Answer:
True
Explanation:
Going even smaller than atoms would get you to subatomic particles such as quarks. From there, it is impossible to distinguish elements. So, yes, atoms are the smallest portions of an element that retains the original characteristic of the element.
Answer:
ni = 2.04e19
Explanation:
we know that in semiconductor like intrinsic, when electron leave the band, it leave a hole in valence band so we have
n = p = ni
from intrinsic carrier concentration
1.7 = ni * 1.6*10^{-19} * (.35 + .17)
ni = 2.014 *10^{19} m^{-3}
ni = 2.04e19
Answer:
hmax = 1/2 · v²/g
Explanation:
Hi there!
Due to the conservation of energy and since there is no dissipative force (like friction) all the kinetic energy (KE) of the ball has to be converted into gravitational potential energy (PE) when the ball comes to stop.
KE = PE
Where KE is the initial kinetic energy and PE is the final potential energy.
The kinetic energy of the ball is calculated as follows:
KE = 1/2 · m · v²
Where:
m = mass of the ball
v = velocity.
The potential energy is calculated as follows:
PE = m · g · h
Where:
m = mass of the ball.
g = acceleration due to gravity (known value: 9.81 m/s²).
h = height.
At the maximum height, the potential energy is equal to the initial kinetic energy because the energy is conserved, i.e, all the kinetic energy was converted into potential energy (there was no energy dissipation as heat because there was no friction). Then:
PE = KE
m · g · hmax = 1/2 · m · v²
Solving for hmax:
hmax = 1/2 · v² / g
Answer:
Explanation:
At constant pressure , work done by gas = P x ΔV where P is pressure and ΔV is change in volume
ΔV = 9.2 - 5.6 = 3.6 L
3.6 L = 3.6 x 10⁻³ m³
ΔV = 3.6 x 10⁻³ m³
P = 3.7 x 10³ Pa
So work done
= 3.7 x 10³ x 3.6 x 10⁻³ J
= 13.32 J .
( c ) is the answer , because work is done by the gas so it will be positive.
Answer: 185.5672566
Explanation: The friction is not relevant
Normal reaction is the force perpendicular to the surface.
this force resists the downwards forces applied which are gravity and a component of the applied force.
