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ehidna [41]
3 years ago
7

One horsepower (hp) is the amount of power required to lift a 75-kg mass a vertical distance of 1 m in 1 s. What is 2 hp equival

ent to in watts (W)?
Physics
1 answer:
Vladimir [108]3 years ago
8 0

Answer:

1470 W

Explanation:

Power: This can be defined as the rate at which work is done or energy is used up. The S.I unit of power is Watt (W).

The expression for power is given as,

P = Energy/time

P = mgh/t ...................... Equation 1

Where P = power, m = mass, h = height, t = time, g = acceleration due to gravity.

Given: m = 75 kg, g =9.8 m/s², h = 1 m, t = 1 s.

Substitute into equation 1

P = (75×1×9.8)/1

P = 735 W.

From the above,

1 hp = 735 W

2 hp = (2×735) W

2 hp = 1470 W.

Hence 2 hp = 1470 W

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3 years ago
An object of mass kg is released from rest m above the ground and allowed to fall under the influence of gravity. Assuming the f
IgorLugansk [536]

Answer:

Explanation:

From, the given information: we are not given any value for the mass, the proportionality constant and the distance

Assuming that:

the mass = 5 kg and the proportionality constant = 50 kg

the distance of the mass above the ground x(t) = 1000 m

Let's recall that:

v(t) = \dfrac{mg}{b}+ (v_o - \dfrac{mg}{b})^e^{-bt/m}

Similarly, The equation of mption:

x(t) = \dfrac{mg}{b}t+\dfrac{m}{b} (v_o - \dfrac{mg}{b}) (1-e^{-bt/m})

replacing our assumed values:

where v_=0 \ and \ g= 9.81

x(t) = \dfrac{5 \times 9.81}{50}t+\dfrac{5}{50} (0 - \dfrac{(5)(9.81)}{50}) (1-e^{-(50)t/5})

x(t) = 0.981t+0.1 (0 - 0.981) (1-e^{-(10)t}) \ m

\mathbf{x(t) = 0.981t-0.981(1-e^{-(10)t}) \ m}

So, when the object hits the ground when x(t) = 1000

Then from above derived equation:

\mathbf{x(t) = 0.981t-0.981(1-e^{-(10)t}) \ m}

1000= 0.981t-0.981(1-e^{-(10)t}) \ m

By diregarding e^{-(10)t} \ m

1000= 0.981t-0.981

1000 + 0.981 = 0.981 t

1000.981 = 0.981 t

t = 1000.981/0.981

t = 1020.36 sec

7 0
3 years ago
Two friends, Al and Jo, have a combined mass of 168 kg. At an ice skating rink they stand close together on skates, at rest and
Keith_Richards [23]

Answer: 91.94 kg

Explanation: Supoose Al's mass is m. The combined mass is 168kg, so Jo's mass is 168 - m.

The Law of Conservation of Momentum states that when two objects collide, the total momentum of both objects before the collision is the same as the total momentum of both objects after the collision.

At the beginning, Al and Jo are united without movement, so their initial momentum is zero.

After the release of their hands, Al goes in one direction and Jo moves to the opposite direction. Suppose the direction Al is moving is positive. Conservation of momentum will be

m_{1}v_{1}_{(f)}+m_{2}v_{2}_{(f)}=m_{1}v_{1}_{(i)}+m_{2}v_{2}_{(i)}

where

index i referes to initial momentum

index f to final momentum

index 1 refers to Al

index 2 to Jo

Calculating:

m(0.91)-[(168-m)(1.1)]=0

0.91m+1.1m-184.8=0

2.01m=184.8

m = 91.94

Al has a mass of 91.94 kg.

6 0
3 years ago
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