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galben [10]
2 years ago
15

The end point of a spring oscillates with a period of 2.0 s when a block with mass m is attached to it. When this mass is increa

sed by 2.0 kg, the period is found to be 3.0 s. Find m.
Physics
1 answer:
Nostrana [21]2 years ago
3 0

Answer: The end point of a spring oscillates with a period of 2.0 s when a block with mass m is attached to it. When this mass is increased by 2.0 kg, the period is found to be 3.0 s.  Then the mass m is 0.625kg.

Explanation: To find the answer, we need to know more about the simple harmonic motion.

<h3>What is simple harmonic motion?</h3>
  • A particle is said to execute SHM, if it moves to and fro about the mean position under the action of restoring force.
  • We have the equation of time period of a SHM as,

                                          T=2\pi \sqrt{\frac{m}{k} }

  • Where, m is the mass of the body and k is the spring constant.
<h3>How to solve the problem?</h3>
  • Given that,

               T_1=2s\\m_1=m\\m_2=m+2kg\\T_2=3s

  • We have to find the value of m,

              T_1=2\pi \sqrt{\frac{m}{k} } \\T_2=2\pi \sqrt{\frac{m+2}{k} } \\\frac{T_1}{T_2} =\sqrt{\frac{m}{m+2} }\\\frac{2}{3} =\sqrt{\frac{m}{m+2} }\\\\

               m=\frac{5}{8} =0.625kg

Thus, we can conclude that, the mass m will be 0.625kg.

Learn more about simple harmonic motion here:

brainly.com/question/28045110

#SPJ4

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Explanation:

time=Distance/speed

t=32/8

t=4 seconds

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Which is an si metric unit of measurement that is used to record the heat transfer of a solution in a classroom investigation?
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Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the required temperature increase to close the gap at C.
Leni [432]

Answer:

T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}

Explanation:

The given data :-

  • Diameter of rod AB ( d₁ ) = 200 mm.
  • Diameter of rod BC ( d₂ ) = 150 mm.
  • The linear co-efficient of thermal expansion of copper ( ∝ ) = 1.6 × 10⁻⁶ /°C
  • The young's modulus of elasticity of copper ( E ) = 120 GPa = 120 × 10³ MPa.
  • Consider the required temperature increase to close the gap at C = T °C
  • Consider the change in length of the rod = бL

Solution :-

\begin{aligned}\sum H& =0\\-R_A+R_C&=0\\R_A&=R_C\\R_A&=R\\R_C&=R\\R_{A}&=\text{reaction\:force\:at\:A}\\R_{C}&=\text{reaction\:force\:at\:C}\\\sigma_{AB}&=\text{axial\:stress\:at\:A}\\\sigma_{BC}&=\text{axial\:stress\:at\:B}\\\sigma_{AB}&=\frac{R}{A_{A}}\\&=\frac{R_{A}}{A_{A}}\\\sigma_{BC}&=\frac{R_{B}}{A_{B}}\\&=\frac{R}{A_{B}}\\\frac{\sigma_{AB}}{\sigma_{BC}}&=\frac{A_{B}}{A_{B}}\\&=\frac{\frac{\pi}{4}\cdot 150^{2}}{\frac{\pi}{4}\cdot 200^{2}}\\&=\frac{9}{16}\end{aligned}

\begin{aligned}\delta L&= (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\0& = (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\&=\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{AB}+\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{BC}\\&=2\:\alpha\:T\:L-\frac{L}{E}(\sigma_{AB}+\sigma_{BC})\\T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}\end{aligned}

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The kinetic energy of a rocket is increased by a factor of eight after its engines are fired, whereas its total mass is reduced
Rudik [331]

The momentum increases by a factor of 2

Explanation:

We can solve this problem by rewriting the momentum of the rocket in terms of the kinetic energy and the mass.

The kinetic energy of the rocket is:

K=\frac{1}{2}mv^2 (1)

where

m is the mass

v is the velocity

The momentum of the rocket is

p=mv (2)

From eq.(1) we get

v=\sqrt{\frac{2K}{m}}

and substituting into (2),

p=\sqrt{2mK}

Now in this problem we have:

- The kinetic energy of the rocket is increased by a factor 8:

K' = 8K

- The mass is reduced by half:

m'=\frac{m}{2}

Substituting, we find the new momentum:

p'=\sqrt{2(\frac{m}{2}(8K)}=\sqrt{4(2mK)}=2\sqrt{2mK}=2p

So, the momentum increases by a factor of 2.

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