All categories

2 years ago

A crane lifts a load of 15000N to a height 40m in 50 seconds. Calculate the power of the crane

Physics

1 answer:

Nataly [62]2 years ago

6 0

Answer:

P = 12000 W

Explanation:

General data:

F = 15000 N
d = 40 m
t = 50 s
P = ?

Work is force times unit of distance. So in order to calculate the power we must first calculate the work.

Formula:

$\boxed{\bold{W=\frac{F}{d}}}$

Replace and solve

$\boxed{\bold{W=\frac{15000\ N}{40\ m}}}$

$\boxed{\bold{W=600000\ J}}$

once the work is found, we proceed to find the power according to the formula:

$\boxed{\bold{P=\frac{W}{t}}}$

$\boxed{\bold{P=\frac{600000\ J}{50\ s}}}$

$\boxed{\boxed{\bold{P=12000\ W}}}$

The power of the crane is 12000 Watts.

Greetings.

You might be interested in

If a net force acting on a moving object causes its speed to increase, what

Klio2033 [76]

Answer:

It increases.

Explanation:

If a force acts in the same direction as the object's motion, then the force speeds the object up. Either way, a force will change the velocity of an object. And if the velocity of the object is changed, then the momentum of the object is changed.

7 0

3 years ago

Use each of the following terms in a separate

Alenkinab [10]

I think the correct answer is going to be A

3 0

3 years ago

Which of the following methods has led to the earliest discoveries of massive planets orbiting near their parent stars a. detect

Mrac [35]

Answer:

c. detecting the gravitational effect of an orbiting planet (The Wobble"") by looking for the Doppler shifts in the star's spectrum

Explanation:

In a solar system the mass of the star and planets affect each other's orbital movements. The center of gravity of a star and a planet is inside the star. This causes the star to be closer and farther from the Earth at different times. Due to this wobble the star appears to be red shifted when it is farther and blue shifted when it is closer.

When the mass of the planet is high, like a hot Jupiter it causes more wobble i.e., change in radial velocity. This makes it easier to detect the planet. The earliest hot Jupiter found by this method is the planet 51 Pegasi b.

3 0

3 years ago

Igneous rocks weather more easily than sedimentary rocks.

IrinaK [193]

Answer:

False because igneous rocks are formed from a volcano and sedimentary never move they stay in one spot

7 0

3 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

kow [346]

Before the engines fail, the rocket's altitude at time t is given by

$y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

and its velocity is

$v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t$

The rocket then reaches an altitude of 1150 m at time t such that

$1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

Solve for t to find this time to be

$t=11.2\,\mathrm s$

At this time, the rocket attains a velocity of

$v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}$

When it's in freefall, the rocket's altitude is given by

$y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

where $g=9.80\frac{\rm m}{\mathrm s^2}$ is the acceleration due to gravity, and its velocity is

$v_2(t)=124\dfrac{\rm m}{\rm s}-gt$

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for $y_2(t)$ to reach 0:

$1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s$

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

${v_f}^2-{v_i}^2=2a\Delta y$

where $v_f$ and $v_i$ denote final and initial velocities, respecitively, $a$ denotes acceleration, and $\Delta y$ the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means $y_2$ will contain the information we need to find the maximum height.

$-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)$

Solve for $y_{\rm max}$ and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to $y_2(t)$ ) to be about 32.6 s. Plug this into $v_2(t)$ to find the velocity before it crashes:

$v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}$

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

3 0

3 years ago

A crane lifts a load of 15000N to a height 40m in 50 seconds. Calculate the power of the crane​

A crane lifts a load of 15000N to a height 40m in 50 seconds. Calculate the power of the crane