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Tema [17]
3 years ago
12

9. The acceleration due to gravity on the

Physics
1 answer:
laiz [17]3 years ago
6 0
W=mg
W=75(1.6)
W=120N
a. 120N
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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
Nostrana [21]

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

3 0
2 years ago
Which is a molecule? Select one: a. graphene b. NaCl c. Ne d. trail mix
Ede4ka [16]
What Is A Molecule? B) NaCi
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3 years ago
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A dog of mass 4 kg runs up a hill of height 8 m. How much gravitational potential energy does the dog gain?
Genrish500 [490]
A. 314 because when you use the formula for the GPE ; GPE=MGH or means mass times gravity time height (4x8x9.8) and thats equivalent to 313.6 which rounds up to 314. Hope it helps 
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3 years ago
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I have an astronomy question... Spinning up the solar nebula. The orbital speed of the material in the solar nebula at Pluto's a
attashe74 [19]
<span>The angular momentum of a particle in orbit is 

l = m v r 

Assuming that no torques act and that angular momentum is conserved then if we compare two epochs "1" and "2" 

m_1 v_1 r_1 = m_2 v_2 r_2 

Assuming that the mass did not change, conservation of angular momentum demands that 

v_1 r_1 = v_2 r_2 

or 

v1 = v_2 (r_2/r_1) 

Setting r_1 = 40,000 AU and v_2 = 5 km/s and r_2 = 39 AU (appropriate for Pluto's orbit) we have 

v_2 = 5 km/s (39 AU /40,000 AU) = 4.875E-3 km/s

Therefore, </span> the orbital speed of this material when it was 40,000 AU from the sun is <span>4.875E-3 km/s.

I hope my answer has come to your help. Thank you for posting your question here in Brainly.
</span>
3 0
3 years ago
HELP ME PLEASE! MOTION CONCEPT MAP!
Zinaida [17]

a. speed b. direction c. Magnitude d. Direction

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