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Dmitriy789 [7]
2 years ago
13

A person who climbs up something (e.g., a hill, a ladder, the stairs) from the ground gains potential energy. a person's weight

(f = mg) is usually measured in pounds. for conversions, you may need to know that the weight of 1 pound is equal to 4.448 newtons. part a how many meters above the ground would a 240 lb person have to climb to gain 3 kilojoules (3000 joules) of potential energy?
Physics
1 answer:
zhannawk [14.2K]2 years ago
3 0

We are given the following values:

weight w = 240 lb = 1,067.52 N

energy E = 3,000 J

 

The formula for potential energy is:

E = w h

where h is the height the person has to climb, therefore:

h = 3000 / 1067.52

<span>h = 2.81 m</span>

<span>
</span>

<span>Therefore he has to climb 2.81 meters</span>

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A 145-g baseball is thrown so that it acquires a speed of 25 m/s. What was the net work done on the ball to make it reach this s
inysia [295]

When the ball has left your hand and is flying on its own, its kinetic energy is

KE = (1/2) (mass) (speed²)

KE = (1/2) (0.145 kg) (25 m/s)²

KE = (0.0725 kg) (625 m²/s²)

<em>KE = 45.3 Joules</em>

If the baseball doesn't have rocket engines on it, or a hamster inside running on a treadmill that turns a propeller on the outside, then there's only one other place where that kinetic energy could come from:  It MUST have come from the hand that threw the ball.  The hand would have needed to do  <em>45.3 J</em>  of work on the ball before releasing it.

6 0
2 years ago
In 1991 four English teenagers built an eletric car that could attain a speed of 30.0m/s. Suppose it takes 8.0s for this car to
ivanzaharov [21]

a=Δv/Δt=(30.0-18.0)/8.0=12.0/8.0=1.5 m/s²


6 0
2 years ago
Read 2 more answers
A ledge on a building is 18 m above the ground. A taut rope attached to a 4.0-kg can of paint sitting on the ledge passes up ove
True [87]

Answer:t=5.07 s

Explanation:

Given

height of Building h=18 m

mass of Paint can m_1=4 kg

mass of second can m_2=3 kg

let T be the Tension in the rope

For  4 kg can

4g-T=4a

T=4(g-a)----1

For 3 kg can

T-3g=3a

T=3(g+a)----2

From 1 and 2

4(g-a)=3(g+a)

g=7a

a=\frac{g}{7}

So time taken to cover 18 m is

h=ut+\frac{at^2}{2}

18=0+\frac{g\cdot t^2}{7\times 2}

t^2=\frac{18\times 2\times 7}{g}

t=5.07 s

5 0
2 years ago
In a charge-free region of space, a closed container is placed in an electric field. Which of the following is a requirement for
galina1969 [7]

Answer:

D. The requirement does not exist -the total electric flux is zero no matter what.

Explanation:

According to Gauss's law , total electric flux over a closed surface is equal to 1 / ε₀ times charge inside.

If charge inside is zero , total electric flux over a closed surface is equal to

zero . It has nothing to do with whether external field is uniform or not. For any external field , lines entering surface will be equal to flux going out.

8 0
2 years ago
Astronauts land on another planet and measure the density of the atmosphere on the planet surface. They measure the mass of a 50
Lana71 [14]

1.6 kg/m^3 is the best estimate of the density of the air on the planet.

Given:

The mass of the conical flask with stopper is 457.23 grams and the volume is 500cm^3.

Mass of conical flask and a stopper after removing the air is 456.43 g

To find:

The density of the air on the planet.

Solution;

Mass of the conical flask and stopper with air on the planet= 457.23 g

Mass of conical flask with a stopper and without air on the planet =  456.43 g

Mass of the air in the conical flask on the planet =m

m = 457.23 g-456.43 g=0.8 g\\\\1 g = 0.001 kg\\\\m =0.8 g =0.8\times 0.001 kg=0.0008 kg

The volume of the conical flask = 500 cm^3

The volume of the air in the conical flask = V = 500cm^3

1 cm^3=10^{-6} m^3\\\\V= 500cm^3= 500\times 10^{-6}m^3=0.0005 m^3

The density of the air on the planet = d

d=\frac{m}{V}\\\\d=\frac{0.0008 kg}{0.0005 m^3}\\\\=1.6 kg/m^3

1.6 kg/m^3 is the best estimate of the density of the air on the planet.

Learn more about density here:

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7 0
1 year ago
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