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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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Why do bones weaken as a person gets older
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6 0
2 years ago
A car is moving at 10 m/s to the right. It accelerates for 10 s after which it is moving at 5 m/s to the left. What was the car'
NNADVOKAT [17]

Answer:

Acceleration, a=-1.5\ m/s^2

Explanation:

It is given that,

Initial velocity of the car, u = 10 m/s (in right)

Final velocity of the car, v = -5 m/s (in left)  

Time taken, t = 10 s

Let a is the acceleration of the car. It can be calculated using the equation of kinematics. The equation is as :

v=u+at

a=\dfrac{v-u}{t}

a=\dfrac{-5-10}{10}    

a=-1.5\ m/s^2

So, the acceleration of the car is -1.5\ m/s^2. Hence, this is the required solution.

6 0
3 years ago
How high does the water rise in the bell after enough time has passed for the air to reach thermal equilibrium
Minchanka [31]

The height risen by water in the bell after enough time has passed for the air to reach thermal equilibrium is 3.8 m.

<h3>Pressure and temperature at equilibrium </h3>

The relationship between pressure and temperature can be used to determine the height risen by the water.

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

where;

  • V₁ = AL
  • V₂ = A(L - y)
  • P₁ = Pa
  • P₂ = Pa + ρgh
  • T₁ = 20⁰C = 293 K
  • T₂ = 10⁰ C = 283 k

\frac{PaAL}{T_1} = \frac{(P_a + \rho gh)A(L-y)}{T_2} \\\\\frac{PaL}{T_1} = \frac{(P_a + \rho gh)(L-y)}{T_2} \\\\L-y = \frac{PaLT_2}{T_1(P_a + \rho gh)} \\\\y = L (1 - \frac{PaT_2}{T_1(P_a + \rho gh)})\\\\y = 4.2(1 - \frac{101325 \times 283}{293(101325\  +\  1000 \times  9.8 \times  100)} )\\\\y = 3.8 \ m

Thus, the height risen by water in the bell after enough time has passed for the air to reach thermal equilibrium is 3.8 m.

The complete question is below:

A diving bell is a 4.2 m -tall cylinder closed at the upper end but open at the lower end. The temperature of the air in the bell is 20 °C. The bell is lowered into the ocean until its lower end is 100 m deep. The temperature at that depth is 10°C. How high does the water rise in the bell after enough time has passed for the air to reach thermal equilibrium?

Learn more about thermal equilibrium here: brainly.com/question/9459470

#SPJ4

3 0
2 years ago
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