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Marina CMI [18]
3 years ago
8

Which of these boxes will not accelerate? A. B. C. D.

Physics
2 answers:
serg [7]3 years ago
7 0

Answer: The correct answer is Image B.

Explanation: For an object to accelerate, there should be unbalanced forces present. An object will move in the direction of net force.

Balanced forces are defined as the forces acting on the same object which are equal in magnitude but act in opposite direction. The net forces are 0.

Unbalanced forces are defined as the forces acting on the same object which are unequal in magnitude. The net force is non-zero.

For the given images:

Image A: This box will accelerate easily because the net force is non-zero and is moving in right direction.

Image B: This box will not accelerate because the net force is zero as all the forces are balancing one another. Hence, the object will stay at rest.

Image C: This box will accelerate easily because the net force is non-zero and  is acting in between the normal and applied force.

Image D: This box will accelerate easily because the net force is non-zero and is moving in right direction.

Hence, the correct option is Image B.

AlladinOne [14]3 years ago
7 0
B. because all around the box the forces are all the same
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A seagull flies at a velocity of 9.00 m/s straight into the wind. (a) if it takes the bird 20.0 min to travel 6.00 km relative t
enot [183]

Here we will the speed of seagull which is v = 9 m/s

this is the speed of seagull when there is no effect of wind on it

now in part a)

if effect of wind is in opposite direction then it travels 6 km in 20 min

so the average speed is given by the ratio of total distance and total time

v_{avg} = \frac{6000}{20*60}

v_{avg} = 5m/s

now since effect of wind is in opposite direction then we can say

V_{net} = v_{bird} - v_{wind}

5 = 9 - v_{wind}

v_{wind}= 4 m/s

Part b)

now if bird travels in the same direction of wind then we will have

v_{net}= v_{bird} + v_{wind}

v_{net} = 9 + 4 = 13 m/s

now we can find the time to go back

time = \frac{distance}{speed}

time = \frac{6000}{13}

time = 7.7 minutes

Part c)

Total time of round trip when wind is present

T = t_1 + t_2

T = 20 + 7.7 = 27.7 min

now when there is no wind total time is given by

T = \frac{6000}{9} + \frac{6000}{9}

T = 22.22 min

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4 0
3 years ago
A man starts walking north at 3 ft/s from a point P. Five minutes later a woman starts walking south at 4 ft/s from a point 500
mrs_skeptik [129]

Answer:

ds/dt = 6.98 ft/s

Explanation:

Given:

- The speed of man due north Vm = 3 ft/s

- The speed of woman due south Vw = 4 ft/s

- Woman starts walking 5 mins later than man

Find:

At what rate are the people moving apart 15 min after the woman starts walking?

Solution:

- The total time for which the man is walking due north from P, is Tm:

                                   Tm = 5 + 15 = 20 mins

- The total distance traveled by man in Tm mins is:

                                   Dm = Tm*Vm

                                   Dm = 20*60*3

                                   Dm = 3,600 ft

- The total time for which the woman is walking due south from 500 ft due east from P, is Tw:

                                   Tw = 15 = 15 mins

- The total distance traveled by man in Tw mins is:

                                   Dw = Tw*Vw

                                   Dw = 15*60*4

                                   Dw = 3,600 ft

- The displacement between man and woman at any instance is (s) which can be related by pythagoras theorem as follows:

                                   s^2 = (dm + dw)^2 + 500^2

Where, dm : Distance travelled by man at any time Tm

            dw : Distance travelled by woman at any time Tw

- Differentiate s with respect to t:

                                   2s*ds/dt = 2*(dm + dw)*(Vm + Vw)

                                   s*ds/dt = (dm + dw)*(Vm + Vw)

                                   ds/dt = [ (dm + dw)*(Vm + Vw) ] / s

- Evaluate the rate of separation of man and woman ds/dt by evaluating at instance Tm = 20 mins and Tw = 15 mins. We have:

                 ds/dt = [ (Dm + Dw)*(Vm + Vw) ] / sqrt ( (Dm + Dw)^2 + 500^2 )

- Plug in the values:

                 ds/dt = [ (3600 + 3600)*(3 + 4) ] / [sqrt ( (3600 + 3600)^2 + 500^2 )]  

                ds/dt = 6.98 ft/s

                 

           

7 0
3 years ago
In the Niagara Falls hydroelectric generating plant, the energy of falling water is converted into electricity. The height of th
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Answer:

2.124 kg of water

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height of the falls is about 48 meters.

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To calculate the energy of the 1kg falling water = Mgh

Energy = 1000*9.81*48

Energy = 470880 joules.

1 megawatt is = 1000000watts

The kilogram of water needed is 1000000/470880 = 2.124 kg of water

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