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kupik [55]
3 years ago
11

a 4kg block is attatched to a vertical sspring constant 800n/m. the spring stretches 5cm down. how much elastic potential energy

is stored in the system
Physics
1 answer:
kow [346]3 years ago
8 0

The Potential energy stored in the system is 1 J

<u>Explanation:</u>

Given-

Mass, m = 4 kg

Spring constant, k = 800 N/m

Distance, x = 5cm = 0.05m

Potential energy, U = ?

We know,

Change in potential energy is equal to the work done.

So,

U = \frac{1}{2} k (x)^2\\\\

By plugging in the values we get,

U = \frac{1}{2} * 800 * (0.05)^2\\ \\U = 400 * 0.0025\\\\U = 1J\\

Therefore, Potential energy stored in the system is 1 J

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When flying in an airplane, you are most likely in which layer of the atmosphere? mesosphere thermosphere stratosphere trosphere
Sergeeva-Olga [200]
Lower stratosphere, this is to avoid turbulence
6 0
3 years ago
Read 2 more answers
A system had 150 kj of work done on it and its internal energy increased by 60 kj. How much energy did the system gain or lose a
mina [271]

Answer:

The system loses 90 kJ of heat

Explanation:

We can answer the question by using the 1st law of thermodynamics, which states that:

\Delta U=Q-W

where

\Delta U is the change in internal energy of the system

Q is the heat absorbed by the system (positive if absorbed, negative if released by the system)

W is the work done by the system (positive if done by the system, negative if done by the surrounding on the system)

In this problem, we have:

W=-150 kJ is the work done (negative, because it is done by the surrounding on the system)

\Delta U=+60 kJ is the increase in internal energy

Using the equation above, we can find Q, the heat absorbed/released by the system:

Q=\Delta U+W=+60 kJ+(-150 kJ)=-90 kJ

And the negative sign means that the system has lost this heat.

8 0
3 years ago
a ball is thrown striaght up in the air and then falls back to earth. if the downward fall takes 2.2s, how fast is the ball trav
lapo4ka [179]

The velocity of the ball when it strikes the ground, given the data is 21.56 m/s

<h3>Data obtained from the question</h3>

From the question given above, the following data were obtained:

  • Time to reach ground from maximum height (t) = 2.2 s
  • Initial velocity (u) = 0 m/s
  • Acceleration due to gravity (g) = 9.8 m/s²
  • Final velocity (v) =?

<h3>How to determine the velocity when the ball strikes the ground</h3>

The velocity of the ball when it strikes the ground can be obtained as illustrated below:

v = u + gt

v = 0 + (9.8 × 2.2)

v = 0 + 21.56

v = 21.56 m/s

Thus, the velocity of the ball when it strikes the ground is 21.56 m/s

Learn more about motion under gravity:

brainly.com/question/22719691

#SPJ1

5 0
1 year ago
A playground carousel is free to rotate about its center on frictionless bearings, and air resistance is negligible. The carouse
Sidana [21]

Answer:

m = 35.98 Kg ≈ 36 Kg

Explanation:

I₀ = 125 kg·m²

R₁ = 1.50 m

ωi = 0.600 rad/s

R₂ = 0.905 m

ωf = 0.800 rad/s

m = ?

We can apply The law of conservation of angular momentum as follows:

Linitial = Lfinal

⇒    Ii*ωi = If*ωf   <em>(I)</em>

where    

Ii = I₀ + m*R₁² = 125 + m*(1.50)² = 125 + 2.25*m

If = I₀ + m*R₂² = 125 + m*(0.905)² = 125 + 0.819025*m

Now, we using the equation <em>(I) </em>we have

(125 + 2.25*m)*0.600 = (125 + 0.819025*m)*0.800

⇒  m = 35.98 Kg ≈ 36 Kg

5 0
3 years ago
How much does the earth/sky weigh?
Talja [164]
Calculated weight (by experimentally) of Earth is 5.972 × 10²⁴ N

Hope this helps!
6 0
2 years ago
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