Answer:
The correct answer to the question is (A)
When it hits the heavy rope, compared to the wave on the string, the wave that propagates along the rope has the same (A) frequency
Explanation:
The speed of a wave in a string is dependent on the square root of the tension ad inversely proportional to the square root of the linear density of the string. Generally, the speed of a wave through a spring is dependent on the elastic and inertia properties of the string
Therefore if the linear density of the heavy rope is four times that of light rope the velocity is halved and since
v = f×λ therefore v/2 = f×λ/2
Therefore the wavelength is halved, however the frequency remains the same as continuity requires the frequency of the incident pulse vibration to be transmitted to the denser medium for the wave to continue as the wave is due to vibrating particles from a source for example
Answer:
áp dụng công thức v = s/t
s là dộ dài qduong
v là vận tốc
t là thời gian
xuyên suốt 2 câu hỏi đều dùng công thức này
Explanation:
Sure !
Start with Newton's second law of motion:
Net Force = (mass) x (acceleration) .
This formula is so useful, and so easy, that you really
should memorize it.
Now, watch:
The mass of the box is 5.25 kilograms, and the box is
accelerating at the rate of 2.5 m/s² .
What's the net force on the box ?
Net Force = (mass) x (acceleration)
= (5.25 kilograms) x (2.5 m/s²)
Net force = 13.125 newtons .
But hold up, hee haw, whoa ! Wait a second !
Bella is pushing with a force of 15.75 newtons, but the box
is accelerating as if the force on it is only 13.125 newtons.
What happened to the rest of Bella's force ? ?
==> Friction is pushing the box in the opposite direction,
and cancelling some of Bella's force.
How much ?
(Bella's 15.75 newtons) minus (13.125 that the box feels)
= 2.625 newtons backwards, applied by friction.
Answer:
The work done by the drag force is given by 29.96 J
Explanation:
Given :
Thrust force 
N
Displacement 
m
Mass of rocket 
Kg
From work energy theorem,
Where 
thrust work 
gravitational work
After cutoff kinetic energy is converted into potential energy,
Put value of KE
Work done by drag force is given by,
J
Therefore, the work done by the drag force is given by 29.96 J
