Answer:
A_resulting = 0.2 m
Explanation:
Let's analyze the impact of the pulse with the pole, this is a fixed obstacle that does not move therefore by the law of action and reluctant, the force that the pole applies on the rope is of equal magnitude to the force of the rope on the pole (pulse), but opposite directional, so the reflected pulse reverses its direction and sense.
With this information we analyze a point on the string where the incident pulse is and each reflected with an amplitude A = 0.1 m, the resulting is
A_res = 2A
A_resultant = 2 .01
A_resulting = 0.2 m
Someone can make an introduction about getting a flu shot because, mostly everyone in the world have gotten or just got a flu shot. One example of a pro is that, a flu shot can prevent the flu, and other deadly sicknesses. On example of a con is that, some flu shots can carry side effects which is something that should be tested before given to adults and especially kids!!! Hope this helps!!! Good luck
Answer:
Mass of bullet is m=0.01kg
Mass of the block is M=4kg
Coefficient=0.25,distance=20m
So, let the speed of the block just after the bullet embedded in it be V and v be the speed of bullet before striking the block,
By applying conservation of momentum,
mv=(m+M)V
V=
M+m
mv
Explanation:
please mark me as the brainliest answer and please follow me for more answers to your questions..
Answer: C
Explanation: Side post terminals need to be removed to inspect them for corrosion.
Over tightening the terminal bolt can damage side post terminals.
The battery terminals and cable ends can corrode especially when the battery or car is not used for a long period of time. Corrosion limits a battery's lifespan and so should be prevented. To inspect the areas where corrosion occur on a side-post battery, you need to remove the terminals.
Also, it is true that over tightening the terminal bolt can damage the side post terminals. The covering on the battery can become twisted, and make the seals on the terminals leak.
Concept:
Frequency- It is defined as the number of oscillations occur in one second.
Its SI unit is Hertz (Hz)
Given: Produced sound vibrations is 18,500 cycles in 0.75 seconds
∵ In 0.75 second, produced sound has oscillations = 18,500 cycles
∴ In 1.0 second, produced sound has oscillations = (18,500 ÷ 0.75) Hz
The frequency of the sound will be ≈ 24,667 Hz
From the study of the given graph, only the animals (c) Cats, (b) Moths and (a) Bats can hear the produced sound because their upper audible frequency range is greater than 24,667 Hz.