<span>Normal
fault. Normal (extensional ) fault is a
displacement of a rock as a result of rock-mass movement and occurs when the
crust is stretching. Because of the stretching the thickness of the crust is
reduced and the crust or horizontally extended. </span>
Answer: -33.3 * 10^9 C/m^2( nC/m^2)
Explanation: In order to solve this problem we have to use the gaussian law, the we have:
Eoutside =0 so teh Q inside==
the Q inside= 4.6 nC/m*L + σ *2*π*b*L where L is the large of the Gaussian surface and b the radius of the shell.
Then we simplify and get
σ= -4.6/(2*π*b)= -33.3 nC/m^2
Answer:
The block will not move.
Explanation:
We'll begin by calculating the frictional force. This can be obtained as follow:
Coefficient of friction (µ) = 0.6
Mass of block (m) = 3 Kg
Acceleration due to gravity (g) = 10 m/s²
Normal reaction (R) = mg = 3 × 10 = 30 N
Frictional force (Fբ) =?
Fբ = µR
Fբ = 0.6 × 30
Fբ = 18 N
From the calculations made above, the frictional force of the block is 18 N. Since the frictional force (i.e 18 N) is bigger than the force applied (i.e 14 N), the block will not move.
Newton's 2nd law of motion:
Force = (mass) x (acceleration)
= (1,127 kg) x (6 m/s² forward)
= (1,127 x 6) newtons forward
= 6,762 newtons forward
______________________________
Momentum = (mass) x (speed)
= (69 kg) x (6 m/s)
= 414 kg-m/s
Answer:
9 meters
Explanation:
Given:
Mass of Avi is,
Spring constant is,
Compression in the spring is,
Let the maximum height reached be 'h' m.
Now, as the spring is compressed, there is elastic potential energy stored in the spring. This elastic potential energy is transferred to Avi in the form of gravitational potential energy.
So, by law of conservation of energy, decrease in elastic potential energy is equal to increase in gravitational potential energy.
Decrease in elastic potential energy is given as:
Now, increase in gravitational potential energy is given as:
Now, increase in gravitational potential energy is equal decrease in elastic potential energy. Therefore,
Therefore, Avi will reach a maximum height of 9 meters.