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3 years ago

Find the work done lifting a 290-pound weight 28 feet in the air.

2 answers:

8 0

Hello there.

Find the work done lifting a 290-pound weight 28 feet in the air.

d.
8,120 foot-pounds

Nataly [62]3 years ago

6 0

The answer is d.8,120 foot-pounds

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You drop a 0.375 kg ball from a height of 1.37 m. It hits the ground and bounces up again to a height of 0.67 m. How much energy

Radda [10]

2.57 joule energy lose in the bounce .

Explanation:

when ball is the height of 1.37 m from the ground it has some gravitational potential energy with respect to hits the ground

Formula for gravitational potential energy given by

Potential Energy = mgh

Where ,

m = mass

g = acceleration due to gravity

h = height

Potential energy when ball hits the ground

m= 0.375 kg

h = 1.37 m

g = 9.8 m/s²

$Potential Energy = 0.375\times9.8\times1.37$

Potential Energy = 5.03 joule

Potential energy when ball bounces up again

h= 0.67 m

$Potential Energy = 0.375\times0.67\times9.8$

Potential Energy = 2.46 joule

Energy loss = 5.03 - 2.46 = 2.57 joule

2.57 joule energy lose in the bounce

6 0

2 years ago

From the gravitational law calculate the weight W (gravitational force with respect to the earth) of a 89-kg man in a spacecraft

zhannawk [14.2K]

Answer:

$W=\frac{773}{4.45}=173.76 l b f$

Explanation:

$W=\frac{G \cdot m_{e} \cdot m}{(R+h)^{2}}$

The law of gravitation

$G=6.673\left(10^{-11}\right) m^{3} /\left(k g \cdot s^{2}\right)$

Universal gravitational constant [S.I. units]

$m_{e}=5.976\left(10^{24}\right) k g$

Mass of Earth [S.I. units]

$m=89 kg$

Mass of a man in a spacecraft [S.I. units]

$R=6371 \mathrm{~km}$

Earth radius [km]

Distance between man and the earth's surface

$h=261 \mathrm{~km} \quad[\mathrm{~km}]$

ESULT $W=\frac{6.673\left(10^{-11}\right) \cdot 5.976\left(10^{24}\right) \cdot 89}{\left(6371 \cdot 10^{3}+261 \cdot 10^{3}\right)^{2}}=773.22 \mathrm{~N}$

$W=\frac{773}{4.45}=173.76 l b f$

4 0

2 years ago

WILL GIVE BRAINLIEST TO THE CORRECT ANSWER

LenKa [72]

Answer:

See the answers below.

Explanation:

We can solve both problems using Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

∑F =m*a

where:

F = force [N] (units of newtons)

m = mass = 1000 [kg]

a = acceleration = 3 [m/s²]

$F = 1000*3\\F=3000[N]$

And the weight of any body can be calculated by means of the mass product by gravitational acceleration.

$W=m*g\\W=1000*9.81\\W=9810 [N]$

4 0

2 years ago

Which transition metals have more in common than other

jeka94

Answer:

A. alkali metals

Explanation:

im not sure about that

8 0

3 years ago

Read 2 more answers

A reservoir located in the mountain 250 m above sea level flows through a pipe to a hydroelectric plant in a town at sea level.

Pavlova-9 [17]

Answer:

v₂ = 70 m / s

Explanation:

For this exercise let's use Bernoulli's equation

where subscript 1 is for the top of the mountain and subscript 2 is for Tuesday's level

P₁ + ½ ρ v₁² + ρ g y₁ = P₂ +1/2 ρ v₂² + ρ g y₂

indicate that the pressure in the two points is the same, y₁ = 250 m, y₂ = 0 m, the water in the upper part, because it is a reservoir, is very large for which the velocity is very small, we will approximate it to 0 (v₁ = 0), we substitute

ρ g y₁ = ½ ρ v₂²

v₂ = $\sqrt {2g \ y_1}$

let's calculate

v₂ = √( 2 9.8 250)

v₂ = 70 m / s

6 0

3 years ago