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3 years ago

A block of mass 2.5kg is pressed against a spring and compresses the spring to a length of

Physics

1 answer:

IrinaK [193]3 years ago

6 0

Answer:

10.032 N.

Explanation:

From the question given above, the following data were obtainedb

Mass (m) = 2.5 Kg

Final length = 10.0 cm

Original length = 21.4 cm

Spring constant (K) = 88 N/m

Force (N) =?

Next, we shall determine the compression of the spring. This can be obtained as follow:

Final length = 10.0 cm

Original length = 21.4 cm

Compression (e) =?

e = Original length – final length

e = 21.4 – 10

e = 11.4 cm

Next, we shall convert 11.4 cm to m. This can be obtained as follow:

100 cm = 1 m

Therefore,

11.4cm = 11.4 cm × 1 m / 100 cm

10 cm = 0.114 m

Finally, we shall determine the force. This can be obtained as illustrated below:

Compression (e) = 0.114 m

Spring constant (K) = 88 N/m

Force (N) =?

F = Ke

F = 88 × 0.114

F = 10.032 N

Thus, the force in the spring is 10.032 N

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Now, we calculate the wavelength for hydrogen:

$\frac{1}{\lambda}=R_H(\frac{1}{2^2}-\frac{1}{3^2})\\\lambda=[R_H(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=[1.0967*10^7m^{-1}(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=6.5646*10^{-7}m=656.46nm$

For deuterium, we have $M=2(1.67*10^{-27}kg)$ :

$R_D=\frac{1.09737*10^7m^{-1}}{(1+\frac{9.11*10^{-31}kg}{2*1.67*10^{-27}kg})}\\R_D=1.09707*10^7m^{-1}\\\\\lambda=[R_D(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=[1.09707*10^7m^{-1}(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=6.5629*10^{-7}=656.29nm$

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