Answer:
0.0483m or 48.3 mm
Explanation:
Let g = 10m/s2
At 8m, the pile has a potential energy of
P = mgh = 3000*10*8 = 240000 J
This energy is converted to kinetic energy once the pile drops down to the bottom:
Taking gravity into consideration, the net force acting on the pile once it hits the ground is
Therefore the net deceleration is:
We can find out how far it's driven into the ground with the following equation of motion:
where v = 0 is the final velocity, which is 0 because it stops, is the initial velocity when it hits the ground, a is the deceleration, and s is the distance it travels into the ground
or 48.3 mm
Answer:
Explanation:
We are given that
d=1.9 cm=
Using 1m=100 cm
We have to find the electric field strength.
Using the formula
Mass of electron,m
Substitute the values
Answer:
Star, hydrogen and energy
Explanation:
A star is a large ball of gas that generates its own energy by fusing hydrogen atoms to make helium. It is held together by its own gravity. This process emits a tremendous amount of energy, and some of the energy is in the form of light.
The energy released from the collapse of the gas into a protostar causes the center of the protostar to become extremely hot. When the core is hot enough, nuclear fusion commences. Fusion is the process where two hydrogen atoms combine to form a helium atom, releasing energy.
Answer:
the kinetic energy of A will be twice that of B
Explanation:
The formula for calculating kinetic energy is expressed using the formula
KE = 1/2mv²
m is the mass of the object
v is the velocity
For object A:
KEA = 1/2mAvA²
For object B:
KEB = 1/2mBvB² ... 1
If object A is twice the mass of object B, then mA = 2mB
From 1:
KEA = 1/2mAvA²
Substitute mA = 2mB
KEA = 1/2(2mB)vA² .... 2
Divide 1 by 2
KEB/KEA = 1/2mBvB²/mBvA²
KEB/KEA = 1/2vB²/vA²
Assuming they have the same velocities then vA ,= VB
The equation becomes:
KEB/KEA = 1/2vB²/vA²
KEB/KEA = 1/2vB²/vB²
KEB/KEA = 1/2
KEA = 2EB
Hence the kinetic energy of A will be twice that of B
Both a molten metallic core and reasonably fast rotation.