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defon
3 years ago
6

A ventilating fan is operated by 0.5hp electric motor. How much work in joules can the fan do in 3 hours? (I need the answer asa

p)
Physics
1 answer:
shtirl [24]3 years ago
3 0

Answer:

Power= 0.5hp = 375W

T = 3hrs = 10800s

Power = Work done/ Time

Work = Power * Time = 375 * 10800 = 4050000J

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while flying a plane parallel to the ground a pilot releases a fuel tank in order to reduce the planes mass. what is the tanks f
stepan [7]

La velocidad vertical del tanque después de caer 10 m es 14 m/seg .

La velocidad vertical del tanque se calcula mediante la aplicación de la fórmula de velocidad , la componente vertical Vfy, del movimiento horizontal como se muestra a continuación :

Vfy=?

  h = 10 m

                              Fórmula de Velocidad vertical Vfy:

                           Vfy²  = 2*g*h

                            Vfy= √(2*9.8m/seg2* 10m )

                            Vfy= 14 m/seg

7 0
3 years ago
Which is the average kinetic energy of particles in an object?
alina1380 [7]
The amount of heat in the body in joule
5 0
3 years ago
If the velocity of an object is doubled, its kinetic energy is ______
swat32
Increased by a factor of 4
4 0
3 years ago
How much power does it take to lift a 30.0 n box 10.0 m high in 5.00 s, if you must apply a 62n force to lift the box?
Illusion [34]
Power is defined as the rate at which the body is doing work:
P=\frac{W}{t}
Work is defined as displacement done by the force times that displacement:
W=F\cdot h
We know that we need 62N to move the box, so when we apply this force along the path of 10m we have done:
W=62N\cdot10m=620J
of work.
Now we just divide that by 5s to get how much power is required:
P=\frac{620J}{5s}=124W
5 0
3 years ago
engineers are using computer models to study train collisions design safer train cars. They start by modeling an elastic collisi
Archy [21]

Answer:

The final velocity of the second car is 57 m/s south.

Explanation:

This is an elastic collision between two train cars. In this case, the total kinetic energy between the two bodies will remain the same.

The formula to apply is :

m_1v_1i +m_2v_2i=m_1v_1f+m_2v_2f

where ;

m_1=mass of object 1\\v_1i=initial  velocity object1\\m_2=mass of object2\\v_2i=initial velocity object 2\\v_1f=final velocity object 1\\v_2f=final velocity object 2

Given in the question that;

m_1=14650kg\\v_1i=18m/s\\m_2=3825kg\\v_2i=11m/s\\v_1f=6m/s\\v_2f=?

Apply the formula as;

m_1v_1i +m_2v_2i=m_1v_1f+m_2v_2f

{14650*18}+{3825*11} = {14650 *6} + {3825 * v₂f}

263700+42075=87900 + 3825v₂f

305775 =87900 + 3825v₂f

305775-87900 = 3825v₂f

217875=3825v₂f

217875/3825 =v₂f

56.96 = v₂f

<u>57 m/s = v₂f { nearest whole number}</u>

4 0
2 years ago
Read 2 more answers
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