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mel-nik [20]
2 years ago
12

Which of the following would produce the most power?

Physics
1 answer:
Fantom [35]2 years ago
6 0

Answer:

A mass of 10 kilograms lifted 10 meters in 5 seconds.

Explanation:

Power can be defined as the energy required to do work per unit time.

Mathematically, it is given by the formula;

Power = \frac {Energy}{time}

But Energy = mgh

Substituting into the equation, we have

Power = \frac {mgh}{time}

Given the following data;

Mass = 10kg

Height = 10m

Time = 5 seconds

We know that acceleration due to gravity is equal to 9.8 m/s²

Power = \frac {10*9.8*10}{5} = 490 Watts

Hence, a mass of 10 kilograms lifted 10 meters in 5 seconds would produce the most power.

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You would have to run a little less than 2 blocks
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A lightweight vertical spring of force constant k has its lower end mounted on a table. You compress the spring by a distance d,
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Answer:

v=d\sqrt{\frac{k}{m}}

Explanation:

In order to solve this problem, we can do an analysis of the energies involved in the system. Basically the addition of the initial potential energy of the spring and the kinetic energy of the mass should be the same as the addition of the final potential energy of the spring and the kinetic energy of the block. So we get the following equation:

U_{0}+K_{0}=U_{f}+K_{f}

In this case, since the block is moving from rest, the initial kinetic energy is zero. When the block loses contact with the spring, the final potential energy of the spring will be zero, so the equation simplifies to:

U_{0}=K_{f}

The initial potential energy of the spring is given by the equation:

U_{0}=\frac{1}{2}kd^{2}

the Kinetic energy of the block is then given by the equation:

K_{f}=\frac{1}{2}mv_{f}^{2}

so we can now set them both equal to each other, so we get:

=\frac{1}{2}kd^{2}=\frac{1}{2}mv_{f}^{2}

This new equation can be simplified if we multiplied both sides of the equation by a 2, so we get:

kd^{2}=mv_{f}^{2}

so now we can solve this for the final velocity, so we get:

v=d\sqrt{\frac{k}{m}}

6 0
3 years ago
Ill give brainliest! 1. Which item below describes a quick change to
stiks02 [169]
The best answer to go with is b
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3 years ago
Read 2 more answers
A large truck collides with a small car. True or False: The truck exerted a greater magnitude force on the car than the car exer
Reptile [31]

Answer:

False.

Explanation:

The forces on the car and truck are equal and opposite. The equal forces cause accelerations of the truck and car inversely proportional to their mass. That is, If the Truck A exerts a force FAB on car B, then the car will exert a force FBA on the truck. Therefore,

FBA = −FAB

However, this can be explained by Newton's second law. Let's say the truck has mass M and the car has mass m. If the magnitude of the force that both vehicles experience is F, then the magnitudes of their respective accelerations are:

atruck = F/M

acar = F/m

and combining these we get:

atruck/acar = m/M

So if the mass of the car is a lot less than the mass of the truck, then the acceleration of the truck is much smaller than the acceleration of the car, and if you were to watch the collision, the truck would pretty much seem like it's motion was unaffected, but the car's motion will change quite a bit.

5 0
3 years ago
Assume that you have a rectangular tank with its top at ground level. The length and width of the top are 14 feet and 7 feet, re
ch4aika [34]

To solve this problem we will use the work theorem, for which we have that the Force applied on the object multiplied by the distance traveled by it, is equivalent to the total work. From the measurements obtained we have that the width and the top are 14ft and 7ft respectively. In turn, the bottom of the tank is 15ft. Although the weight of the liquid is not given we will assume this value of 62 lb / ft ^ 3 (Whose variable will remain modifiable until the end of the equations subsequently presented to facilitate the change of this, in case be different). Now the general expression for the integral of work would be given as

W = \gamma A * \int_0^15 dy

Basically under this expression we are making it difficult for the weight of the liquid multiplied by the area (Top and widht) under the integral of the liquid path to be equivalent to the total work done, then replacing

W = (62)(14*7)\int^{15}_0 (15-y)dy

W = (14*7*62)\big [15y-\frac{y^2}{2}\big ]^{15}_0

W = (14*7*62)[15(15)-\frac{(15)^2}{2}]

W = 683550ft-lbs

Therefore the total work in the system is 683550ft-lbs

6 0
3 years ago
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