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Oksanka [162]
4 years ago
14

I need help, please answer

Physics
1 answer:
Burka [1]4 years ago
7 0

This being a perfect collision means no energy is lost during the collision. Because this question asks for speed and not velocity, the speed will be the same because the final energy is the same. The speed after the collision would therefore be 1.27 m/s.

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During an experiment, a toy car accelerates forward for a total time of 5 s. Which of the following procedures could a student u
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Answer: B

Explanation: The motion sensor will measure the speed (velocity) of the car. Since the mass of the car has been measured, we can use formulas to calculate the average force.

Average net force = mass × acceleration

The mass of the toy car is measured first and noted

To calculate the velocity,

the car starts from rest since the velocity is associated with the distance and time after 5s.

Acceleration = velocity/time

With that the acceleration can be found.

acceleration is defined as change in velocity per unit time.

Then,

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Option B is the best answer

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As Earth moves around the Sun, its _____ is approximately perpendicular to the force of gravity exerted by the Sun.
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3 years ago
A boat of mass 250 kg is coasting, with its engine in neutral, through the water at speed 1.00 m/s when it starts to rain with i
Vera_Pavlovna [14]

Answer:

Explanation:

1. We use the conservation of momentum for before the raining and after. And also we take into account that in 0.5h the accumulated water is

100kg/h*0.5h = 50kg

p_{b}=p_{a}\\M_{b}v_{b}=(M_{b}+m_{r})v_{a}\\v_{a}=\frac{M_{b}v_{b}}{M_{b}+m_{r}}=\frac{250kg*1\frac{m}{s}}{250kg+50kg}=0.83\frac{m}{s}

2. the momentum does not conserve because the drag force of water makes that the boat loses velocity

3. If we assume that the force of the boat before the raining is

F=ma=m\frac{v-v_{0}}{t-t_{0}}=250kg\frac{1m}{s^{2}}=250N

where we have assumed that the acceleration of the boat is 1m/s{2} just before the rain starts

And if we take the net force as

F_{net}=M_{b}a_{net}=F-F_{d}=250N-bv^{2}\\F_{net}=250N-0.5N\frac{s^{2}}{m^{2}}(1\frac{m}{s})^{2}=249.5N\\a_{net}=\frac{249.5N}{M_{b}}=\frac{249.5N}{250kg}=0.99\frac{m}{s^{2}}

where we take v=1m/s because we are taking into account tha velocity just after the rain stars.

I hope this is useful for you

regards

5 0
4 years ago
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