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Veseljchak [2.6K]
3 years ago
7

Hooke’s law states that the distance that a spring is stretched by hanging object varies directly as the mass of the object. If

the distance is 140 cm when the mass is 21 kg, what is the distance when the mass is 5 kg?
Physics
1 answer:
trasher [3.6K]3 years ago
8 0

Answer:

d_{2} = 33.33 cm

Explanation:

Given:

When mass, m_{1} =21 kg

          distance travelled is  d_{1}  = 140 cm

When mass, m_{2} =5 kg

         distance travelled is  d_{2}  = ?

Hooke's law state that within elastic limit, when an external force is applied to a body, the body gets deformed and when the force is released the gets back to its original form.

Therefore according to the question,

\frac{d_{1}}{m_{1}}=\frac{d_{2}}{m_{2}}

\frac{140}{21}=\frac{d_{2}}{5}

d_{2} = 33.33 cm

Distance travelled is 33.33 cm when mass is 5 kg.

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Refer to Concept Simulation 4.4 for background relating to this problem. The drawing shows a large cube (mass = 28.9 kg) being a
Usimov [2.4K]

Answer:

smallest magnitud is P=33.3 N

Explanation:

We are analyze the situation as an external force is applied and there is a friction force. We have a problem with Newton's second law.

          F = ma

As the two blocks go together they must have the same acceleration, so we can calculate this for the entire system

        P = (m1 + m2) a

        a = P / (m1 + m2)

In this case there is no friction force because the small block does not touch the ground.

In order to calculate the friction force, we must analyze each system component separately.

The large block on the X axis has an applied P force and as it moves feels a force from the small block.  In the Y axis has the weight (W1) and the reaction to normal (N1)

For the small block on the X axis, the force it feels is the thrust of the large block, note that this is an action and reaction force between the two blocks, it is the same definition we have of the normal one, so we can call this force (N)

Y axis it has the weight (W2) down, the force of friction (fr) that opposes the movement, so it is directed upwards. we write these equations

       N = m2 a

       fr -W2 = 0    

       fr = W2

       

The definition of friction force is

       fr = μ N

       

Let's replace and calculate

       μ (m2 a) = m2 g

       μ (P / (m1 + m2)) = g

       P = g /μ  (m1 + m2)

Let's calculate the value of this force

       P = 9.8 / 0.710 (28.9 +4.4)

       P = 13.80 (33.3)

       P = 33.3 N

This is the minimum friction force that prevents the block from sliding down

6 0
3 years ago
A wind turbine takes in energy from wind with the goal of converting it into electrical energy. Much of the wind energy is also
Shalnov [3]
German physicist  Albert Betz  (in 1919) demonstrated  that the highest efficiency you can achieve with a wind turbine is around 59%

We would have to analyze the design of an specific turbine to determine its efficiency, however it is unlikely to achieve 50% , as todays turbines have an average efficiency in the 20-35%

The answer would be around 25%

6 0
3 years ago
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It has been argued that power plants should make use of off-peak hours to generate mechanical energy and store it until it is ne
sdas [7]

Answer:

Explanation:

90 rpm = 90 / 60 rps

= 1.5 rps

= 1.5 x 2π rad /s

angular velocity of flywheel

ω = 3π rad /s

Let I be the moment of inertia of flywheel

kinetic energy = (1/2) I ω²

(1/2) I ω² = 10⁷ J

I = 2 x  10⁷ / ω²

=2 x  10⁷ / (3π)²

= 2.2538 x 10⁵ kg m²

Let radius of wheel be R

I = 1/2 M R² , M is mass of flywheel

= 1/2 πR² x t x d x R² , t is thickness , d is density of wheel .

1/2 πR⁴ x t x d = 2.2538 x 10⁵

R⁴ = 2 x 2.2538 x 10⁵ / πt d

= 4.5076 x 10⁵ / 3.14 x .1 x 7800

= 184

R= 3.683 m .

diameter = 7.366 m .

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= ω² R

= 9π² x 3.683

= 326.816 m /s²

3 0
3 years ago
PLEASE HELP
Mashutka [201]

Answer:

It cannot be constant because if it does not change and each time it increases its strength and speed.

Explanation:

4 0
3 years ago
Determine which law is appropriate for solving the following problem.
Nostrana [21]

Charles Law

Explanation:

Step 1:

It is given that the original volume of the gas is 250 ml at 300 K temperature and 1 atmosphere pressure. We need to find the volume of the same gas when the temperature is 350 K and 1 atmosphere pressure.

Step 2:

We observe that the gas pressure is the same in both the cases while the temperature is different. So we need a law that explains the volume change of a gas when temperature is changed, without any change to the pressure.

Step 3:

Charles law provides the relationship between the gas volume and temperature, at a given pressure

Step 4:

Hence we conclude that Charles law can be used.

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