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goldfiish [28.3K]
3 years ago
7

A weightlifter curls a 30 kg bar, raising it each time a distance of 0.50 m .How many times must he repeat this exercise to burn

off the energy in one slice of pizza? Assume 25% efficiency. Energy content of one slice of pizza is 1260 kJ.
Physics
1 answer:
ludmilkaskok [199]3 years ago
4 0

Answer:

N = 2141 times

Explanation:

Each time the work done to raise a given mass is

W = mgh

here we know that

m = 30 kg

h = 0.50 m

now we have

W = (30 kg)(9.81 m/s^2)(0.50)

W = 147.15 J

since it is just 25% of actual energy consumed as we know its efficiency is 25%

so we have total energy consumed in this way

E_{total} = \frac{147.15}{0.25}

E_{total} = 588.6 J

now if it took N number of times so burn the fat of a pizza then

N(588.6) = 1260 \times 10^3

N = 2141 times

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Zarrin [17]

Answer:

Gamma radiation or Cathode rays

Explanation:

by striking incident gamma or cathode rays onto the solid when placed on a photographic plate

5 0
3 years ago
A space vehicle of mass m has a speed v. At some instant, it separates into two pieces, each of mass 0.5m. One of the pieces is
Damm [24]

Answer:

W = ½ m v²

Explanation:

In this exercise we must solve it in parts, in a first part we use the conservation of the moment to find the speed after the separation

We define the system formed by the two parts of the rocket, therefore the forces during internal separation and the moment are conserved

initial instant. before separation

        p₀ = m v

final attempt. after separation

       p_{f} = m /2  0 + m /2 v_{f}

       p₀ = p_{f}

       m v = m /2 v_{f}

       v_{f}= 2 v

this is the speed of the second part of the ship

now we can use the relation of work and energy, which establishes that the work is initial to the variation of the kinetic energy of the body

     

initial energy

         K₀ = ½ m v²

final energy

        K_{f} = ½ m/2  0 + ½ m/2 v_{f}²

        K_{f} = ¼ m (2v)²

        K_{f} = m v²

         

the expression for work is

         W = ΔK = K_{f} - K₀

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5 0
3 years ago
A 3600 kg rocket traveling at 2900 m/s is moving freely through space on a journey to the moon. The ground controllers find that
Nana76 [90]

Answer:

m=417.24 kg

Explanation:  

Given Data

Initial mass of rocket  M = 3600 Kg

Initial velocity of rocket vi = 2900 m/s  

velocity of gas vg = 4300  m/s

Θ = 11° angle in degrees

To find

m = mass of gas  

Solution

Let m = mass of gas    

first to find Initial speed with angle given

So

Vi=vi×tanΘ...............tan angle

Vi= 2900m/s × tan (11°)

Vi=563.7 m/s

Now to find mass

m = (M ×vi ×tanΘ)/( vg + vi tanΘ)

put the values as we have already solve vi ×tanΘ

m = (3600 kg ×563.7m/s)/(4300 m/s + 563.7 m/s)

m=417.24 kg

7 0
3 years ago
The drum of a washing machine spins at 200 rpm the drum had a radius of 0.5m what will the centripetal force experienced by a to
miskamm [114]

The centripetal force experienced by the towel is 55 N.

The given parameters;

  • angular speed of the washing machine, ω = 200 rpm
  • radius of the machine' drum, r = 0.5 m
  • mass of the towel, m = 0.25 kg

The centripetal force experienced by the towel spinning along the walls of the drum is calculated as follows;

Fc = mrω²

where;

<em>Fc is the centripetal force</em>

<em>ω is angular speed in rad/s</em>

The angular speed in rad/s is calculate as;

\omega = \frac{200 \ rev}{\min} \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1\min}{60 \ s} \\\\\omega = 20.95 \ rad/s

The centripetal force experienced by the towel is calculated as;

F_c = mr\omega ^2\\\\F_c = 0.25 \times 0.5 \times (20.95)^2\\\\F_c = 54.9 \ N \ \approx 55 \ N

Thus, the centripetal force experienced by the towel is 55 N.

Learn more here: brainly.com/question/20905151

5 0
3 years ago
Suppose that on earth you can throw a ball vertically upward a distance of 1.20 m. Given that the acceleration of gravity on the
tatuchka [14]

Answer:

7.04 m

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u = Initial velocity

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a = Acceleration due to gravity Earth= 9.81 m/s²

Accelration going up is considered as negetive

Initial Velocity of the ball

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Assuming that the ball is thrown with the same velocity on the Moon, displacement of the ball is

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-4.85^2}{2\times -1.67}\\\Rightarrow s=7.04\ m

The displacement of the ball on the moon is 7.04 m

6 0
3 years ago
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