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SpyIntel [72]
2 years ago
15

What effect does the time taken to lift the mass have on power output?

Physics
1 answer:
sineoko [7]2 years ago
4 0
The longer the time taken, the lower the power output.

This is since power is calculated through

Energy transferred / time

Increasing the denominator will lead to a lower value overall
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IF a car trip 400 km in four hours what is the rate of speed that a car is traveling?
Andreas93 [3]

Speed = Distance/ Time

Speed = 400 / 4

Speed = 100 km/hr.

100 km per hour.
8 0
3 years ago
Which equation represents an equilibrium system??
MrRissso [65]

A. 2 S O 2, gas, plus O 2, gas, in equilibrium with 2 S O 3, gas.


7 0
3 years ago
Need help in this question ???? Hurry please
Digiron [165]

Answer:

how quickly or slowly the object is moving

Hope this helps

7 0
3 years ago
A weight lifter is trying to do a bicep curl with a weight of 300 N. At the "sticking point", the moment arm of this weight is 3
lesantik [10]

Answer:

The weight lifter would not get past this sticking point.

Explanation:

Generally torque applied on the weight is mathematically represented as

             T =  F z

To obtain Elbow torque we substitute 4000 N for F (the force ) and 2cm = \frac{2}{100} = 0.02m for z the perpendicular distance

So Elbow Torque is   T_e= 4000 * 0.02

                                   = 80Nm

 To obtain the torque required we substitute 300 N for F and 30cm =\frac{30}{100} = 0.3 m

  So the Required Torque is T_R = 300 *0.3

                                                     =90Nm

Now since   T_e < T_R it mean that the weight lifter would not get past this sticking point

                                   

7 0
3 years ago
A real object is 10.0 cm to the left of a thin, diverging lens having a focal length of magnitude 16.0 cm. What is the location
amm1812

Answer:

A)6.15 cm to the left of the lens

Explanation:

We can solve the problem by using the lens equation:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}

where

q is the distance of the image from the lens

f is the focal length

p is the distance of the object from the lens

In this problem, we have

f=-16.0 cm (the focal length is negative for a diverging lens)

p=10.0 cm is the distance of the object from the lens

Solvign the equation for q, we find

\frac{1}{q}=\frac{1}{-16.0 cm}-\frac{1}{10.0 cm}=-0.163 cm^{-1}

q=\frac{1}{-0.163 cm^{-1}}=-6.15 cm

And the sign (negative) means the image is on the left of the lens, because it is a virtual image, so the correct answer is

A)6.15 cm to the left of the lens

6 0
3 years ago
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