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2 years ago

How much work does a 65 kg person climbing a 2000 m high cliff do?

Physics

1 answer:

qaws [65]2 years ago

3 0

The answer for the following question is explained below.

Therefore the work done is 130 kilo Joules.

Explanation:

Work:

A force causing the movement or displacement of an object.

Given:

mass of the person (m) = 65 kg

height of the cliff (h) = 2000 m

To calculate:

work done (W)

We know;

According to the formula:

W = m × g × h

Where;

m represents mass of the person

g represents the acceleration due to gravity

where the value of g is;

g = 10 m/ s²

h represents the height of the cliff

From the above formula;

W = 65 × 10 × 2000

W = 130,000 J

W = 130 Kilo Joules

Therefore the work done is 130 kilo Joules.

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When the metal wire in an incandescent lightbulb glows when the light is switched on and stops glowing when it is switched off, this is an example of resistance, which provides light and heat.

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2 years ago

A 0.5 kg basketball moving 5 m/s to the right collides with a 0.05 kg tennis

Natali5045456 [20]

Answer:

A. 1.4 m/s to the left

Explanation:

To solve this problem we must use the principle of conservation of momentum. Let's define the velocity signs according to the direction, if the velocity is to the right, a positive sign will be introduced into the equation, if the velocity is to the left, a negative sign will be introduced into the equation. Two moments will be analyzed in this equation. The moment before the collision and the moment after the collision. The moment before the collision is taken to the left of the equation and the moment after the collision to the right, so we have:

$M_{before} = M_{after}$

where:

M = momentum [kg*m/s]

M = m*v

where:

m = mass [kg]

v = velocity [m/s]

$(m_{1} *v_{1} )-(m_{2} *v_{2})=(m_{1} *v_{3} )+(m_{2} *v_{4})$

where:

m1 = mass of the basketball = 0.5 [kg]

v1 = velocity of the basketball before the collision = 5 [m/s]

m2 = mass of the tennis ball = 0.05 [kg]

v2 = velocity of the tennis ball before the collision = - 30 [m/s]

v3 = velocity of the basketball after the collision [m/s]

v4 = velocity of the tennis ball after the collision = 34 [m/s]

Now replacing and solving:

(0.5*5) - (0.05*30) = (0.5*v3) + (0.05*34)

1 - (0.05*34) = 0.5*v3

- 0.7 = 0.5*v

v = - 1.4 [m/s]

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2 years ago

A woman falls to the ground while wearing a parachute. The air resistance on the parachute of the parachute is 500N. If the woma

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The gravitational force on the woman is A) 500 N

Explanation:

There are two forces acting on the woman during her fall:

The force of gravity, $F_G$ , acting downward
The air resistance, $F_D$ , acting upward

According to Newton's second law, the net force acting on the woman is equal to the product between the woman's mass and her acceleration:

$\sum F=ma$

where m is the mass of the woman and a her acceleration.

The net force can be written as

$\sum F = F_G - F_D$

Also, we know that the woman falls at a constant velocity (5 m/s), this means that her acceleration is zero:

$a=0$

Combining the equations together, we get:

$F_G-F_D = 0$

which means that the magnitude of the gravitational force is equal to the magnitude of the air resistance:

$F_G=F_D=500 N$

Learn more about forces and Newton's second law:

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2 years ago

Read 2 more answers

Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103

son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

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As M=m, then

$PE = -\frac{Gm^2}{R}$

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

$U = -\frac{2Gm^2}{R}$

$dU = -\frac{2Gm^2}{R}$

$dKE = -dU$

Therefore replacing we have that,

$\frac{1}{2}mv^2 =\frac{Gm^2}{2R}$

Re-arrange to find v,

$v= \sqrt{\frac{Gm}{R}}$

$v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}$

$v = 816.7m/s$

Therefore the velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

$2r = 2*10^3m$

Therefore

$dU = Gm^2(\frac{1}{R}-\frac{1}{2r})$

$v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}$

$v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}$

$v = 1.83*10^7m/s$

Therefore the velocity when they are about to collide is $1.83*10^7m/s$

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3 years ago

If the car ha a mass of 1000 kilograms, what is its momentum (v=35m/s)

Black_prince [1.1K]

Answer:

The momentum of the car is 35,000 kg.m/s

Explanation:

Momentum

Momentum is often defined as mass in motion.

Since all objects have mass, if it's moving, then it has momentum. It can be calculated as the product of the mass by the velocity of the object:

$\vec p = m\vec v$

If only magnitudes are considered:

p = mv

The car has a mass of m=1,000 kg and travels at v=35 m/s. Calculating its momentum:

p = 1,000 kg * 35 m/s

p = 35,000 kg.m/s

The momentum of the car is 35,000 kg.m/s

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2 years ago