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3 years ago

Calculate the power in kilowatts required to lift a 500 kg mass 12 meters in one second

1 answer:

8 0

The weight of the load is (mass) x (gravity).

The work done to lift it is (weight) x (distance) = (mass) x (gravity) x (distance)

The power is (work)/(time) =

   (mass) x (gravity) x (distance) / (time) =

   (500 kg) x (9.8 m/s²) x (12 m) / (1 sec) =

(500 x 9.8 x 12 / 1) (kg-m / sec²) (m) / (sec) =

   (500 x 9.8 x 12 / 1) newton - meter / sec =

(500 x 9.8 x 12 / 1) joule    / sec =

   58,800 watts

   = <em>58.8 kW</em>

   =    about 78.8 horsepower

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How much work in joules is required to lift a 23 kg box up from the ground to your waist that is 1.0 meters high, carry it 6 met

PSYCHO15rus [73]

Answer:

2682

Explanation:

Work done is given by :

Work = Force x distance

= mg x d

So, work done in lifting the box of 23 kg up to my waist of 1 m high is :

W = mg x d

= 23 x 9.18 x 1

= 211.14

Now work done carrying the box horizontally 6 meters across the room is

W = mg x d

= 23 x 9.18 x 6

= 1266.84

Work done in placing the box on the shelf that is 5.7 m above the ground is

W = mg x d

= 23 x 9.18 x 5.7

= 1203.49

So the total work done is = 211.14 + 1266.84 + 1203.49

= 2681.47

= 2682 (rounding off)

5 0

3 years ago

Can u Anwser this Plzzzz

nata0808 [166]

Answer:

-0. 75m/s^2

Explanation:

use formula of acceleration

5 0

3 years ago

Examine the scenario. Object A has 5 protons and 5 electrons. Object B has 5 protons and 7 electrons. Which option most accurate

Travka [436]

<span>Last choice on the list:
Object A has a net charge of 0 because the positive and negative
charges are balanced.
Object B has a net charge of –2 because there is an imbalance of
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4 0

3 years ago

You are standing in a building whose height (40m) you throw a ball downward at a angle of -30 at a speed of (10m/s) acceleration

jeyben [28]

Answer: 3.41 s

Explanation:

Assuming the question is to find the time $t$ the ball is in air, we can use the following equation:

$y=y_{o}+V_{o}sin \theta t-\frac{1}{2}gt^{2}$

Where:

$y=0m$ is the final height of the ball

$y_{o}=40 m$ is the initial height of the ball

$V_{o}=10 m/s$ is the initial velocity of the ball

$t$ is the time the ball is in air

$g=9.8 m/s^{2}$ is the acceleration due to gravity

$\theta=30\°$

Then:

$0 m=40 m+(10 m/s)(sin(30\°))t-\frac{1}{2}9.8 m/s^{2}t^{2}$

$0 m=40 m+5m/s t-4.9 m/s^{2}t^{2}$

Multiplying both sides of the equation by -1 and rearranging:

$4.9 m/s^{2}t^{2}-5m/s t-40 m=0$

At this point we have a quadratic equation of the form $at^{2}+bt+c=0$ , which can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

Where:

$a=4.9$

$b=-5$

$c=-40$

Substituting the known values:

$t=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(4.9)(-40)}}{2(4.9)}$

Solving the equation and choosing the positive result we have:

$t=3.41 s$ This is the time the ball is in air

5 0

4 years ago

Swinging a golf club or baseball bat are examples of ______________ stretching.

kari74 [83]

Answer:

Static stretching is the answer.

Explanation:

Static stretching is the most common form that greatly improves flexibility. However, static stretches does little to contract the muscles needed to generate powerful golf swings. Dynamic stretches help improve your range of motion while reducing muscle stiffness.

6 0

4 years ago