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3 years ago

An 800 N man climbs 5 m up a ladder. How much gravitational potential energy does he gain?

Physics

1 answer:

Artemon [7]3 years ago

5 0

Answer:

4000J

Explanation:

Given parameters:

Weight of the man = 800N

Height of ladder = 5m

Unknown:

Gravitational potential energy gained = ?

Solution:

The gravitational potential energy is due to the position of a body.

Gravitational potential energy = weight x height

Now insert the parameters;

Gravitational potential energy = 800 x 5 = 4000J

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In refraction, when a wave travels from one medium to another, it

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In refraction, when a wave travels from one medium to another it changes speed.

Correct option is A.

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3 years ago

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An object on a planet has a mass of 243 kg. What is the acceleration of the

Vesna [10]

Answer:

The gravitational acceleration of a planet of mass M and radius R

a = G*M/R^2.

In this case we have:

G = 6.67 x 10^-11 N (m/kg)^2

R = 2.32 x 10^7 m

M = 6.35 x 10^30 kg

Now we can compute:

a = (6.67*6.35/2.32^2)x10^(-11 + 30 - 2*7) m/s^2 = 786,907.32 m/s^2

The acceleration does not depend on the mass of the object.

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3 years ago

What force is necessary to accelerate a 2500 kg care from rest to 20 m/s over 10 seconds?

EleoNora [17]

Force = mass x acceleration
force = 2500kg x (20m/s / 10m/s)
force = 2500kg x 2m/s^2
force = 5000kg m/s^2 = 5kN

i hope this is right (^^)

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2 years ago

A 300 g ball and a 600 g ball are connected by a 40-cm-lon massless, rigid rod. The structure rotates about its center of me at

Readme [11.4K]

Answer:

KE = 1.75 J

Explanation:

given,

mass of ball, m₁ = 300 g = 0.3 Kg

mass of ball 2, m₂ = 600 g = 0.6 Kg

length of the rod = 40 cm = 0.4 m

Angular speed = 100 rpm= $100\times \dfrac{2\pi}{60}$

=10.47\ rad/s

now, finding the position of center of mass of the system

r₁ + r₂ = 0.4 m.....(1)

equating momentum about center of mass

m₁r₁ = m₂ r₂

0.3 x r₁ = 0.6 r₂

r₁ = 2 r₂

Putting value in equation 1

2 r₂ + r₂ = 0.4

r₂ = 0.4/3

r₁ = 0.8/3

now, calculation of rotational energy

$KE = \dfrac{1}{2}I_1\omega^2+\dfrac{1}{2}I_2\omega^2$

$KE = \dfrac{1}{2}\omega^2 (I_1 +I_2)$

$KE = \dfrac{1}{2}\omega^2 (m_1r_1^2 +m_2r^2_2)$

$KE = \dfrac{1}{2}\times 10.47^2(0.3\times (0.8/3)^2 +0.6\times (0.4/3)^2)$

KE = 1.75 J

the rotational kinetic energy is equal to 1.75 J

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3 years ago

According to “The Cyclone,” who invented the Fourth of July?

Vilka [71]

A. Pecos Bill

According to “The Cyclone,” who invented the Fourth of July is PECOS BILL
NOT:
B. Alexander Hamilton
C. Sam Houston 
D. George Washington

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