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4 years ago

calculate the spring constant if a weight of 250N is added to a spring which increases in length by 20cm

Physics

1 answer:

ZanzabumX [31]4 years ago

8 0

Since, F = k . ∆x

Therefore, k = F / ∆x = 250 / 0.2 = 1250 N/m

(ps: convert 20 cm into 0.2 m)

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A train starts from rest and travels for 5.0 s with a uniform acceleration of 1.5 m/s2. What is the final velocity of the train?

alexandr1967 [171]

Answer:

Final speed of the train is 7.5 m/s

Explanation:

It is given that,

Uniform acceleration of the train is, a = 1.5 m/s²

It starts from rest and travels for 5.0 s. We have to find the final velocity of the train. By using first equation of motion as :

$v=u+at$

Here, train starts from rest so, u = 0

$v=0+1.5\ m/s^2\times 5\ s$

v = 7.5 m/s

So, the final velocity of the train is 7.5 m/s. Hence, this is the required solution.

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3 years ago

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A 2.50 kg ball moving at 7.50 m/s is caught by a 70.0 kg man while the man is standing on ice. What is the common velocity of th

yarga [219]

Answer:

V = 2.5*7.0 / ( 2.5 + 70 )

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3 years ago

A pipe of length 10.0 m increases in length by 1.5 cm when its temperature is increased by 90°F. What is its coefficient of line

azamat

The coefficient of linear expansion, given that the length of the pipe increased by 1.5 cm is 1.67×10¯⁵ /°F

<h3>How to determine the coefficient of linear expansion</h3>

From the question given above, the following data were obtained

Original diameter (L₁) = 10 m
Change in length (∆L) = 1.5 cm = 1.5 / 100 = 0.015 m
Change in temperature (∆T) = 90 °F
Coefficient of linear expansion (α) =?

The coefficient of linear expansion can be obtained as illustrated below:

α = ∆L / L₁∆T

α = 0.015 / (10 × 90)

α = 0.015 / 900

α = 1.67×10¯⁵ /°F

Thus, we can conclude that the coefficient of linear expansion is 1.67×10¯⁵ /°F

Learn more about coefficient of linear expansion:

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1 year ago

An ordinary egg can be approximated as a 5.5-cm diameter sphere. The egg is initially at a uniform temperature of 8°C and is dro

kupik [55]

Answer:

a) $Q_{in} = 13.742\,kW$ , b) $\Delta S = 370.15\,\frac{kJ}{K}$

Explanation:

a) The heat transfered to the egg is computed by the First Law of Thermodynamics:

$Q_{in} +U_{sys,1} - U_{sys,2} = 0$

$Q_{in} = U_{sys,2} - U_{sys,1}$

$Q_{in} = \rho_{egg}\cdot \left(\frac{4\pi}{3}\cdot r^{3}\right)\cdot c \cdot (T_{2}-T_{1})$

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$Q_{in} = 13.742\,kW$

b) The amount of entropy generation is determined by the Second Law of Thermodynamics:

$\Delta S = \frac{Q_{in}}{T_{in}}$

$\Delta S = \frac{13.742\,kJ}{370.15\,K}$

$\Delta S = 370.15\,\frac{kJ}{K}$

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4 years ago

At what point does the external energy enter the system?

Phoenix [80]

The correct answer as the first one above !

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4 years ago