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Kipish [7]
3 years ago
11

Elias stretches a rubber band and let's it go. The rubber band flies across the room. Elias says this demonstrates the transform

ation of kinetic energy to elastic potential energy. Is Elias correct? Explain.

Physics
1 answer:
notka56 [123]3 years ago
7 0

Answer:

This is not correct, actually it shows the transformation of elastic potential energy into kinetic energy.

Explanation:

We will explain this with and example and some numerical values to demostrate this energy transformation.

The elastic band has a behavior similar to a spring which is compressed and when released it transforms its energy from elastic to kinetic, we're going to choose a system which consists of a body 1 [kg], which is compressed against a spring with a constant of 20[kN/m], the maximum point where it is compressed is taken as the reference point where the energy potential is zero. When the package is released determine the maximum height the package reaches when released. The spring has been compressed 50 [mm].

In the attached image we can see a sketch of the mass and the spring, and the conditions of the block at maximum height.

The energy when we compress the spring will be:

U_{1-2} =\frac{1}{2} *k*dx\\where\\k=spring constant [kN/m]\\dx=spring change [m]

The energy when the block is in the highest position will be defined only by the potential energy with respect to the reference point. In this point the velocity is zero.

U_{1-2} =\frac{1}{2} *(20000)*(50*10^-3)\\U_{1-2}= 500[J]\\Elastic energy = potential energy\\U_{1-2}=Ep\\Ep=m*g*h\\500 = 1*9.81*h\\h= 50 [m]

We can see that the mass goes 50 [m] in the vertical direction, as a result of the transformation of elasticity energy into potential.

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A proton is initially moving west at a speed of 1.10 106 m/s in a uniform magnetic field of magnitude 0.281 T directed verticall
kifflom [539]

Answer:

so here it will move in circle with radius 4.06 cm

Explanation:

As we know that proton is moving towards west while the magnetic field is vertically upwards

So here the force on the proton must be perpendicular to the velocity

So here we have

F = q(\vec v \times \vec B)

so here we have

F = qvB sin90

since force is perpendicular to the velocity so here it must be centripetal force

here we have

\frac{mv^2}{R} = qvB

so we have

R = \frac{mv}{qB}

R = \frac{(1.66 \times 10^{-27})(1.10 \times 10^6)}{(1.6 \times 10^{-19})(0.281)}

R = 4.06 cm

so here it will move in circle with radius 4.06 cm

8 0
3 years ago
Karen has a mass of 51.9 kg as she rides the up escalator at Woodley Park Station of the Washington D.C. Metro. Karen rode a dis
kifflom [539]

Answer:

28852 J

Explanation:

When a force applied in a body produces a displacement in it, the force realized a work. The force that moves Karen is contrary to her weight and must be equal to it.

The work (W) is:

W = F.d.cos(θ), where F is the force, d is the displacement, and θ is the angle.

Knowing that cos(26°) = 0.899, and F = m*g

W = 51.9*9.8*63.1*0.899

W = 28852 J

4 0
3 years ago
How much acceleration does a car need if it is to go a distance of 620 ft. In a time of 20 sec while it is speeding up
yan [13]
31ft/second . 620/20
3 0
3 years ago
Water flows from a large drainage pipe at a rate of 950 gal/min. What is this volume rate of flow in (a) m3/s , (b) liters/min,
Paladinen [302]

Answer:

0.05997\ m^3/s

3596.1395\ L/min

2.11647\ ft^3/s

Explanation:

1\ gal/min=\dfrac{1}{264\times 60}\\\Rightarrow 950\ gal/min=950\times \dfrac{1}{264\times 60}\\\Rightarrow 950\ gal/min=0.05997\ m^3/s

Volume rate of flow is 0.05997\ m^3/s

1\ gal/min=3.78541\ L/min\\\Rightarrow 950\ gal/min=950\times 3.78541=3596.1395\ L/min

Volume rate of flow is 3596.1395\ L/min

1\ gal/min=\dfrac{1}{7.481\times 60}\\\Rightarrow 950\ gal/min=950\times \dfrac{1}{7.481\times 60}\\\Rightarrow 950\ gal/min=2.11647\ ft^3/s

Volume rate of flow is 2.11647\ ft^3/s

6 0
3 years ago
A rescue pilot drops a survival kit while her plane is flying at an altitude of 1.5 km with a forward velocity of 100.0 m/s. If
tresset_1 [31]

1750 meters.  
First, determine how long it takes for the kit to hit the ground. Distance over constant acceleration is: 
d = 1/2 A T^2
 where
 d = distance
 A = acceleration
 T = time 
 Solving for T, gives
 d = 1/2 A T^2
 2d = A T^2
 2d/A = T^2
 sqrt(2d/A) = T 
 Substitute the known values and calculate.
 sqrt(2d/A) = T
 sqrt(2* 1500m / 9.8 m/s^2) = T
 sqrt(3000m / 9.8 m/s^2) = T
 sqrt(306.122449 s^2) = T
 17.49635531 s = T 
 Rounding to 4 significant figures gives 17.50 seconds. Since it will take
17.50 seconds for the kit to hit the ground, the kit needs to be dropped 17.50
seconds before the plane goes overhead. So just simply multiply by the velocity. 
17.50 s * 100 m/s = 1750 m
3 0
3 years ago
Read 2 more answers
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