Answer:
0.35
Explanation:
According to Newton's second law;
\sum Fx = ma
Fm - Ff =ma
Fm is the moving force = Wsin theta
Fm = 4(9.8)sin55
Fm = 32.1N
Ff is the frictional force = nmgcos theta
Ff = n(4)(9.8)cos55
Ff = 22.48n
Acceleration a = 6.0m/s²
Substitute the given values into the formula and get the coefficient of friction
32.11-23.48n = 4(6)
32.11-24= 23.48n
8.11 = 23.48
n = 8.11/23.48
n = 0.35
Hence the coefficient of friction is 0.35
To develop this problem we will apply the concepts related to the potential energy per unit volume for which we will obtain an energy density relationship that can be related to the electric field. From this formula it will be possible to find the electric field required in the problem. Our values are given as
The potential energy, 
The volume, 
The potential energy per unit volume is defined as the energy density.



The energy density related with electric field is given by

Here, the permitivity of the free space is

Therefore, rerranging to find the electric field strength we have,



Therefore the electric field is 2.21V/m
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 
Answer:
6.378 * 10^6 m
Explanation:
Radius = 6378 km
Changing in metres
6378 km = 6378000 m
Now in scientific notation
= 6.378 * 10 ^ 6 m