Answer:
a. A = 0.1656 m
b. % E = 1.219
Explanation:
Given
mB = 4.0 kg , mb = 50.0 g = 0.05 kg , u₁ = 150 m/s , k = 500 N / m
a.
To find the amplitude of the resulting SHM using conserver energy
ΔKe + ΔUg + ΔUs = 0
¹/₂ * m * v² - ¹/₂ * k * A² = 0
A = √ mB * vₓ² / k
vₓ = mb * u₁ / mb + mB
vₓ = 0.05 kg * 150 m / s / [0.050 + 4.0 ] kg = 1.8518
A = √ 4.0 kg * (1.852 m/s)² / (500 N / m)
A = 0.1656 m
b.
The percentage of kinetic energy
%E = Es / Ek
Es = ¹/₂ * k * A² = 500 N / m * 0.1656²m = 13.72 N*0.5
Ek = ¹/₂ * mb * v² = 0.05 kg * 150² m/s = 1125 N
% E = 13.72 / 1125 = 0.01219 *100
% E = 1.219
Answer:
The compression in the spring is 5.88 meters.
Explanation:
Given that,
Mass of the car, m = 39000 kg
Height of the car, h = 19 m
Spring constant of the spring, 
We need to find the compression in the spring in stopping the ore car. It can be done by balancing loss in gravitational potential energy and the increase in elastic energy. So,
x is the compression in spring
So, the compression in the spring is 5.88 meters.
Answer:
- The energy that must be added to the electron to move it to the third excited state is -1.153 eV
- The energy that must be added to the electron to move it to the fourth excited state is -1.181 eV
Explanation:
Given;
Energy of electron in ground state (n = 1 ) = 1.23 eV
E₁ = 1.23 eV
Eₙ = E₁ /n²
where;
E₁ is the energy of the electron in ground state
n is the energy level,
For third excited state, n = 4
E₄ = E₁ /4²
E₄ = (1.23 eV) / 16
E₄ = 0.077 eV
Change in energy level, = E₄ - E₁ = 0.077 eV - 1.23 eV = -1.153 eV
The energy that must be added to the electron to move it to the third excited state is -1.153 eV
For fourth excited state, n = 5
E₅ = E₁ /5²
E₄ = (1.23 eV) / 25
E₄ = 0.049 eV
Change in energy level, = E₅ - E₁ = 0.049 eV - 1.23 eV = -1.181 eV
The energy that must be added to the electron to move it to the fourth excited state is -1.181 eV
Answer:
D. I'm guessing
The gold-foil experiment showed that the atom consists of a small, massive, positively charged nucleus with the negatively charged electrons being at a great distance from the centre.
Answer:
Time = 0.55 s
Height = 8.3 m
Explanation:
The ball is dropped and therefore has an initial velocity of 0. Its acceleration, g, is directed downward in the same direction as its displacement, 
.
The dart is thrown up in which case acceleration, g, acts downward in an opposite direction to its displacement, 
. Both collide after travelling for a time period, t. Let the height of the dart from the ground at collision be and the distance travelled by the ball measured from the top be .
It follows that 
.
Applying the equation of motion to each body (h = v_0t + 0.5at^2),
Ball:
(since 
.)
Dart:
(the acceleration is opposite to the displacement, hence the negative sign)
But
The height of the collision is the height of the dart above the ground, .
