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Anna11 [10]
3 years ago
14

How much power does it take to lift 30.0 N 10.0 m high in 5.00 s?

Physics
2 answers:
Rudik [331]3 years ago
7 0

Power = work/time = (Force times distance)/time

= (30N *10.0m)/5.00s = 300/5 = 60 Watts

Anna007 [38]3 years ago
5 0

Answer: P = 60 W

Explanation: Power is the ratio of energy per unit time. Energy can also be substituted by work.

P = W / t

= 30 N ( 10 m) / 5.00 s

= 60 W

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The wetter the surface, the friction
allochka39001 [22]
Answer: Water can either increase or decrease the friction between surfaces.
7 0
3 years ago
Which statement describes a scientific theory?
PSYCHO15rus [73]

Answer:

A theory changes based on new observations and testing.

Explanation:

A scientific theory is a product of multiple trials and repeated experiments. It usually follows after carefully conducting and testing the validity of the hypothesis.

A scientific theory provides an explanation into how something behaves.

A law just states a finding will not explain it.

Most theories are tenable and can be improved upon when new observations and testing are carried out.

4 0
3 years ago
An automatic dryer spins wet clothes at an angular speed of 5.2 rad/s. Starting from rest, the dryer reaches its operating speed
Dvinal [7]
<h2>Time taken by dryer to come up to speed is 1.625 seconds.</h2>

Explanation:

We have equation of motion v = u + at

     Initial velocity, u =  0 rad/s

     Final velocity, v = 5.2 rad/s    

     Time, t = ?

     Acceleration, a = 3.2 rad/s²

     Substituting

                      v = u + at  

                      5.2 = 0 + 3.2 x t

                      t = 1.625 s

Time taken by dryer to come up to speed is 1.625 seconds.

6 0
3 years ago
What is the efficiency (w/qh) of an ideal carnot heat engine operating between a hot region at t= 400 k and a cold one at t= 300
Vinvika [58]
The efficiency of an ideal Carnot heat engine can be written as:
\eta = 1-  \frac{T_{cold}}{T_{hot}}
where
T_{cold} is the temperature of the cold region
T_{hot} is the temperature of the hot region

For the engine in our problem, we have T_{cold}=300 K and T_{hot}=400 K, so the efficiency is
\eta= 1 - \frac{300 K}{400 K}=0.25
4 0
3 years ago
Consider a small car of mass 1200 kg and a large sport utility vehicle (SUV) of mass 4000 kg. The SUV is traveling at the speed
Karolina [17]

Answer:

63.9 m/s

Explanation:

Parameters given:

Mass of small car, m = 1200 kg

Mass of SUV, M = 4000 kg

Speed of SUV, V = 35 m/s

Their kinetic energy of the small car is equal to the kinetic energy of the SUV, hence:

0.5 * m * v² = 0.5 * M * V²

=> 0.5 * 1200 * v² = 0.5 * 4000 * 35²

600 * v² = 2450000

v² = 2450000/600

v² = 4083.3

=> v = 63.9 m/s

The speed of the small car is 63.9 m/s.

6 0
3 years ago
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