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2 years ago

6

A weight trainer lifts a 90.0-kg barbell from a stand 0.90 m high and raises it to a height of 1.75 m. What is the increase in t

he potential energy of the barbell-Earth system?

1 answer:

nirvana33 [79]2 years ago

7 0

Answer:

∆PE = 749.7 J

At 0.9 m high, PE = 793.8 J

At 1.75 m high, PE = 1543.5 J

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A vehicle of mass 100kg has a kinetic energy of 5000 J at an instant. The velocity at that instant is

snow_lady [41]

Answer:

$\bf\pink{10\:m/s}$

Explanation:

<h2><u>Given</u> :-</h2>

$\sf\red{Mass \ of \ vehicle \ (m) = 100 \ kg}$
$\sf\orange{Kinetic \ energy \ (K.E.) = 5000 \ J}$

<h2><u>To Find</u> :-</h2>

$\sf\green{Velocity\: of\: the \:vehicle \:at \:the \:instant}$

<h2><u>Formula to be used</u> :-</h2>

$\sf\blue{K.E. = \dfrac{1}{2} mv^{2}}$

Where,

K.E. = Kinetic energy possessed by the body
M = Mass of the body
V = Velocity of the body

<h2><u>Solution</u> :-</h2>

$\to\:\:\sf\red{K.E. = \dfrac{1}{2}mv^{2}}$

$\to\:\:\sf\orange{5000 = \dfrac{1}{2} \times 100 \times v^{2}}$

$\to\:\:\sf\green{5000 = 50 \times v^{2}}$

$\to\:\:\sf\blue{\dfrac{5000}{50} = v^{2}}$

$\to\:\:\sf\purple{100 = v^{2}}$

$\to\:\:\sf\red{\sqrt{100} = v}$

$\to\:\:\sf\orange{ 10 = v}$

$\to\:\:\bf\pink{v = 10\:m/s}$

Velocity of the vehicle at the instant is $\bf\green{10\:m/s}$

7 0

3 years ago

Read 2 more answers

35) Kinetic energy in flowing water drives

Paha777 [63]

Answer:

C) Turbines

Explanation:

C. because as the water flows, its kinetic energy is used to turn a turbine.

8 0

3 years ago

Why doesn't every planet have a moon?

GalinKa [24]

Answer:

Up first are Mercury and Venus. Neither of them has a moon. Because Mercury is so close to the Sun and its gravity, it wouldn't be able to hold on to its own moon. Any moon would most likely crash into Mercury or maybe go into orbit around the Sun and eventually get pulled into it.

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3 years ago

In a choir practice room, two parallel walls are 5.70 m apart. The singers stand against the north wall. The organist faces the

ololo11 [35]

Answer:

4.98 m

Explanation:

Given that

Width of the mirror, d = 0.6 m

Organist distance to the mirror, s = 0.78 m

Distance between the singer and the organist, S = 5.7 + 0.78 = 6.48 m

Width of north wall, D?

Using the simple relationship

D/S = d/s, on rearranging

D = dS /s

D = (0.6 * 6.48) / 0.78

D = 3.888 / 0.78

D = 4.98 m

Therefore, we can conclude that the Width of north wall is 4.98 m

7 0

3 years ago

A coin is dropped in a 15.0 m deep well.

labwork [276]

Answer:

t = 1.75

t = 0.04

Explanation:

a)

For part 1 we want to use a kenamatic equation with constant acceleration:

X = 1/2*a*t^2

isolate time

t = sqrt(2X / a)

Plugin known variables. Acceleration is the force of gravity which is 9.8 m/s^2

t = sqrt(2*15m / 9.8m/s^2)

t = 1.75 s

b)

The speed of sound travels at a constant speed therefore we don't need acceleration and can use the equation:

v = d / t

isolate time

t = d / v

plug in known variables

t = 15m / 340m/s

t = 0.04 s

7 0

3 years ago