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yarga [219]
3 years ago
12

The massless spring of a spring gun has a force constant k=12~\text{N/cm}k=12 N/cm. When the gun is aimed vertically, a 15-g pro

jectile is shot to a height of 5.0 m above the end of the expanded spring. (See below.) How much was the spring compressed initially?
Physics
1 answer:
ASHA 777 [7]3 years ago
4 0

Answer:

0.011 m.

Explanation:

Energy stored in the spring = Energy of the projectile.

1/2ke² = mgh ................ Equation 1

Where k = spring constant, e = extension or compression, m = mass of the projectile, g = acceleration due to gravity, h = height.

make e the subject of the equation

e = √(2mgh/k)............................. Equation 2

Given: k = 12 N/cm = 1200 N/m, m = 15 g = 0.015 kg, h = 5.0 m

Constant: g = 9.8 m/s²

Substitute into equation 2

e = √(2×0.015×5/1200)

e = √(0.15/1200)

e = √(0.000125)

e = 0.011 m.

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Both substance will melt and become a solution.

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5 0
3 years ago
Determine the potential difference between two charged parallel plates that are 0.50 cm apart and have an electric field strengt
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3 years ago
A diver stands on a diving platform 10.0 m above the surface of a pool and leaps upward with an initial speed of 2.5 m/s. how fa
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7 0
3 years ago
How many electrons does it take to make up 4.33 C of charge?
Dovator [93]

Answer:

Number of electrons, n=2.7\times 10^{19}

Explanation:

It is given that,

Charge, q = 4.33 C

We need to find the number of electrons that make 4.33 C of charge. According to quantization of charge as :

q=ne

n = number of electrons

e = electron's charge

n=\dfrac{q}{e}

n=\dfrac{4.33\ C}{1.6\times 10^{-19}\ C}

n=2.7\times 10^{19}

So, the number of electrons are 2.7\times 10^{19} Hence, this is the required solution.  

7 0
3 years ago
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sp2606 [1]

Power = \frac{Work}{Time}

Delilah: 170J/30s = 5.66 W

Adam: 260J/20s = 13 W

6 0
3 years ago
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