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3 years ago

A mass of 2000 kg is raised 5.0 m in 10 seconds. What is the power output in horsepower to raise the object?

Physics

1 answer:

Lapatulllka [165]3 years ago

3 0

Answer:

The power output in horsepower to raise the object is, P = 13.14 Hp

Explanation:

Given data,

The mass of the object, m = 2000 kg

The object is raised to a height, h = 5.0 m

The time duration of the work, t = 10 s

The force acting on the mass,

F = mg

= 2000 x 9.8

19600 N

The work done on the object is,

W = F x h

= 19600 x 5

= 98000 J

The rate of doing work is defined as the power. It is given by the formula,

P = W / t

= 98000 /10

= 9800 Watts

1 Hp = 745.7 Watts

Therefore,

9800 watts = 13.14 Hp

Hence, the power output in horsepower to raise the object is, P = 13.14 Hp

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Jenise is buying a car for $13,085. She wants $9,400 in upgrades to the car.

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Answer:

$55.6

Explanation:

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3 years ago

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What is the kinetic energy of a toy truck with a mass of 0.75 kg and a velocity of 4 m/s? (Formula: )

Dvinal [7]

Answer: Option (b) is the correct answer.

Explanation:

Kinetic energy is the energy of an object or particle when it is in motion.

Mathematically, Kinetic energy = $\frac{1}{2}mass \times (velocity)^{2}$

It is given that toy truck has mass = 0.75 kg, and velocity = 4 m/s.

Calculate the kinetic energy as follows.

Kinetic energy = $\frac{1}{2}mass \times (velocity)^{2}$

= $\frac{1}{2}0.75 kg \times (4 m/s)^{2}$

= $\frac{1}{2}0.75 kg \times 16 m^{2}/s^{2}$

= $0.75 kg \times 8 m^{2}/s^{2}$

= $6 kg m^{2}/s^{2}$

or, Kinetic energy = 6 joules

Thus, we can conclude that kinetic energy of toy truck is 6 J.

6 0

4 years ago

Read 2 more answers

If someone throws a 3 gram fry accelerating at 5 meter/second2, what is the fry’s force?

Arada [10]

The force on the fry is $0.015 N$

Explanation:

We can find the force acting on the fry by using Newton's second law:

$F=ma$

where

F is the net force on the fry

m is its mass

a is its acceleration

For the fry in this problem,

$m=3 g = 0.003 kg$

$a=5.0 m/s^2$

Therefore, the force exerted on the fry is

$F=(0.003)(5)=0.015 N$

Learn more about Newton's second law here:

brainly.com/question/3820012

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3 0

3 years ago

A rope is wrapped around the rim of a large uniform solid disk of mass 325 kg and radius 3.00 m. The horizontal disk is made to

Naily [24]

Answer:

The angular speed is 0.13 rev/s

Explanation:

From the formula

$\tau = I\alpha$

Where $\tau$ is the torque

$I$ is the moment of inertia

$\alpha$ is the angular acceleration

But, the angular acceleration is given by

$\alpha = \frac{\omega}{t}$

Where $\omega$ is the angular speed

and $t$ is time

Then, we can write that

$\tau = \frac{I\omega}{t}$

Hence,

$\omega = \frac{\tau t}{I}$

Now, to determine the angular speed, we first determine the Torque $\tau$ and the moment of inertia $I$ .

Here, The torque is given by,

$\tau = rF$

Where r is the radius

and F is the force

From the question

r = 3.00 m

F = 195 N

∴ $\tau = 3.00 \times 195$

$\tau = 585$ Nm

For the moment of inertia,

The moment of inertia of the solid disk is given by

$I = \frac{1}{2}MR^{2}$

Where M is the mass and

R is the radius

∴ $I = \frac{1}{2} \times 325 \times (3.00)^{2}$

$I = 1462.5$ kgm²

From the question, time t = 2.05 s.

Putting the values into the equation,

$\omega = \frac{\tau t}{I}$

$\omega = \frac{585 \times 2.05}{1462.5}$

$\omega = 0.82$ rad/s

Now, we will convert from rad/s to rev/s. To do that, we will divide our answer by 2π

0.82 rad/s = 0.82/2π rev/s

= 0.13 rev/s

Hence, the angular speed is 0.13 rev/s,

6 0

3 years ago

When point charges q = +8.4 uC and q2 = +5.6 uC are brought near each other, each experiences a repulsive force of magnitude 0.6

Bezzdna [24]

Answer:

Distance between the charges, r = 0.8 meters

Explanation:

Given that,

Charge 1, $q_1=+8.4\ \mu C=+8.4\times 10^{-6}\ C$

Charge 2, $q_2=+5.6\ \mu C=+5.6\times 10^{-6}\ C$

Repulsive force between charges, F = 0.66 N

Let r is the distance between charges. The formula for the electrostatic force is given by :

$F=k\dfrac{q_1q_2}{r^2}$

$r=\sqrt{\dfrac{kq_1q_2}{F}}$

$r=\sqrt{\dfrac{9\times 10^9\times 8.4\times 10^{-6}\times 5.6\times 10^{-6}}{0.66}}$

r = 0.8009 meters

r = 0.8 meters

So, the distance between the charges i 0.8 meters. Hence, this is the required solution.

4 0

3 years ago