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Tom [10]
3 years ago
8

A(n) 131 g ball is dropped from a height

Physics
1 answer:
larisa [96]3 years ago
7 0

Answer:

26.59 N/m

Explanation:

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

Mass (m) = 131 g

Extention (e) = 4.82755 cm

Acceleration due to gravity (g) = 9.8 m/s²

Spring constant (K) =?

Next, we shall convert 131 g to Kg. This can be obtained as follow:

1000 g = 1 Kg

Therefore,

131 g = 131 g × 1 Kg / 1000 g

131 g = 0.131 Kg

Thus, 131 g is equivalent to 0.131 Kg.

Next, we shall the force exerted by the ball on the spring. This can be obtained as follow:

Mass (m) = 0.131 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Force (F) =?

F = ma

F = 0.131 × 9.8

F = 1.2838 N

Next, we shall convert 4.82755 cm to metre (m)

This can be obtained as follow:

100 cm = 1 m

Therefore,

4.82755 cm = 4.82755 cm × 1 m / 100 cm

4.82755 cm = 0.0482755 m

Thus, 4.82755 cm is equivalent to 0.0482755 m

Finally, we shall determine the spring constant as follow:

Force (F) = 1.2838 N

Extention (e) = 0.0482755 m

Spring constant (K) =?

F = Ke

1.2838 = K × 0.0482755

Divide both side by 0.0482755

K = 1.2838 / 0.0482755

K = 26.59 N/m

Thus the spring constant is 26.59 N/m

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lapo4ka [179]

The equations to find the acceleration are the suvat equations:

v=u+at\\s=ut+\frac{1}{2}at^2\\s=vt-\frac{1}{2}at^2\\v^2-u^2=2as\\s=(\frac{u+v}{2})t

Explanation:

The acceleration of an object is the rate if change in velocity of the object. It is calculated as

a=\frac{v-u}{t}

where

v is the final velocity of the object

u is the initial velocity

t is the time elapsed

For an object moving in a straight line at constant acceleration, there are several equations that can be used to find the acceleration: they are called suvat equations. They are the following:

v=u+at\\s=ut+\frac{1}{2}at^2\\s=vt-\frac{1}{2}at^2\\v^2-u^2=2as\\s=(\frac{u+v}{2})t

where

u is the initial velocity

v is the final velocity

t is the time

s is the distance covered

a is the acceleration

Therefore, any of the above equations can be used to  calculate the acceleration.

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5 0
3 years ago
When a crown of mass 14.7 kg is submerged in water, an accurate scale reads only 13.4 kg. What material is the crown made of?
Bogdan [553]

Answer:

  ρ = 1.13 10⁴  km/m³

Explanation:

For this exercise we use Newton's equilibrium equation

        B –W + W_scale = 0

Where B is the thrust and W_scale is the balance reading

The push is given by Archimedes' law

        B = ρ_water g V

        B = W- W_scale

        B = m g - m_scale g  

       

Let's calculate

         B = 14.7 9.8 - 13.4 9.8

         B = 12.74 N

         ρ_water g V = 12.74

         V = 12.74 / ρ_water g

         V = 12.74 / 1000 9.8

         V = 0.0013 m³

Let's use density

          ρ = m / V

           

We replace

         ρ = 14.7 / 0.0013

         ρ = 1.13 10⁴  km/m³

3 0
3 years ago
(a) If the maximum acceleration that is tolerable for passengers in a subway train is 1.34 m/s² and subway stations are located
andreev551 [17]

Answer:

The correct answer is 32.9 m/s

Explanation:

To solve this, we list out the known and the unknown variables as follows

Maximum allowable acceleration = 1.34 m/s²

Distance between sttions = 806 m

Therefore from the equation of motion

v = ut + 0.5·×at²

Where v = final velocity

u = initial velocity

S = distance covered

t  = time

a = acceleration

Also v² = u² + 2·a·S

where u is the initial velocity, which we can take as u = 0, then

v² = 2·1.34·S = 2.68S m²/s²  then

Also the train has to decelerate from maximum speed to stop at the next tran station wherev = 0, thus v² = u² -2·1.34·Z,  so u² = 2.68Z

since u² = 2.68S from the previous calculation, then for v = 0

2.68S = 2.68Z thus S = Z which and to reach the next subway station S + Z must be = 806 m, then S = 806 m ÷ 2 = 403 m

and v² = 2.68S m²/s² = 1080.04 m²/s²

v = 32.9 m/s

The maximum speed a subway train can attain between stations is 32.9 m/s

3 0
3 years ago
A basket has a mass of 5.5 kg. Find the magnitude of the normal force if the basket is at rest on a ramp inclined above the hori
storchak [24]
Here is the correct answer of the given problem above.
Given that the basket has a mass of 5.5kg, the magnitude of the normal force if the basket is at rest on a ramp inclined above the horizontal is at 12 degrees. The solution is simple: 
<span>Fn at rest = lmgl </span>
<span>= 5.5kg (9.80N/kg) 
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8 0
4 years ago
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A spring has a spring constant of 20 N/m. How much potential energy is stored in the spring as it stretched 0.20 m
makkiz [27]
The elastic potential energy (Ep) is given by Ep = \frac{1}{2}*k*x^2

Data: 
Ep = ? (Joule)
k = 20 N/m
x (displacement) = 0.20 m

Solving:
Ep = \frac{1}{2}*k*x^2
Ep = \frac{1}{2}*20*0.20^2
Ep =  \frac{20*0.04}{2}
Ep =  \frac{0.8}{2}
\boxed{\boxed{Ep = 0.4\:J}}\end{array}}\qquad\quad\checkmark
7 0
3 years ago
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