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sveta [45]
4 years ago
11

A spring with a mass on the end of it hangs in equilibrium a distance of 0.4200 m above the floor. The mass is pulled down a dis

tance 0.0600 m below the original position, released, and allowed to oscillate. How high above the floor is the mass at the highest point in its oscillation
Physics
1 answer:
den301095 [7]4 years ago
3 0

Answer:

0.48 m

Explanation:

I'm assuming that this takes place in an ideal situation, where we neglect a host of factors such as friction, weight of the spring and others

If the mass is hanging from equilibrium at 0.42 m above the floor, from the question, and it is then pulled 0.06 m below that particular position. This pulling is a means of adding more energy into the spring, when it is released, the weight compresses the spring and equals its distance (i.e, 0.06 m) above the height.

0.42 m + 0.06 m = 0.48 m

At the highest point thus, the height is 0.48 m above the ground.

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What unit is used for the value of G in Newton's Law of Universal<br> Gravitation?<br> *
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Answer:

In SI units, its value is approximately 6.674×10−11 m3⋅kg−1⋅s−2. The modern notation of Newton's law involving G was introduced in the 1890s by C. V. Boys. The first implicit measurement with an accuracy within about 1% is attributed to Henry Cavendish in a 1798 experiment.

Explanation:

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3 0
3 years ago
how does the composition of a mixture of hydrogen and oxygen differ of hydrogen and oxygen differ from the composition of a comp
mafiozo [28]
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3 years ago
11. Which of the following phenomena is taking place when sound waves are reflected from a surface along parallel lines? A. Refr
allsm [11]
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4 0
3 years ago
Energy Conservation With Conservative Forces: If a spring-operated gun can shoot a pellet to a maximum height of 100 m on Earth,
crimeas [40]

Answer:

h' = 603.08 m

Explanation:

First, we will calculate the initial velocity of the pellet on the surface of Earth by using third equation of motion:

2gh = Vf² - Vi²

where,

g = acceleration due to gravity on the surface of earth = - 9.8 m/s² (negative sign due to upward motion)

h = height of pellet = 100 m

Vf = final velocity of pellet = 0 m/s (since, pellet will momentarily stop at highest point)

Vi = Initial Velocity of Pellet = ?

Therefore,

(2)(-9.8 m/s²)(100 m) = (0 m/s)² - Vi²

Vi = √(1960 m²/s²)

Vi = 44.27 m/s

Now, we use this equation at the surface of moon with same initial velocity:

2g'h' = Vf² - Vi²

where,

g' = acceleration due to gravity on the surface of moon = 1.625 m/s²

h' = maximum height gained by pellet on moon = ?

Therefore,

2(1.625 m/s²)h' = (44.27 m/s)² - (0 m/s)²

h' = (1960 m²/s²)/(3.25 m/s²)

<u>h' = 603.08 m</u>

4 0
3 years ago
Alice and Tom dive from an overhang into the lake below. Tom simply drops straight down from the edge, but Alice takes a running
crimeas [40]

Answer:c

Explanation:

Given

Alice launches with horizontal velocity u=25\ m/s

Tom simply drops straight down from the edge

Time taken by both the person is same as they have same initial vertical velocity i.e. zero so the time taken to reach the ground is zero.

Although Alice will travel more horizontal distance compared to Tom.

Thus option c is correct

3 0
4 years ago
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