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vodka [1.7K]
3 years ago
14

One end of a 7-cm-long spring is attached to the ceiling. When a 5.4 kg mass is hung from the other end, the spring is stretched

by a length of 4.3 cm. How long is the spring when a 3.3 kg mass is suspended from it in cm?
Physics
1 answer:
mash [69]3 years ago
8 0

Answer:

2.63 cm

Explanation:

Hooke's law gives that the force F is equal to cy where c is spring constant and x is extension

Making c the subject of the formula then

c=\frac {F}{y}

Since F is gm but taking the given mass to be F

c=\frac {5.4 kg}{4.3 cm}=1.2558139534883720930232558139534883720930

By substitution now considering F to be 3.3 kg

y=\frac {3.3 kg}{1.2558139534883720930232558139534883720930}=2.6277777777777 cm\approx 2.63 cm

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Answer:

at T = 0ºC the change of state is from the solid state to the gaseous state

Explanation:

In this exercise we are asked about the changes of state, from the data we will assume that the material is water.

Water can exist in three solid states, liquid and gas, in a graph of pressure ℗ against temperature (T) there is a point called triple at T = 0.01ºC, below this point the curve has two states at high pressure solid and low pressure gas.

As a result of the previous ones at T = 0ºC the change of state is from the solid state to the gaseous state

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Which of the following paraphrases Hubble Law?Select one:A. The greater the distance to a galaxy, the greater its redshift. B. T
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~A. The greater the distance to a galaxy, the greater its redshift.

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4 0
3 years ago
Horseshoe bats (genus Rhinolophus) emit sounds from their nostrils, then listen to the frequency of the sound reflected from the
Triss [41]

Answer:

Check the explanation

Explanation:

This is the step by step explanation to the above question:

v_i = v [ f_L *(v - v_b) - f_s*(v + v_b)] / [f_L * (v - v_b) + f_s*(v +v_b)]

= v * (83.1 * (v-4.3) - 80.7 ( v+4.3))/ [83.1 *(v - 4.3) + 80.7*(v + 4.3)]

v = 344 m/s

vi = 344 * ( 83.1* (344-4.3) - 80.7*(344+4.3) ) / (83.1 *(344 - 4.3) + 80.7*(344 + 4.3))

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8 0
3 years ago
. During a collision with a wall, the velocity of a 0.200-kg ball changes from 20.0 m/s toward the wall to 12.0 m/s away from th
mixer [17]

Answer:

106.7 N

Explanation:

We can solve the problem by using the impulse theorem, which states that the product between the average force applied and the duration of the collision is equal to the change in momentum of the object:

F \Delta t = m (v-u)

where

F is the average force

\Delta t is the duration of the collision

m is the mass of the ball

v is the final velocity

u is the initial velocity

In this problem:

m = 0.200 kg

u = 20.0 m/s

v = -12.0 m/s

\Delta t = 60.0 ms = 0.06 s

Solving for F,

F=\frac{m(v-u)}{\Delta t}=\frac{(0.200 kg) (-12.0 m/s-20.0 m/s)}{0.06 s}=-106.7 N

And since we are interested in the magnitude only,

F = 106.7 N

5 0
3 years ago
Read 2 more answers
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