Answer:
d) It will be cut to a fourth of the original force.
Explanation:
The magnitude of the electrostatic force between the charged objects is

where
k is the Coulomb's constant
q1 and q2 are the charges of the two objects
r is the separation between the two objects
In this problem, the initial distance is doubled, so
r' = 2r
Therefore, the new electrostatic force will be

So, the force will be cut to 1/4 of the original value.
Answer: Part(a)=0.041 secs, Part(b)=0.041 secs
Explanation: Firstly we assume that only the gravitational acceleration is acting on the basket ball player i.e. there is no air friction
now we know that
a=-9.81 m/s^2 ( negative because it is pulling the player downwards)
we also know that
s=76 cm= 0.76 m ( maximum s)
using kinetic equation

where v is final velocity which is zero at max height and u is it initial
hence


now we can find time in the 15 cm ascent


using quadratic formula

t=0.0409 sec
the answer for the part b will be the same
To find the answer for the part b we can find the velocity at 15 cm height similarly using

where s=0.76-0.15
as the player has traveled the above distance to reach 15cm to the bottom


when the player reaches the bottom it has the same velocity with which it started which is 3.861
hence the time required to reach the bottom 15cm is

t=0.0409
For this case we have that by definition, the momentum is given by:

Where,
- <em>m: mass
</em>
- <em>v: speed
</em>
Therefore, replacing values we have:

From here, we clear the value of the speed:

Answer:
The magnitude of velocity is:

Answer:
0.173 m.
Explanation:
The fundamental frequency of a closed pipe is given as
fc = v/4l .................. Equation 1
Where fc = fundamental frequency of a closed pipe, v = speed of sound l = length of the pipe.
Making l the subject of the equation,
l = v/4fc ................ Equation 2
also
v = 331.5×0.6T ................. Equation 3
Where T = temperature in °C, T = 18.0 °c
Substitute into equation 3
v = 331.5+0.6(18)
v = 331.5+10.8
v = 342.3 m/s.
Also given: fc = 494 Hz,
Substitute into equation 2
l = 342.3/(4×494)
l = 342.3/1976
l =0.173 m.
Hence the length of the organ pipe = 0.173 m.
Yes , increased tension suggests increased molecular attraction between the molecules of the ropes which affect the increase in the speed of wave.