The answer is P-waves and S-waves
Answer:
The right response will be "450 volts".
Explanation:
The given values are:
R1 = 4.00 cm
R2 = 6.00 cm
q1 = +6.00 nC
q2 = −9.00 nC
As we know,
The potential difference between the two shell's difference will be:
⇒ ![\Delta V=K[(\frac{q1}{R1}+\frac{q2}{R2})-(\frac{q1}{R1} +(\frac{q2}{R2}))]](https://tex.z-dn.net/?f=%5CDelta%20V%3DK%5B%28%5Cfrac%7Bq1%7D%7BR1%7D%2B%5Cfrac%7Bq2%7D%7BR2%7D%29-%28%5Cfrac%7Bq1%7D%7BR1%7D%20%2B%28%5Cfrac%7Bq2%7D%7BR2%7D%29%29%5D)
![=K[\frac{q1}{R2}-\frac{q1}{R1} ]](https://tex.z-dn.net/?f=%3DK%5B%5Cfrac%7Bq1%7D%7BR2%7D-%5Cfrac%7Bq1%7D%7BR1%7D%20%5D)
On substituting the values, we get
Δ 
The image is missing (however it's not necessary to solve the problem).
The correct answer is A) decreases, because the gravitational force is inversely proportional to the square of the distance. In fact, the magnitude of the gravitational force between two object of mass M and m, at a distance d one from each other, is

where G is the gravitational constant. As can be seen from the formula, if the distance d between the two object increases, the intensity of the force decreases.
Answer:
0.37sec
Explanation:
Period of oscillation of a simple pendulum of length L is:
T
=
2
π
×
√
(L
/g)
L=length of string 0.54m
g=acceleration due to gravity
T-period
T = 2 x 3.14 x √[0.54/9.8]
T = 1.47sec
An oscillating pendulum, or anything else in nature that involves "simple harmonic" (sinusoidal) motion, spends 1/4 of its period going from zero speed to maximum speed, and another 1/4 going from maximum speed to zero speed again, etc. After four quarter-periods it is back where it started.
The ball will first have V(max) at T/4,
=>V(max) = 1.47/4 = 0.37 sec
The formula for this problem that we will be using is:
F * cos α = m * g * μs where:F = 800m = 87g = 9.8
cos α = m*g*μs/F= 87*9.8*0.55/800= 0.59 So solving the alpha, find the arccos above.
α = arccos 0.59 = 54 ° is the largest value of alpha