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brilliants [131]
2 years ago
12

A spring-loaded toy gun is used to shoot a ball straight up inthe air. The ball reaches a maximum height H, measuredfrom the equ

ilibrium position of the spring. The same ball is shot straight up a second time from the same gun, but this time the spring is compressed only half as far before firing. How far up does the ball go this time?
Physics
1 answer:
rewona [7]2 years ago
6 0

Answer:

H' = H/4

Explanation:

By applying the law of conservation of energy to this problem, we know that:

Elastic Potential Energy Stored by Spring = Gravitational Potential Energy of ball

(1/2)kx² = mgH

H = (1/2)kx²/mg   -------------- equation (1)

where,

H = Height reached by the ball

x = compression of spring

k = stiffness of spring

m = mass of ball

g = acceleration due to gravity

Now, if we make the compression to half of its value:

x' = x/2

then:

H' = (1/2)k(x/2)²/mg

H' = (1/4)(1/2)kx²/mg

using equation (1), we get:

<u>H' = H/4</u>

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h

Explanation:

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3 years ago
A laser used to dazzle the audience at a rock concert emits blue light with a wavelength of 463 nm . calculate the frequency of
ZanzabumX [31]

The wavelength of light is given as 463 nm or can also be written as 463 x 10^-9 m. [wavelength = ʎ]

We know that the speed of light is 299 792 458 m / s or approximately 3 x 10^8 m / s. [speed of light = c]

 

Given the two values, we can calculate for the frequence (f) using the formula:

f = c / ʎ

 

Substituting the given values:

f = (3 x 10^8 m / s) / 463 x 10^-9 m

f = 6.48 x 10^14 / s = 6.48 x 10^14 s^-1

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3 years ago
You are trying to decide between two new stereo amplifiers. One is rated at 130 W per channel and the other is rated at 200 W pe
NARA [144]

Answer:

The sound level will be 1.870 dB louder.

Explanation:

Given that,

Power = 130 W

Power = 200 W

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I_{dB}=10\log(\dfrac{I}{I_{0}})

For one amplifier,

I_{1}=10\log(\dfrac{130}{I_{0}})...(I)

For other amplifier,

I_{2}=10\log(\dfrac{200}{I_{0}})...(II)

For difference in dB levels

I_{2}-I_{1}=10\log(\dfrac{200}{I_{0}})-10\log(\dfrac{130}{I_{0}})

I_{2}-I_{1}=10\log(\dfrac{200}{I_{0}}\times\dfrac{I_{0}}{130})

I_{2}-I_{1}=10\log(\dfrac{200}{130})

I_{2}-I_{1}=1.870\ dB

Hence, The sound level will be 1.870 dB louder.

7 0
3 years ago
What happen to the salt and the sand when water is added and the mixtures is stirred?
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3 years ago
A horse has a momentum of 25 kg*m/s and a velocity of 2.5 m/s. What is the horse mass?
Ne4ueva [31]
<h3>Answer:  10 kg</h3>

==================================================

Work Shown:

We have these variables

  • p = momentum = 25 kg*m/s
  • m = mass = unknown
  • v = velocity = 2.5 meters per second

Solving for the mass gets us...

p = m*v

25 = m*2.5

25/2.5 = m

10 = m

m = 10

The mass of the horse is 10 kg.

6 0
2 years ago
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