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Anna35 [415]
3 years ago
7

A toy gun uses a spring to project a 4.5-g soft rubber sphere horizontally. The spring constant is 8.0 N/m, the barrel of the gu

n is 13 cm long, and a constant frictional force of 0.035 N exists between barrel and projectile. With what speed does the projectile leave the barrel if the spring was compressed 5.8 cm for this launch? (Assume the projectile is in contact with the barrel for the full 13 cm.)
Physics
2 answers:
marishachu [46]3 years ago
4 0

Answer:

1.93 m/s

Explanation:

Parameters given:

Mass = 4.5g = 0.0045kg

Spring constant = 8.0 N/m

Length of barrel = 13 cm = 0.013m

Frictional force = 0.035N

Compression = 5.8 cm = 0.058m

First, we find the P. E. stored in the spring:

P. E. = ½*k*x²

P. E. = ½ * 8 * 0.058² = 0.013J

Then, we find the work done by the frictional force while the sphere is leaving the barrel of the gun:

Work = Force * distance

The distance here is the length of the barrel.

Work = 0.035 * 0.13 = 0.0046 J

The kinetic energy of the sphere can now be found:

K. E. = P. E. - Work done

K. E. = 0.013 - 0.0046 = 0.0084J

We can now find the speed using the formula for K. E.:

K. E. = ½*m*v²

0.0084 = ½ * 0.0045 * v²

v² = 0.0084/0.00255 = 3.733

=> v = 1.93 m/s

marshall27 [118]3 years ago
3 0

Answer: v = 4.97 m/s

Explanation: mass of spring (m) = 4.5g = 4.5/1000 =0.0045 kg

Spring constant (k) = 8.0 N/m

Length of barrel = distance traveled when projectile was launched = 13cm = 0.13 m

Frictional force = 0.035 N

Extension (x) = 5.8cm = 0.058 m

Assuming the motion is of a constant acceleration, newton's laws of motion is valid

F - fr = ma

Where F = applied force that propelled the projectile = kx = 8×0.058 = 0.464 N

a = acceleration of object.

By substituting the parameters, we have

0.464 - 0.035 = 0.0045 a

0.429 = 0.0045 a

a = 0.429/0.0045

a = 95.3 m/s²

The bullet started from rest, hence initial velocity (u) equals zero.

To get the final velocity v, we use

v² = u² + 2aS, where S = length of barrel or distance covered by projectile.

v ² = 0² + 2(95.3)(0.13)

v² = 2(95.3)(0.13)

v² = 24.778

v = √24.778

v = 4.97 m/s

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2 years ago
A 100 N force is applied to move an object a horizontal distance of 5 meters at constant speed in 10 seconds. How much power is
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Answer:

50 W

Explanation:

<h3><u>Given :</u></h3>

  • Force applied = 100 N
  • Distance covered = 5 metres
  • Time = 10 seconds

<h3><u>To find :</u></h3>

Power

<h3><u>Solution :</u></h3>

For calculating power, we first need to know about the work done.

\bf \boxed{Work = Force \times displacement}

Now, substituting values in the above formula;

Work = 100 × 5

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\bf \boxed{Power=\dfrac{Work\:done}{Time\: taken}}

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3 years ago
A current of 9 A flows through an electric device with a resistance of 43 Ω. What must be the applied voltage in this particular
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Answer:
387 volts

Explanation:
Ohm's law is used to relate voltage, current and resistance.
The formula is as follows:V = I * R
where:
V is the applied voltage (measured in volts)
I is the current flowing (measured in amperes)
R is the resistance (measured in ohm)

In the given, we have:
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