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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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And the object is moving toward the origin as the distance of the object motion is found to decrease with increase of time as per the graph. But the slope of the graph is found to be almost constant, this indicates that the velocity of the object is constant. Thus, the object represented by this graph is moving toward the origin at constant velocity.

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The formation of volcanic arcs, trenches and subsidence leads to the production of volcanic eruptions.

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A train is approaching a town at a constant speed of 12 m/s. The town is 1.0 km distant. After 30 seconds, the conductor applies
Yuliya22 [10]

To solve this problem we will apply the linear motion kinematic equations. To determine the position in which the braking starts we will start from the definition of distance as a function of speed and time, that is

x = x_0 - vt

Here,

x_0 = Initial position

v = Velocity

t = time

Replacing we have that

x = 1000-12*30

x = 640m

Now the acceleration is given by the function,

v_f^2=v_0^2 +ax

Here,

v_f= Final velocity

v_0= Initial velocity

a = Acceleration

x = Displacement

Replacing we have that

0 = 12^2+2a(640)

a = -0.1125m/s^2

Therefore the acceleration necessary to bring the train to rest is -0.1125m/s^2

4 0
3 years ago
A spring (70 N/m ) has an equilibrium length of 1.00 m. The spring is compressed to a length of 0.50 m and a mass of 2.0 kg is p
madreJ [45]

Answer:

Part a)

L = 0.68 m

Part b)

L = 0.63 m

Explanation:

Part a)

As we know that there is no friction in the path

So here we can use energy conservation to find the distance moved by the mass

Initial spring energy = final gravitational potential energy

so we will have

\frac{1}{2}kx^2 = mgL sin\theta

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8.75 = 12.87 L

L = 0.68 m

Part b)

Now if spring is connected to the block then again we can use energy conservation

so we will have

\frac{1}{2}kx^2 = mg(x + x')sin\theta + \frac{1}{2}kx'^2

so we will have

\frac{1}{2}(70)(0.5^2) = 2(9.81) (0.50 + x') sin41 + \frac{1}{2}(70)x'^2

8.75 = 6.43 + 12.87 x' + 35 x'^2

x' = 0.13 m

so total distance moved upwards is

L = 0.5 + 0.13 = 0.63 m

8 0
3 years ago
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