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2 years ago

15

A simple watermelon launcher is designed as a spring with a light platform for the watermelon. When an 8.00 kg watermelon is put

on the launcher, the launcher spring compresses by 10.0 cm. The watermelon is then pushed down by an additional 30.0 cm and it’s ready to go. Just before the launch, how much energy is stored in the spring?

1 answer:

olga_2 [115]2 years ago

6 0

To solve this problem it is necessary to apply the concepts related to the Force from Hooke's law, the force since Newton's second law and the potential elastic energy.

Since the forces are balanced the Spring force is equal to the force of the weight that is

$F_s = F_g$

$kx = mg$

Where,

k = Spring constant

x = Displacement

m = Mass

g = Gravitational Acceleration

Re-arrange to find the spring constant

$k = \frac{mg}{x}$

$k = \frac{8*9.8}{0.1}$

$k = 784N/m$

Just before launch the compression is 40cm, then from Potential Elastic Energy definition

$PE = \frac{1}{2} kx^2$

$PE =\frac{1}{2} 784*0.4^2$

$PE = 63.72J$

Therefore the energy stored in the spring is 63.72J

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Does a compressed spring transfer elastic energy to its surroundings?

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Answer:

Yes

Explanation:

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2 years ago

1 A 75-g ball is projected from a height of 1.6 m with a horizontal velocity of 2 m/s and bounces from a 400-g smooth plate supp

Tanzania [10]

Answer with explanation:

We are given that

Mass of ball, $m_1=$ 75 g= $\frac{75}{1000}=0075kg$

1 kg=1000 g

Height, $h_1=1.6 m$

$h_2=0.6 m$

Horizontal velocity, $v_x=2 m/s$

Mass of plate $m_2=400 g=\frac{400}{1000}=0.4 kg$

a.Initial velocity of plate, $u_2=0$

Velocity before impact= $u_1=\sqrt{2gh_1}=\sqrt{2\times 9.8\times 1.6}=5.6m/s$

Where $g=9.8 m/s^2$

Velocity after impact, $v_1=\sqrt{2gh_2}=\sqrt{2\times 9.8\times 0.6}=3.4m/s$

According to law of conservation of momentum

$m_1u_1+m_2u_1=-m_1v_1+m_2v_2$

Substitute the values

$0.075\times 5.6+0=-0.075\times 3.4+0.4v_2$

$0.4v_2=0.075\times 5.6+0.075\times 3.4$

$v_2=\frac{0.075\times 5.6+0.075\times 3.4}{0.4}=1.69 m/s$

Velocity of plate=1.69 m/s

b.Initial energy= $\frac{1}{2}m_1v^2_x+m_1gh_1=\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 1.6=1.326 J$

Final energy= $\frac{1}{2}m_1v^2_x+m_1gh_2+\frac{1}{2}m_2v^2_2$

Final energy= $\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 0.6+\frac{1}{2}(0.4)(1.69)^2=1.162 J$

Energy lost due to compact=Initial energy-final energy=1.326-1.162=0.164 J

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Answer:

No

Explanation:

The reason why no current is produced are basically that, the wavelengths of light in the Balmer transition are reflected, not absorbed in solar panels, hence no current is produced.

The Balmer series consists of lines in the visible spectrum. It corresponds to emission of a photon of light when electrons descend from higher energy levels to the n=2 level in the hydrogen spectrum. The various wavelengths in the Balmer series can be separated by a prism since they are all in the visible region of the electromagnetic spectrum.

In solar panels, light corresponding to the wavelengths in the Balmer series is merely reflected by the panel and not absorbed. Since light is not absorbed, no current can be produced when the panel is irradiated with light corresponding to the wavelengths in the Balmer series.

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2 years ago

Can the velocity of a body revese the direction when acceleration is constant?

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Answer:

Yes, the velocity of the object can reverse direction when its acceleration is constant. For example consider that the velocity of any object at any time t is given as: ... At At t = 0 sec, the magnitude of velocity is 2m/s and is moving in the forward direction i.e.v (t) = -2.

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2 years ago

A 873-kg (1930-lb) dragster, starting from rest completes a 401.4-m (0.2509-mile) run in 4.945 s. If the car had a constant acce

Delvig [45]

To solve this problem it is necessary to apply the kinematic equations of motion.

By definition we know that the position of a body is given by

$x=x_0+v_0t+at^2$

Where

$x_0 =$ Initial position

$v_0 =$ Initial velocity

a = Acceleration

t= time

And the velocity can be expressed as,

$v_f = v_0 + at$

Where,

$v_f = Final velocity$

For our case we have that there is neither initial position nor initial velocity, then

$x= at^2$

With our values we have $x = 401.4m, t=4.945s$ , rearranging to find a,

$a=\frac{x}{t^2}$

$a = \frac{ 401.4}{4.945^2}$

$a = 16.41m/s^2$

Therefore the final velocity would be

$v_f = v_0 + at$

$v_f = 0 + (16.41)(4.945)$

$v_f = 81.14m/s$

Therefore the final velocity is 81.14m/s

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2 years ago