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2 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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What is a nebula? <br> please help I am clueless

Rainbow [258]

A nebula is gas, dust, and other things that form and form some sort of clould like thing. It also looks cool because of the colors it produces. I'm sure you could do a web search to find more about it, I don't take astronomy hahaha

4 0

2 years ago

How many milliliters of water at 25.0°C with a density of 0.997 g/mL must be mixed with 162 mL of coffee at 94.6°C so that the r

Andre45 [30]

Answer:

The volume of water is 139 mL.

Explanation:

Due to the Law of conservation of energy, the heat lost by coffee is equal to the heat gained by the water, that is, the sum of heats is equal to zero.

$Q_{coffee} + Q_{water} = 0\\Q_{coffee} = - Q_{water}$

The heat (Q) can be calculated using the following expression:

$Q=c \times m \times \Delta T$

where,

c is the specific heat of each substance

m is the mass of each substance

ΔT is the difference in temperature for each substance

The mass of coffee is:

$162mL.\frac{0.997g}{mL} = 162g$

Then,

$Q_{coffee} = - Q_{water}\\c_{c} \times m_{c} \times \Delta T_{c} = -c_{w} \times m_{w} \times \Delta T_{w}\\m_{c} \times \Delta T_{c} = -m_{w} \times \Delta T_{w}\\m_{w} = \frac{m_{c} \times \Delta T_{c}}{-\Delta T_{w}} \\m_{w}=\frac{162g \times (62.5 \°C - 94.6 \°C ) }{-(62.5 \°C - 25.0 \°C)} \\m_{w} = 139 g$

The volume of water is:

$139g.\frac{1mL}{0.997g} =139mL$

7 0

3 years ago

when an object is charged by contact, what kind of charge does the object have compared with the charge on the on the object giv

sveta [45]

Answer:

the charge that is given by the object is positive charge and the object which is taking the charge is negetively charged

Explanation:

5 0

2 years ago

A mother and her 35.0 -kg child are riding an escalator to the third level of a shopping mall. If the child's gravitational pote

notka56 [123]

The increase in potential energy of his mother if her mass is 56.0 kg will be 6031.97 J.

<h3>What is gravitational potential energy?</h3>

The energy that an item has due to its location in a gravitational field is known as gravitational potential energy.

The potential energy increases by 3773 J

PE₂-PE₁=mg(h₂-h₁)

3773 J = 35.0 × 9.81 × (h₂-h₁)

(h₂-h₁) = 10.98

Case 2 ;

ΔPE =?

ΔPE=mg(h₂-h₁)

ΔPE=56.0 × 9.81 ×10.98

ΔPE=6031.97 J.

Hence, the increase in potential energy of his mother if her mass is 56.0 kg will be 6031.97 J.

To learn more about the gravitational potential energy, refer;

brainly.com/question/3884855#SPJ1

#SPJ1

8 0

1 year ago

A double-slit experiment uses light of wavelength 650 nm with a slit separation of 0.100 mm and a screen placed 4.0 m away. a) W

dezoksy [38]

Answer:

Explanation:

a ) Slit separation d = .1 x 10⁻³ m

Screen distance D = 4 m

wave length of light λ = 650 x 10⁻⁹ m

Width of central fringe = λ D / d

= $\frac{650\times10^{-9}\times4}{.1\times10^{-3}}$

= 26 mm

b ) Distance between 1 st and 2 nd bright fringe will be equal to width of dark fringe which will also be equal to 26 mm

c ) Angular separation between the central maximum and 1 st order maximum will be equal to angular width of fringe which is equal to

λ / d

= $\frac{650\times10^{-9}}{.1\times10^{-3}}$

= 6.5 x 10⁻³ radian.

8 0

2 years ago