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A. The potential energy stored in the compressed spring of a dart gun, with a spring constant of 62.00 N/m, is 0.940 J. Find by

valina [46]

a. The spring is compressed by 0.174 m.

b. The vertical distance the dart travels from its position when the spring is compressed to its highest position is 9.6 m.

c. The horizontal velocity of the dart at that time is 13.74 m/s.

d. The horizontal distance from the equilibrium position at which the dart hits the ground is 19.236 m.

<h3>What is potential energy?</h3>

The energy by virtue of its position is called the potential energy.

a. Given is the potential energy stored in the compressed spring of a dart gun, with a spring constant of 62.00 N/m, is 0.940 J.

PE of spring = 1/2 kx²

Put the values, we get

P.E = 0.940 = 1/2 x 62 x²

x = 0.174 m

Thus, the spring is compressed by 0.174 m.

b. Given is a 0.010 kg dart is fired straight up.

The vertical height is find out by

0.940 J = (0.010 kg) (9.8 m/s²) h

h = 9.6 m

Thus, the vertical distance the dart travels from its position is 9.6 m

c. From the conservation of energy principle, total mechanical energy is conserved.

1/2 mv² =mgh

v = √2gh

Plug the values, we get

v = √2 x 9.8x 9.6

v = 13.74 m/s

Thus, the horizontal velocity is 13.74 m/s.

d. Time that dart spends in air, t = √2h/g

t = √(2x9.6)/9.81

t = 1.4 s

The horizontal distance from the equilibrium position at which the dart hits the ground.

Horizontal distance = (Velocity on x direction) x time

Horizontal distance = 13.74 m/s x 1.4s

Horizontal distance = 19.236 m

Thus, the horizontal distance is 19.236 m.

Learn more about potential energy.

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7 0

2 years ago

A sound wave has a frequency of 265 hz and a wavelength measured at 1.3

pochemuha

Answer:

V = f λ speed of wave in terms of frequency and wavelength

t = S / V time for wave to travel a distance S

t = 91.4 m / 344.5 m/s = .265 sec time to travel 91.4 m

8 0

2 years ago

A 0.750 kg block is attached to a spring with spring constant 17.5 N/m. While the block is sitting at rest, a student hits it wi

Dmitriy789 [7]

Answer:

a

$A = 0.081 \ m$

b

The value is $u = 0.2569 \ m/s$

Explanation:

From the question we are told that

The mass is $m = 0.750 \ kg$

The spring constant is $k = 17.5 \ N/m$

The instantaneous speed is $v = 39.0 \ cm/s= 0.39 \ m/s$

The position consider is x = 0.750A meters from equilibrium point

Generally from the law of energy conservation we have that

The kinetic energy induced by the hammer = The energy stored in the spring

So

$\frac{1}{2} * m * v^2 = \frac{1}{2} * k * A^2$

Here a is the amplitude of the subsequent oscillations

=> $A = \sqrt{\frac{m * v^ 2 }{ k} }$

=> $A = \sqrt{\frac{0.750 * 0.39 ^ 2 }{17.5} }$

=> $A = 0.081 \ m$

Generally from the law of energy conservation we have that

The kinetic energy by the hammer = The energy stored in the spring at the point considered + The kinetic energy at the considered point

$\frac{1}{2} * m * v^2 = \frac{1}{2} * k x^2 + \frac{1}{2} * m * u^2$

=> $\frac{1}{2} * 0.750 * 0.39^2 = \frac{1}{2} * 17.5* 0.750(0.081 )^2 + \frac{1}{2} * 0.750 * u^2$

=> $u = 0.2569 \ m/s$

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2 years ago

What is the compound oc2 called?

MAXImum [283]

The answer to your question is dioxygen carbide

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How do you think cotton, tobbaco, and other crops were shipped to their destination

Effectus [21]

I think it was by boat and coming from over seas

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