv^2-0

on substituting the values in the above equation, we get

3 × 9.8 × 2 + W_f = \frac{1}{2}\times 3\times2^2

or

58.8 + W_f = 6

or

W_f = -52.8 J

here negative sign depicts that the work is done against the motion of the mass

also,

Power = (Work done)/time

or

Power = -52.8/6 = -8.8 W ≈ 9 J

Hence, option d) -9 J is correct

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Answer:

89.53    North

Explanation:

Total N-S displacement

   23.72  -  39.25 + 105.06   =  + 89.53

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1 year ago
Two workers pull horizontally on a heavy box. but one pulls twice as hard as the other. The larger pull is directed at 21.0° wes
pantera1 [17]

Answer:

The  magnitude of F1 is

|F1|=358.74 \ N

The magnitude of F2 is

|F2|=179.37\ N

And the direction of F2 is

\alpha = 44.214^o

Explanation:

<u>Net Force </u>

Forces are represented as vectors since they have magnitude and direction. The diagram of forces is shown in the figure below.  

The larger pull F1 is directed 21° west of north and is represented with the blue arrow. The other pull F2 is directed to an unspecified direction (red arrow). Since the resultant Ft (black arrow) is pointed North, the second force must be in the first quadrant. We must find out the magnitude and angle of this force.  

Following the diagram, the sum of the vector components in the x-axis of F1 and F2 must be zero:

\displaystyle -2F\ sin21^o+F\ cos\alpha =0

The sum of the vertical components of F1 and F2 must equal the total force Ft

\displaystyle -2F\ cos21^o+F\ sin\alpha =460

Solving for \alpha in the first equation

\displaystyle cos\alpha =\frac{2F\ sin21^o}{F}=2sin21^o

\displaystyle cos\alpha =0.717=>\alpha =44.214^o

\displaystyle F(2cos21^o+sin\alpha)=460

\displaystyle F=\frac{460}{2cos21^o+sin\alpha}

\displaystyle F=\frac{460}{2cos21^o+sin44.214^o}

\displaystyle F=179.37\ N

The  magnitude of F1 is

|F1|=2*F=358.74 \ N

The magnitude of F2 is

|F2|=179.37\ N

And the direction of F2 is

\alpha = 44.214^o

4 0
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10 points
enot [183]

Answer:

B

Explanation:

I just took the test

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Does kinetic energy decrease as speed increase
leva [86]
Potential energy increases as speed decreases. Kinetic increases when speed increases.
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B) What are the only two sources of energy that produce all the energy we use on Earth? What form
Zina [86]

Answer:

Solar and nuclear power generate more than 99 percent of our civilization's energy. Every other important source of energy is a combination of these two. The majority of them are solar in nature. We discharge previously collected solar energy when we burn wood.

and

Nuclear energy, fossil energy (oil, coal, and natural gas), and renewable energy (wind, solar, geothermal, and hydropower) are all examples of primary energy sources.

Explanation:

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