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2 years ago

8

A rubber band has potential energy of 5 J. If the spring constant of the rubber band is 50 N/m, what is the displacement of the

rubber band?

2 answers:

Valentin [98]2 years ago

7 0

To determine the displacement, since we are given the potential energy, we use the equation for potential energy. For a spring, it is one-half the product of the spring constant and the square of the displacement. We do as follows:

PE = kx^2/2
5 Nm = 50N/m (x^2)
x = 0.32 m

Therefore, the displacement would be 0.32 m.

Allisa [31]2 years ago

7 0

The correct answer is 0.4 m on Edge.

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In the middle of the night you are standing a horizontal distance of 14.0 m from the high fence that surrounds the estate of you

olchik [2.2K]

PART A)

horizontal distance that will be moved = 14 m

Height of the fence = 5.0 m

height from which it is thrown = 1.60 m

angle of projection = 54 degree

So here we can say that stone will travel vertically up by distance

$\Delta y = 5 - 1.6 = 3.40 m$

now we will have displacement in horizontal direction

$\Delta x = 14 m$

now we know that

$v_x = vcos54$

$v_y = vsin54$

now we will have

$\Delta x = v_x t$

$14 = (vcos54)t$

also for y direction

$\Delta y = v_y t + \frac{1}{2}at^2$

$3.40 = (vsin54)t - \frac{1}{2}(9.8) t^2$

now from the two equations we will have

$3.40 = (vsin54)(\frac{14}{vcos54}) - 4.9 t^2$

$3.40 = 14 tan54 - 4.9 t^2$

$3.40 = 19.3 - 4.9 t^2$

$t = 1.8 s$

now from above equations

$14 = vcos54 (1.8)$

$v = 13.2 m/s$

So the minimum speed will be 13.2 m/s

Part B)

Total time of the motion after which it will land on the ground will be "t"

so its vertical displacement will be

$\Delta y = -1.60 m$

now we will have

$-1.60 = v_y t + \frac{1}{2}at^2$

$-1.60 = (13.2sin54)t - \frac{1}{2}(9.8)t^2$

$4.9 t^2 - 10.7t - 1.60 = 0$

$t = 2.3 s$

Now the time after which it will reach the fence will be t1 = 1.8 s

so total time after which it will fall on other side of fence

$t_2 = t - t_1$

$t_2 = 2.3 - 1.8 = 0.5 s$

now the displacement on the other side is given as

$\Delta x = (vcos54) t_2$

$\Delta x = (13.2 cos54)(0.5)$

$\Delta x = 3.88 m$

4 0

2 years ago

A large ebony wood log, totally submerged, is rapidly floating down a flooded river. If the mass of the log is 165 kg, what is t

Sedbober [7]

Answer:

F = 1618.65[N]

Explanation:

To solve this problem we use the following equation that relates the mass, density and volume of the body to the floating force.

We know that the density of wood is equal to 750 [kg/m^3]

density = m / V

where:

m = mass = 165[kg]

V = volume [m^3]

V = m / density

V = 165 / 750

V = 0.22 [m^3]

The floating force is equal to:

F = density * g * V

F = 750*9.81*0.22

F = 1618.65[N]

3 0

2 years ago

If the cold temperature reservoir of a Carnot engine is held at a constant 306 K, what temperature should the hot reservoir be k

Paraphin [41]

Efficiency η of a Carnot engine is defined to be:
<span>η = 1 - Tc / Th = (Th - Tc) / Th </span>
<span>where </span>
<span>Tc is the absolute temperature of the cold reservoir, and </span>
<span>Th is the absolute temperature of the hot reservoir. </span>

<span>In this case, given is η=22% and Th - Tc = 75K </span>
<span>Notice that although temperature difference is given in °C it has same numerical value in Kelvins because magnitude of the degree Celsius is exactly equal to that of the Kelvin (the difference between two scales is only in their starting points). </span>

<span>Th = (Th - Tc) / η </span>
<span>Th = 75 / 0.22 = 341 K (rounded to closest number) </span>
<span>Tc = Th - 75 = 266 K </span>

<span>Lower temperature is Tc = 266 K </span>
<span>Higher temperature is Th = 341 K</span>

6 0

2 years ago

In an experiment you are performing, your lab partner has measured the distance a cart has traveled: 28.4inch. You need the dist

alexgriva [62]

As per the question the distance travelled by a car is 28.4 inch.

we are asked to determine the conversion factor in centimeter which when multiplied with 28.4 inch will give a unit.

we know that one inch =2.54 centimeter.

Hence 28.4 inch = 2.54 ×28.4 cm

=72.136 cm.

Now we have to determine the conversion factor .The multiplication factor is calculated as $P =\frac{28.4 inch*2.54cm}{1 inch}$

$p= 72.136 cm$ [p is the multiplication factor.]

Hence the multiplication factor is 72.137 cm which will give unit conversion when multiplied with 28.4 inch.

4 0

2 years ago

Read 2 more answers

A 0.28-kg stone you throw rises 34.3 m in the air. The magnitude of the impulse the stone received from your hand while being th

goldfiish [28.3K]

Answer:

7.3 kg m/s

Explanation:

First of all, let's calculate the gravitational potential energy of the stone as it reaches its highest point:

$U=mgh=(0.28 kg)(9.8 m/s^2)(34.3 m)=94.1 J$

For the law of conservation of energy, this is equal to the initial kinetic energy of the stone at ground level (where the potential energy is zero), just after the stone leaves your hand:

$K=\frac{1}{2}mv^2=94.1 J$

From this equation we can find the velocity of the stone as it leaves your hand:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(94.1 J)}{0.28 kg}}=25.9 m/s$

The initial velocity of the stone (before leaving your hand) is zero:

$u=0$

The impulse received by the stone is equal to its change in momentum, so:

$I=\Delta p=m\Delta v=m(v-u)=(0.28 kg)(25.9 m/s-0)=7.3 kg m/s$

5 0

2 years ago