Answer:
Equal to change in momentum of larger mass.
Explanation:
We are given that
Two difference masses .
Force act on both masses for the same length of time.
We have to find the change in momentum of the smaller mass.
Let M and m are two masses
M>m
We know that
Change in momentum for large mass=
Change in momentum for small mass=
Because Force and length of time are same for both masses .
Hence, the change in momentum of smaller mass is equal to change in momentum of larger mass.
Answer:
the first one answer is no
This is the period in a simple harmonic motion which is 2 seconds in this question.
<h3>
What is Period ?</h3>
The period of an oscillatory object can be defined as the total time taken by a vibrating body to make one complete revolution about a reference point.
We are given the below question
2×3.14√(1.0m/(9.8〖ms〗^(2) )= T
This question can as well be expressed as
2π√(L/g) which is equal to period T.
In a nut shell, Period T = 2×3.14√(1.0m/9.8)
T = 6.28√0.102
T = 6.28 × 0.32
T = 2.006 s
Therefore, the period T of the oscillation is 2 seconds approximately.
Learn more about Period here: brainly.com/question/12588483
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Answer:
oxygen silicon aluminun iron
Answer:
Explanation:
Given that,
5J work is done by stretching a spring
e = 19cm = 0.19m
Assuming the spring is ideal, then we can apply Hooke's law
F = kx
To calculate k, we can apply the Workdone by a spring formula
W=∫F.dx
Since F=kx
W = ∫kx dx from x = 0 to x = 0.19
W = ½kx² from x = 0 to x = 0.19
W = ½k (0.19²-0²)
5 = ½k(0.0361-0)
5×2 = 0.0361k
Then, k = 10/0.0361
k = 277.008 N/m
The spring constant is 277.008N/m
Then, applying Hooke's law to find the applied force
F = kx
F = 277.008 × 0.19
F = 52.63 N
The applied force is 52.63N
