Answer :
(a). The wavelength of electron is 26.22 μm.
(b).The wavelength of car is
Explanation :
Given that,
Speed = 100 km/hr
Mass of car = 1 ton
(a). We need to calculate the wavelength of electron
Using formula of wavelength
Put the value into the formula
(II). We need to calculate the wavelength of car
Using formula of wavelength again
The wavelength of the electron is greater than the dimension of electron and the wavelength of car is less than the dimension of car.
Therefore, electron is quantum particle and car is classical.
Hence, (a). The wavelength of electron is 26.22 μm.
(b).The wavelength of car is .
So lets fill out what we have first:
Vi or initial velocity = 20 m/s
Acceleration or a = 4 m/s^2
Time for the motion = 10s
Now, using the four main kinematic equations we can deduce that the best kinematic equation to use in these terms is:
Δx = Vi(t) + 0.5at²
Plug all of our information in:
Δx = (20)(10) + (0.5)(4)(100)
Δx = 400 m
Answer:
400 m
Explanation:
answer on ed
Write the velocity vectors in component form.
• initial velocity:
v₁ = 4 m/s at 45º N of E
v₁ = (4 m/s) (cos(45º) i + sin(45º) j)
v₁ ≈ (2.83 m/s) i + (2.83 m/s) j
• final velocity:
v₂ = 4 m/s at 10º N of E
v₂ = (4 m/s) (cos(10º) i + sin(10º) j)
v₂ ≈ (3.94 m/s) i + (0.695 m/s) j
The average acceleration over this 3-second interval is then
a = (v₂ - v₁) / (3 s)
a ≈ (0.370 m/s²) + (-0.711 m/s²)
with magnitude
||a|| = √[(0.370 m/s²)² + (-0.711 m/s²)²] ≈ 0.802 m/s²
and direction θ such that
tan(θ) = (-0.711 m/s²) / (0.370 m/s²) ≈ -1.92
→ θ ≈ -62.5º
which corresponds to an angle of about 62.5º S of E, or 27.5º E of S. To use the notation in the question, you could say it's E 62.5º S or S 27.5º E.
Answer:
a) x = 5.48 10⁻² m and b) 0.05 m
Explanation:
a) For a system in oscillatory motion the mechanical energy conserves and is described by the equation
Em = ½ k A²
Where k is the spring constant and at the amplitude of the movement
When the spring has the greatest extent, the kinetic energy is zero
Em = U = ½ k x²
Therefore, the amplitude of the movement is the same amplitude of the spring
Let's calculate
A = √ (2Em / k)
A = √ (2 0.12 / 80)
A = 0.0548 m = 5.48 10⁻² m
b) In this case the spring has kinetic energy that becomes elastic potential energy, let's calculate the mechanical energy before and after compressing the spring
Initial
Em = K = ½ m v²
Final
Em = Ke = ½ k x²
½ m v² = ½ k x²
x = √(m/k) v
x = 2 √(0.50 /800.0)
x = 0.05 m
Answer:
a) The greatest extension of the spring is 0.055 m
b) The spring compress 0.05 m
Explanation:
Please look at the solution in the attached Word file
Complete question:
An electron is a subatomic particle (m = 9.11 x 10-31 kg) that is subject to electric forces. An electron moving in the +x direction accelerates from an initial velocity of +6.18 x 105 m/s to a final velocity of 2.59 x 106 m/s while traveling a distance of 0.0708 m. The electron's acceleration is due to two electric forces parallel to the x axis: F₁ = 8.87 x 10-17 N, and , which points in the -x direction. Find the magnitudes of (a) the net force acting on the electron and (b) the electric force F₂.
Answer:
(a) The net force of the electron, ∑F = 4.07 x 10⁻¹⁷ N
(b) the electric force, F₂ = 4.8 x 10⁻¹⁷ N
Explanation:
Given;
initial velocity of the electron, = +6.18 x 10⁵ m/s
final velocity of the electron, = 2.59 x 10⁶ m/s
the distance traveled by the electron, d = 0.0708 m
The first electric force,
(a) The net force of the electron is given as;
∑F = F₁ - F₂ = ma
where;
a is the acceleration of the electron
∑F = ma = (9.11 x 10⁻³¹ kg)(4.468 x 10¹³)
∑F = 4.07 x 10⁻¹⁷ N
(b) the electric force, F₂ is given as;
∑F = F₁ - F₂
F₂ = F₁ - ∑F
F₂ = 8.87 x 10⁻¹⁷ - 4.07 x 10⁻¹⁷
F₂ = 4.8 x 10⁻¹⁷ N
The problem involves calculating the acceleration of an electron, then using Newton's second law to find the net force on the electron. This is used to find the magnitude of a second electric force acting on the electron.
First, we can calculate the acceleration of the electron using the formula a = Δv/Δt, where 'a' is acceleration, 'Δv' is the change in velocity, and 'Δt' is the change in time. In this case, Δv = vf - vi = 2.59 x 106 m/s - 6.18 x 105 m/s = 1.972 x 106 m/s. The time taken by the electron to travel 0.0708 m can be found using the equation d = vi t + 0.5 a t₂. We use these values to get Δt which we use to find 'a'.
Next, let's use Newton's second law F = ma to find the net force acting on the electron. The only forces acting on the electron are electric forces, and we know one them is 8.87 x 10-17 N. If we designate this known force as F₁ then the total force F total = F₁ + F₂ where F₂ is the unknown electric force.
Finally, we can find F₂ = F total - F₁. This gives the magnitude of the second electric force.
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7500 J were released in the explosion, how much kinetic energydid
each piece acquire?
Answer:
4500 J and 3000 J
Explanation:
According to conservation of momentum
Given that m_2 = 1.5 m_1 , so
the kinetic energy of each piece is
substituting the value of V1 in the above equation
Given that
K_1 + k_2 = 7500 J
1.5 K_2 + K_2 = 7500
K_2 = 7500 / 2.5
= 3000 J
this is the KE of heavier mass
K_1 = 7500 - 3000 = 4500 J
this is the KE of lighter mass
The question is about finding the kinetic energy acquired by each of two pieces of an object following an internal explosion, using principles of conservation of energy and momentum in physics.
The student has asked about an internal explosion that breaks an object into two pieces with different masses, releasing a certain amount of kinetic energy in the process. This question involves applying the principle of conservation of energy and momentum to find the kinetic energy acquired by each piece post-explosion.
Assuming piece 1 has a mass of m and piece 2 has a mass of 1.5m, the total mass of the system is 2.5m. Since 7500 J of energy was released in the explosion, to find the kinetic energy of each piece, we can use the fact that the total kinetic energy is equal to the energy released during the explosion. Let the kinetic energy of the smaller piece be K1 and of the larger piece be K2. Because the object was initially at rest and momentum must be conserved, the momenta of the two pieces must be equal and opposite. This relationship allows us to derive the ratio of the kinetic energies. We can solve for K1 and K2 proportionally. Finally, because the kinetic energy is a scalar quantity, adding the kinetic energies of the two pieces will equal the total energy released.
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