This is false. I hope this helps.
Answer:
2.2nC
Explanation:
Call the amount by which the spring’s unstretched length L,
the amount it stretches while hanging x1
and the amount it stretches while on the table x2.
Combining Hooke’s law with Newton’s second law, given that the stretched spring is not accelerating,
we have mg−kx1 =0, or k = mg /x1 , where k is the spring constant. On the other hand,
applying Coulomb’s law to the second part tells us ke q2/ (L+x2)2 − kx2 = 0 or q2 = kx2(L+x2)2/ke,
where ke is the Coulomb constant. Combining these,
we get q = √(mgx2(L+x2)²/x1ke =2.2nC
Answer:
76.74 Hz
Explanation:
Given:
Wave velocity ( v ) = 330 m / sec
wavelength ( λ ) = 4.3 m
We have to calculate Frequency ( f ):
We know:
v = λ / t [ f = 1 / t ]
v = λ f
= > f = v / λ
Putting values here we get:
= > f = 330 / 4.3 Hz
= > f = 3300 / 43 Hz
= > f = 76.74 Hz
Hence, frequency of sound is 76.74 Hz.
Answer:
v = 10 m/s
Explanation:
Given that,
Distance covered by a sprinter, d = 100 m
Time taken by him to reach the finish line, t = 10 s
We need to find his average velocity. We know that velocity is equal to the distance covered divided by time taken. So,
v = d/t
Hence, his average velocity is 10 m/s.
