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3 years ago

The power of a motor is 60kW. At what speed can it raise a load of 50000

1 answer:

8 0

Well right off the bat, you've handed us a serious problem. You left the unit off of the 50,000, so we don't know the single most important thing that we need to know about the load.

Is the load 50,000 kilograms ? 50,000 Newtons ? 50,000 pounds ? Each of these will have a different answer.

Since you didn't specify the unit, I can make it anything I want, and I can pick the unit to make the problem easy to solve.

So I'm going to say that the load weighs 50,000 Newtons, and now I shall proceed to solve the problem that I have invented.

If you're using a motor that's marked "60 kW", that number is the maximum safe power the motor can deliver without overheating and breaking down. In order to raise our 50,000N load as fast as possible, we'll run the motor at its maximum rated power of 60 kW. That means it'll do 60,000 Joules of work for us every second.

Power = 60,000 Joules/second

Power = 60,000 Newton-meters/second

60,000 Newton-meters/sec = (50,000 Newton) x S meters/sec.

Divide each side by 50,000 Newtons:

S = (60,000 Newton-m/sec) / (50,000 Newton)

S = (60,000/50,000) meter/sec

<em>Speed = 1.2 meters/sec </em>

Our 60 kW motor can raise the load at the speed of 1.2 m/s or any slower, but no faster than that.

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A plane flying against the wind covers the 900-kilometer distance between two aerodromes in 2 hours. The same plane flying with

Mashcka [7]

Answer:

The speed of the wind is 25 km/hr.

Explanation:

Let us call $v_p$ the speed of the plane and $v_w$ the speed of the wind. When the plane is flying against the wind, it covers the distance of 900-km in 2 hours (120 minutes); therefore;

(1). $v_p - v_w = \dfrac{900km}{120min}$

And when the plane is flying with the wind, it covers the same distance in 1 hour 48 minutes (108 minutes)

(2). $v_p+v_w= \dfrac{900km}{108min}$

From equation (1) we solve for $v_p$ and get:

$v_p = \dfrac{900km}{120min}+v_w$ ,

and by putting this into equation (2) we get:

$\dfrac{900km}{120min}+v_w+v_w= \dfrac{900km}{108min}$

$2v_w= \dfrac{900km}{108min}-\dfrac{900km}{120min}$

$2v_w = 8.3km/min - 7.5km/min$

$2v_w = 0.83km/min$

$v_w = 0.4165km/min$

or in km/hr this is

$\boxed{v_w= 25km/hr }$

4 0

3 years ago

Between 1911 and 1990, the top of the leaning bell tower at Pisa, Italy, moved toward the south at an average rate of 1.2 mm/y.T

Anit [1.1K]

Answer:

Angular speed, $\omega=6.90\times 10^{-13}\ rad/s$

Explanation:

It is given that,

The top of the leaning bell tower at Pisa, Italy, moved toward the south at an average rate of, v = 1.2 mm/yr

$1\ mm/yr=3.171\times 10^{-11}\ m/s$

Velocity, $v=3.80\times 10^{-11}\ m/s$

Height of the tower, h = 55 m

The height of the tower is equivalent to the radius. Let $\omega$ is the angular speed of the tower’s top about its base. The relation between the angular speed and the angular speed is given by :

$v=r\omega$

$\omega=\dfrac{v}{r}$

$\omega=\dfrac{3.80\times 10^{-11}\ m/s}{55\ m}$

$\omega=6.90\times 10^{-13}\ rad/s$

So, the average angular speed of the tower’s top about its base is $6.90\times 10^{-13}\ rad/s$ . Hence, this is the required solution.

6 0

3 years ago

An antigen is a protein made by your body to respond to a specific foreign molecule.

Tom [10]

hope this helps it's F

8 0

3 years ago

What kind of energy is in a rock at the edge of a cliff.

soldi70 [24.7K]

Answer:

kinetic energy

Explanation:

8 0

2 years ago

While on a playground, you and your niece take turns sliding down a frictionless slide. Your mass is 75 kg while your little nie

Olenka [21]

Answer:

Explanation:

In absence of friction, total mechanical energy must be conserved.
This means that the change in gravitational potential energy (in magnitude) must be equal to the change in kinetic energy:

$\Delta K = \Delta U \\ \\ \frac{1}{2} * m* v^{2} = m*g*h$

As it can be seen, the mass m is on the both sides of the equation, which means that it can be simplified.
We can solve this equation for v (speed at the bottom) as follows:

$v = \sqrt{2*g*h}$

As it can be seen, the mass has no part in the equation, so, due both are starting from the same height, both people have the same speed at the bottom.
The option c is the right one.

6 0

3 years ago