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Irina-Kira [14]
3 years ago
10

Suppose two children push horizontally, but in exactly opposite directions, on a third child in a wagon. The first child exerts

a force of 75.0N, the second child exerts a force of 90.0 N, friction is 12.0 N, and the most of the third child plus wagon is 23.0 kga)what is the system of interest if the acceleration of the child in the wagon is to be calculated
Physics
1 answer:
Naddika [18.5K]3 years ago
8 0

Answer:

Explanation:

75 N and 90 N are acting in opposite direction so net force = 90 - 75 = 15 N .

Friction force will act in the direction opposite to the direction of net force .

So friction force will act in the direction in which 75 N is acting .

Total force acting in the direction of 75 =  75 + 12 = 87 N

Net force acing on the third child = 90 - 87 = 3 N  

Its direction will be that in the direction of 90 N .

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A flywheel turns through 26 rev as it slows from an angular speed of 2.9 rad/s to a stop. (a) Assuming a constant angular accele
Misha Larkins [42]

a) The wheel takes 112.7 s to comes to rest

b) The angular acceleration of the wheel is -0.0257 rad/s^2

c) The wheel takes 33.0 s to complete 13 revolutions

Explanation:

a)

The rotational motion of the wheel is at constant angular acceleration, so we can use the equivalent of the suvat equations for rotational motion. In particular, to find the time the wheel takes to come to a stop, we use the following:

\theta = (\frac{\omega_f+\omega_i}{2})t

where:

\theta is the angular displacement

\omega_i is the initial angular speed

\omega_f is the final angular speed

t is the time

For the wheel in this problem:

\omega_i = 2.9 rad/s

\omega_f = 0

\theta = 26 rev \cdot (2\pi)= 163.4 rad

And solving for t, we find the time:

t=\frac{2\theta}{\omega_i + \omega_f}=\frac{2(163.4)}{0+2.9}=112.7 s

b)

We can find the angular acceleration by using the equation:

\alpha = \frac{\omega_f - \omega_i}{t}

where here we have:

\omega_f = 0 is the final angular speed

\omega_i = 2.9 rad/s is the initial angular speed

t = 112.7 s is the time

And substituting,

\alpha = \frac{0-2.9}{112.7}=-0.0257 rad/s^2

And the negative sign is because the wheel is decelerating.

c)

In order to find the time taken to complete 13 revolutions, we use the following suvat equation:

\theta=\omega_i t+ \frac{1}{2}\alpha t^2

where:

\omega_i = 2.9 rad/s is the initial angular speed

\theta = 13 rev \cdot (2\pi)=81.7 rad is the angular displacement

\alpha=-0.0257 rad/s^2 is the angular acceleration

Substituting into the equation:

81.7 = 2.9 t +\frac{1}{2}(-0.0257)t^2\\0.0129t^2-2.9t+81.7 = 0

which has two solutions:

t = 33.0 s

t = 191.8 s

And here we have to take the first solution (33.0 s), because it corresponds to the first time at which the angular displacement of the wheel is equal to 13 revolutions (the second solution is the time at which the angular displacement of the wheel returns to 13 revolutions, after the wheel has came to rest and it started to rotate in the opposite direction).

Learn more about rotational motions:

brainly.com/question/9575487

brainly.com/question/9329700

#LearnwithBrainly

5 0
4 years ago
The drawing shows a triple jump on a checkerboard, starting at the center of square A and ending on the center of square B. Each
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Answer:

Each jump covers one center of the square to the other. If you can imagine the scenario, a single jump would cover a length of the side of one whole square. So, if it took three jumps, the total distance would be

d = 3 * 5 cm

d = 15 cm

Therefore, the magnitude or distance is 15 cm. The displacement depends on the direction of motion. 

3 0
4 years ago
Assume biological substances are 90 percent water and the density of water is 1.0 ✕ 103 kg/m3.
Mariulka [41]
I think its 1x^103 (kg/m.3) I hope this help
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3 years ago
The lowest note on a grand piano has a frequency of 28.1 Hz. It is a fixed string oscillating in its fundamental (longest wavele
tangare [24]

Answer:

2210.91 N

Explanation:

f = v/∧ = 1/2 √ T/ μ

where f= 28.1 Hz , T= tension ,

L= 2m

mass density =  μ = 350÷1000/2.00

= 0.175kg/m

from  f = 1/2L √ T/ μ

make T the subject of the formula

T= (f×2L)² ₓ  μ

T= (28.1×2×2)² ×0.175

T=12633.76×0.175

=2210.91 N

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Which objects are inclined planes? Check all that apply.
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Wheel chair ramp, slide, and staircase
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