Which Scientist discovered <br>atomic Theory?

The point beneath the surface where rock breaks and an earthquake is produced is known as the ______.

posledela

B. Focus is the answer.

8 0

4 years ago

A rod 14.0 cm long is uniformly charged and has a total charge of -22.0 μC. Determine the magnitude and direction of the net ele

lesya692 [45]

Explanation:

It is given that,

Length of the rod, l = 14 cm = 0.14 m

Charge on the rod, $q=-22\ \mu C=-22\times 10^{-6}\ C$

We need to find the magnitude and direction of the net electric field produced by the charged rod at a point 36.0 cm to the right of its center along the axis of the rod, z = 36 cm = 0.36 m

Electric field at the axis of the rod is given by :

$E=\dfrac{\lambda}{2\pi \epsilon_o z}$

Where

$\lambda$ is the linear charge density of the rod,

$\lambda=\dfrac{q}{l}=\dfrac{-22\times 10^{-6}\ C}{0.14\ m}=-0.00015\ C/m$

$E=\dfrac{-0.00015}{2\pi\times 8.85\times 10^{-12}\times 0.36}$

E = -7493170.57 N/C

or

$E=-7.49\times 10^6\ N/C$

Negative sign shows that the electric field is acting in inwards direction. Hence, this is the required solution.

4 0

3 years ago

what will be the effect on the acceleration due to gravity of the earth if it is compressed to a the size the moon?

GuDViN [60]

Answer:

it gravitional pull on earth will increased becauste it is compress to a form of moon which is comperatively smaller so the gravitonal pull on per cm of earth will incrased so we can say that there will be change in acceleration due to gravity

7 0

3 years ago

Read 2 more answers

What is the deceleration of the rocket sled if it comes to rest in 1.1 s from a speed of 1000 km/h? (such deceleration caused on

Wewaii [24]

The deceleration of the rocket sled if it comes to rest in 1.1 s from a speed of 1000 km/h is $252.52\ m/s^2$ .

The acceleration in opposite direction is known as the deceleration. Basically the deceleration is negative value of the acceleration since the negative sign depicts its opposite in direction.

The given data:

time, t = 1.1 s

initial speed, u = 1000 km/h = $\frac{2500}{9}\ m/s$