(a) The average speed between is approximately 7.44 m/s.(b) The instantaneous speed is approximately 19.60 m/s and 28.96 m/s. (c) The average acceleration is approximately 7.20 m/s². (d) The instantaneous acceleration is approximately 7.20 m/s². (e) the object is at rest at approximately t = 0.28 s
a) For determining the average speed between t = 3.00 s and t = 4.30 s, need to find the change in position and divide it by the change in time. The change in position can be obtained by subtracting the position at t = 3.00 s from the position at t = 4.30 s:
Δx = x(4.30 s) - x(3.00 s)
[tex]= (3.60 * (4.30^2) - 2.00 * 4.30 + 3.00) - (3.60 * (3.00^2) - 2.00 * 3.00 + 3.00)\\\approx 13.08 m - 2.70 m\\\approx 10.38 m[/tex]
The change in time is simply 4.30 s - 3.00 s = 1.30 s. Therefore, the average speed is:
Average Speed = Δx / Δt
= 10.38 m / 1.30 s
≈ 7.44 m/s.
b) To determine the instantaneous speed at t = 3.00 s, can differentiate the position function with respect to time:
v(t) = dx/dt
= 2 * 3.60t - 2.00
= 7.20t - 2.00
Substituting t = 3.00 s into the above equation:
v(3.00 s) = 7.20 * 3.00 - 2.00
≈ 21.60 m/s - 2.00 m/s
≈ 19.60 m/s.
Similarly, find the instantaneous speed at t = 4.30 s by substituting t = 4.30 s into the equation:
v(4.30 s) = 7.20 * 4.30 - 2.00
≈ 30.96 m/s - 2.00 m/s
≈ 28.96 m/s.
c) To find the average acceleration between t = 3.00 s and t = 4.30 s, can differentiate the velocity function with respect to time:
a(t) = dv/dt
= 7.20
Since the acceleration is constant, the average acceleration is equal to the instantaneous acceleration:
Average Acceleration = Instantaneous Acceleration = 7.20 m/s².
d) To determine the instantaneous acceleration at t = 3.00 s, can use the same acceleration function:
a(3.00 s) = 7.20 m/s².
Similarly, substituting t = 4.30 s into the equation, find the instantaneous acceleration at t = 4.30 s:
a(4.30 s) = 7.20 m/s².
e) To find when the object is at rest, need to determine the time(s) when the velocity is zero. Setting the velocity function equal to zero and solving for t:
7.20t - 2.00 = 0
7.20t = 2.00
t ≈ 0.28 s.
Therefore, the object is at rest at approximately t = 0.28 s.
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A Coast Guard ship is traveling at a constant velocity of 4.38 m/s, due east, relative to the water. On his radar screen the navigator detects an object that is moving at a constant velocity. The object is located at a distance of 2600 m with respect to the ship, in a direction 25.0
∘
south of east. Six minutes later, he notes that the object's position relative to the ship has changed to 1110 m,55.0
∘
south of west. What are (a) the magnitude and (b) direction of the velocity of the object relative to the water? Express the direction as an angle with respect to due west. (a) Number Units (b) Number Units
(a) The magnitude of the velocity of the object relative to the water is 70.8 m/s.
(b) The direction of the velocity of the object relative to the water with respect to due west is -64.5 °.
Explanation: First, we can draw the diagram as shown below. The ship's velocity can be represented as an arrow pointing due east with a magnitude of 4.38 m/s. The object's velocity can be represented by an arrow connecting its initial and final positions relative to the ship.
We can use the Law of Cosines to determine the magnitude of the velocity of the object relative to the water.The Law of Cosines states that for any triangle ABC, where a, b, and c are the side lengths opposite their respective angles A, B, and C, then: c2=a2+b2−2abcosC.
where, C is the angle opposite side c.We can label the sides of the triangle as shown in the diagram. c is the magnitude of the velocity of the object relative to the water, a is the distance from the ship to the initial position of the object, and b is the distance from the ship to the final position of the object.
We know that:a=2600 mb=1110 m.
We can use the Law of Cosines to solve for c:
c2=26002+11102−2(2600)(1110)cos(155°)c=70.8 m/s.
Therefore, the magnitude of the velocity of the object relative to the water is 70.8 m/s.
Next, we can find the direction of the velocity of the object relative to the water. We can use the Law of Sines to solve for the angle θ opposite side b.The Law of Sines states that for any triangle ABC, where a, b, and c are the side lengths opposite their respective angles A, B, and C, then. a/sinA=b/sinB=c/sinC.
We can use the Law of Sines to solve for sin(θ + 55°):sin(θ + 55°)/1110=sin(155°)/70.8sin(θ + 55°)=0.223sinθ=0.179θ=10.4°
We can find the angle with respect to due west by subtracting θ from 90°:90° - θ = 79.6°.
Finally, we can subtract the angle between the ship's velocity and the x-axis from 79.6°.
To find the direction of the velocity of the object relative to the water with respect to due west: 79.6° - 90° - 25° = -64.5°.
Therefore, the direction of the velocity of the object relative to the water with respect to due west is -64.5 °.
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(13\%) Problem 2: You are given a 45 V battery and two resistors of resistances 22.5Ω and 98.5Ω. a 25% Part (a) Find the current in A when these resistors are connected in series with the battery. I
3
= Hints:
05
deduction per hint. Hints remaining: 3 Feedback: 0% deduction per feedback. A 25% Part (b) Find the power dissipated in W by the 22.5Ω resistor when connected in series with the rest of the circuit. a 25\% Part (c) Find the power dissipated in W by the 98.5Ω resistor when connected in series with the rest of the circuit. a 25% Part (d) Find the current in A being drawn from the battery when the resistors are connected in parallel with the battery. A 25% Part (a) Find the current in A when these resistors are connected in series with the battery. A 25% Part (b) Find the power dissipated in W by the 22.5Ω resistor when connected in series with the rest of the circuit. P
l
= A 25% Part (a) Find the current in A when these resistors are connected in series with the battery. a 25% Part (b) Find the power dissipated in W by the 22.5Ω resistor when connected in series with the rest of the circuit. A 25% Part (c) Find the power dissipated in W by the 98.5Ω resistor when connected in series with the rest of the circuit. P
2
= Q 25% Part (a) Find the current in A when these resistors are connected in series with the battery. 225% Part (b) Find the power dissipated in W by the 22.5Ω resistor when connected in series with the rest of the circuit. a 25% Part (c) Find the power dissipated in W by the 98.5Ω resistor when connected in series with the rest of the circuit.
The answers for each part of the problem are Current when the resistors are connected in series with the battery is 0.3719 A.
Voltage, V = 45 V
Resistor 1, R1 = 22.5 Ω
Resistor 2, R2 = 98.5 Ω
Total resistance, RT = R1 + R2 = 22.5 + 98.5 = 121 Ω
Question 1: Find the current when the resistors are connected in series with the battery
The total current in the circuit is given by;
I = V/RT
where V = 45 V and RT = 121 ΩI = 45 / 121I = 0.3719 A
Thus, the current when the resistors are connected in series with the battery is 0.3719 A.
Question 2: Find the power dissipated by the 22.5 Ω resistor
The power dissipated by the resistor R1 is given by;P1 = I²R1
Where I = 0.3719 A and R1 = 22.5 Ω.P1 = (0.3719)² x 22.5P1 = 3.27 W
Thus, the power dissipated by the 22.5 Ω resistor is 3.27 W.
Question 3: Find the power dissipated by the 98.5 Ω resistor
The power dissipated by the resistor R2 is given by;P2 = I²R2
Where I = 0.3719 A and R2 = 98.5 Ω.P2 = (0.3719)² x 98.5P2 = 13.9 W
Thus, the power dissipated by the 98.5 Ω resistor is 13.9 W.
Question 4: Find the current drawn from the battery when the resistors are connected in parallel with the battery
When the resistors are connected in parallel with the battery, the equivalent resistance is given by;1/RT = 1/R1 + 1/R2 = 1/22.5 + 1/98.5RT = 18.19 Ω
The current drawn from the battery when the resistors are connected in parallel with the battery is given by;I = V/RT
Where V = 45 V and RT = 18.19 ΩI = 45 / 18.19I = 2.47 A
Thus, the current drawn from the battery when the resistors are connected in parallel with the battery is 2.47 A.Answer:
Therefore, the answers for each part of the problem are:Current when the resistors are connected in series with the battery is 0.3719 A.
Power dissipated by the 22.5 Ω resistor is 3.27 W.Power dissipated by the 98.5 Ω resistor is 13.9 W.
Current drawn from the battery when the resistors are connected in parallel with the battery is 2.47 A.
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Two parallel metal plates are connected to a 12 Volt battery. If the plates are separated by 0.5 m, how much energy would it take to move a 2 Coulomb charged object from one plate to the other? 6 Joules B) 12 Joules (C) 24 joules D 48 Joules (E) 96 Joule
the energy required to move the 2 Coulomb charged object from one plate to the other is 24 Joules.
The correct answer is C.
To calculate the energy required to move a charged object from one plate to the other in a parallel plate capacitor, we can use the formula:
Energy = Q * V
Where:
Q is the charge of the object
V is the potential difference (voltage) between the plates
In this case, the charge Q is given as 2 Coulombs, and the potential difference V is given as 12 Volts.
Plugging in these values into the formula, we have:
Energy = 2 C * 12 V = 24 Joules
Therefore, the energy required to move the 2 Coulomb charged object from one plate to the other is:
The correct answer is C. 24 Joules.
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I have decided to focus on radio waves, which have the lowest frequency on the electromagnetic spectrum. Radio waves are any kind of electromagnetic wave formed by a current in a wire or a circuit. AM (amplitude modulation) and FM (frequency modulation) differ in how they carry information and what their transmission sites look like. AM radio waves have varying amplitudes and their antennas are constructed to be quite tall (so they can transmit waves at their preferred size of λ/2). Due to their larger amplitudes, they can often travel over hills and other obstructions.
FM radio waves are carrier waves with audio signals of identical amplitudes added to them in order to send information about sound. In television, the FM audio signals are extremely important and often much more complex than they would be for radio, which requires a large range of frequencies to be transmitted. Satellite dishes, cable, and HD formats are carried at much higher frequencies to be able to carry even more information about the television audio. FM transmission towers are normally smaller than AM transmission towers because their amplitudes do not vary, but they are elevated to achieve a direct ‘line of sight’ between the antenna and the receiver.
In responding to the above posts, aim to restate your understanding of how this EM instrumentation works. Include an outside citation in at least one of your responses.
Electromagnetic (EM) instrumentation works by utilizing Electromagnetic (EM) instrumentation, which are a type of electromagnetic wave that has the lowest frequency on the electromagnetic spectrum. They're formed by a current in a wire or a circuit. AM (amplitude modulation) and FM (frequency modulation) are two different methods of carrying information and have varying transmission sites.
Radio waves have a frequency range of 3 kHz to 300 GHz, which is the lowest frequency range on the electromagnetic spectrum. Radio waves are utilized in numerous applications, including radio and television broadcasting, GPS, radar, and wireless communication. Radio waves are utilized to transfer information in the form of analog or digital signals in a variety of ways, including amplitude modulation (AM) and frequency modulation (FM).
AM and FM waves are utilized in radio communication. AM radio waves have varying amplitudes and their antennas are constructed to be quite tall so they can transmit waves at their preferred size of λ/2. Due to their larger amplitudes, they can often travel over hills and other obstructions.
On the other hand, FM radio waves are carrier waves with audio signals of identical amplitudes added to them to send information about sound. The FM audio signals used in television are extremely important and are often much more complex than those utilized in radio, which requires a wide range of frequencies to be transmitted.
EM instrumentation utilizes radio waves, which have the lowest frequency on the electromagnetic spectrum, to transfer information in various applications. Amplitude modulation (AM) and frequency modulation (FM) are two different methods of carrying information. Due to their larger amplitudes, AM radio waves can travel over hills and other obstructions. On the other hand, FM radio waves have identical amplitudes, and the FM transmission towers are usually smaller than the AM transmission towers because their amplitudes do not vary, but they are elevated to achieve a direct ‘line of sight’ between the antenna and the receiver.
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A long coaxial cable has an inner conductor with a radius of 0.75 mm and an outer conductor with an inner radius of 3.00 mm. Assume the space between has the same permittivity as free space. a) If these conductors are charged with +5.00nC/m(inner) and −5.00nC/m (outer), what is the voltage difference between them? b) What is the capacitance per length of this cable? Please answer in pF/m c) How much energy is stored in 10 m of this cable? Please answer in μJ.
If the conductors are charged with +5.00nC/m(inner) and −5.00nC/m, the voltage difference would be -0.061 V. The capacitance will be 0.488 pF/m. The energy stored in 10 m of the cable is 0.015 μJ.
Given:
Inner conductor radius (r₁) = 0.75 mm = 0.75 × 10^(-3) m,
Outer conductor inner radius (r₂) = 3.00 mm = 3.00 × 10^(-3) m,
Inner conductor charge density (λ₁) = +5.00 nC/m = +5.00 × 10^(-9) C/m,
Outer conductor charge density (λ₂) = -5.00 nC/m = -5.00 × 10^(-9) C/m,
Permittivity of free space (ε₀) = 8.854 × 10^(-12) F/m,
Length of the cable (length) = 10 m.
(a) Calculating the voltage difference:
V = (1 / (4πε₀)) * ((λ₂ - λ₁) / ln(r₂ / r₁))
= (1 / (4π × 8.854 × 10^(-12) F/m)) * ((-5.00 × 10^(-9) C/m - 5.00 × 10^(-9) C/m) / ln(3.00 × 10^(-3) m / 0.75 × 10^(-3) m))
= -0.061 V (rounded to three decimal places)
The voltage difference between the inner and outer conductors is approximately -0.061 V.
(b) Calculating the capacitance per length:
C = (2πε₀) / ln(r₂ / r₁)
= (2π × 8.854 × 10^(-12) F/m) / ln(3.00 × 10^(-3) m / 0.75 × 10^(-3) m)
= 0.488 pF/m (rounded to three decimal places)
The capacitance per length of the cable is approximately 0.488 pF/m.
(c) Calculating the energy stored in the cable:
E = (1/2) * C * V² * length
= (1/2) * (0.488 pF/m) * (-0.061 V)² * 10 m
= 0.015 μJ (rounded to three decimal places)
The energy stored in 10 m of the cable is approximately 0.015 μJ.
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Find the magnitude of the linear momentum (a). for a 14.0 g bullet moving with speed of 500 m/s (b). for a 75.0 kg runner running with a speed of 10.m/s Hint: the momentum of an object is calculated as p=mv - m= mass - v= speed (a). 7 kg⋅m/s (b). 75 kg⋅m/s (a). 75 kg⋅m/s (b). 7.5 kg⋅m/s (a). 7 kg⋅m/s (b). 750 kg⋅m/s (a). 700 kg⋅m/s (b). 75 kg⋅m/s (a). 17 kg⋅m/s (b). 175 kg⋅m/s (a). 72 kg⋅m/s (b). 15 kg⋅m/s
The magnitude of the linear momentum for the given scenarios is as follows:
(a) For a 14.0 g bullet moving with a speed of 500 m/s, the magnitude of the linear momentum is 7 kg⋅m/s.
To calculate the momentum, we first need to convert the mass of the bullet to kilograms: 14.0 g = 0.014 kg. Then we multiply the mass by the speed to obtain the momentum:
Momentum (p) = Mass (m) × Speed (v)
= 0.014 kg × 500 m/s
= 7 kg⋅m/s
(b) For a 75.0 kg runner running with a speed of 10 m/s, the magnitude of the linear momentum is 750 kg⋅m/s.
Again, we use the formula:
Momentum (p) = Mass (m) × Speed (v)
= 75.0 kg × 10 m/s
= 750 kg⋅m/s
Therefore, the magnitude of the linear momentum for the bullet is 7 kg⋅m/s, and for the runner, it is 750 kg⋅m/s.
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An unknown material has a mass of 0.447 kg and its temperature increases by 2.87 ∘ C when 943 J of heat are added. What is the specific heat of the material? ( Unit =J/(kg8c)0|
The specific heat of the unknown material is approximately 3.98 J/(kg·°C).
To find the specific heat of the material, we can use the equation:
Q = mcΔT
where Q is the heat added, m is the mass of the material, c is the specific heat, and ΔT is the change in temperature.
Given:
Q = 943 J
m = 0.447 kg
ΔT = 2.87 °C
Converting the change in temperature to Kelvin:
ΔT = 2.87 °C + 273.15 = 276.02 K
Substituting the given values into the equation:
943 J = (0.447 kg) * c * (276.02 K)
Simplifying the equation:
c = 943 J / (0.447 kg * 276.02 K)
c ≈ 3.98 J/(kg·°C)
Therefore, the specific heat of the unknown material is approximately 3.98 J/(kg·°C).
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A cart of mass 400 kg traveling at a speed of 15 m/s is struck by a 50-L/s jet of water flowing at 40 m/s.
a. If the water jet is going in the same direction as the cart, what is the applied acceleration force?
b. What is the relative speed of the cart?
a. The applied acceleration force on the cart cannot be determined without the time interval over which the water jet acts.
b. The relative speed of the cart is 25 m/s.
a. To determine the applied acceleration force on the cart, we can use Newton's second law of motion:
F = m * a
where F is the force, m is the mass of the cart, and a is the acceleration.
Given:
mass of the cart (m) = 400 kg
initial velocity of the cart (u) = 15 m/s
velocity of the water jet (v) = 40 m/s
Since the water jet is going in the same direction as the cart, the resulting velocity (v') of the cart after being struck by the water jet can be calculated using the equation:
v' = (m * u + M * v) / (m + M)
where M is the mass flow rate of the water jet, which can be calculated by multiplying the jet velocity (v) by the density of water (ρ).
Given:
density of water (ρ) = 1000[tex]kg/m^3[/tex]
flow rate of water (Q) = 50 L/s = 0.05[tex]m^3[/tex]/s
M = ρ * Q
Substituting the given values into the equation, we have:
M = (1000 kg/[tex]m^3[/tex]) * (0.05 [tex]m^3[/tex]/s)
Now we can substitute the values of m, u, v, M, and v' into Newton's second law equation to solve for the force (F):
F = m * (v' - u) / t
where t is the time interval over which the water jet acts on the cart. Since the time interval is not given, we cannot determine the exact force value.
b. The relative speed of the cart can be calculated by taking the difference between the initial velocity of the cart and the velocity of the water jet:
Relative speed = |u - v|
Substituting the given values into the equation:
Relative speed = |15 m/s - 40 m/s| = 25 m/s
Therefore, the relative speed of the cart is 25 m/s.
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Find the wavelength of an oscillator having a vertical displacement given by:
y(x, t) = (6.00 mm)cos[(3.25m–¹)x-(7.22s–¹)t)]
Group of answer choices
For the given vertical displacement, the wavelength of the oscillator is 1.94m.
The given equation is y(x, t) = (6.00 mm) cos[(3.25m⁻¹)x - (7.22s⁻¹)t], where the oscillator has a vertical displacement given by y(x, t), and cos is the cosine function.
Therefore, the wavelength of the oscillator can be calculated using the formula; wavelength = (2π/k)where k is the wavenumber. The wavenumber can be determined from the equation; k = 2π/λ where λ is the wavelength. The vertical displacement y(x, t) can be written in the form; y(x, t) = Acos(kx - ωt)Comparing this with the given equation, we get; A = 6.00 mmω = 7.22 s⁻¹k = 3.25 m⁻¹From the equation; k = 2π/λ. We can rearrange and solve for λ;λ = 2π/kλ = 2π/3.25 m⁻¹λ = 1.94 m. Therefore, the wavelength of the oscillator is 1.94 m.
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A small glass marble hanging from a string is charged to +8.0nC. A 4.0 g plastic bead is gently released 4.0 cm beneath the marble. What charge in nC must the plastic bead have to hang suspended in space beneath the marble? Express your answer in nanocoulombs.
The charge required for the plastic bead to hang in space beneath the marble is q₂ = 7.84 nC. Charge on a small glass marble = q1 = +8.0 nC. Coordination of a 4.0 g plastic bead with respect to the marble = d = 4.0 cm = 4 × 10⁻² m.
We need to calculate the charge required for the plastic bead to hang in space beneath the marble.
Let this charge be q2.Step 1: Find the force of attraction between the two charges.
The formula to find the force of attraction between two charged particles is:F = k * q₁ * q₂ / r² where k = 9 × 10⁹ Nm²/C² is Coulomb's constant q₁ and q₂ are the charges of the particles r is the distance between the two particles.
For the plastic bead to hang suspended in space beneath the marble, the force of attraction between the two charged particles must balance the weight of the plastic bead.
Mathematically,F = mg where m is the mass of the plastic bead and g is the acceleration due to gravity.
F = 4.0 g × 9.8 m/s² = 39.2 × 10⁻³ N.
Thus,k * q₁ * q₂ / r² = 39.2 × 10⁻³ NK * 8.0 nC * q₂ / (0.04 m)² = 39.2 × 10⁻³ NK * 8.0 nC * q₂ / 1.6 × 10⁻³ m² = 39.2 × 10⁻³ Nq₂ = (39.2 × 10⁻³ N * 1.6 × 10⁻³ m²) / (8.0 nC)q₂ = 7.84 nC.
The charge required for the plastic bead to hang in space beneath the marble is q₂ = 7.84 nC.
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How long does it take an automobile traveling in the left lane of a highway at 45.0 km/h to overtake (become even with) another car that is traveling in the right lane at 35.0 km/h when the cars' front bumpers are initially 75 m apart?
It will take 27 seconds for the automobile traveling in the left lane of a highway at 45.0 km/h to overtake another car that is traveling in the right lane at 35.0 km/h when the cars' front bumpers are initially 75 m apart.
In order to determine the time that it takes an automobile traveling in the left lane of a highway at 45.0 km/h to overtake another car that is traveling in the right lane at 35.0 km/h when the cars' front bumpers are initially 75 m apart, we will have to use the formula for relative velocity which is:
V= Va - Vb
Where;V= relative velocity
Va = Velocity of object A.
Vb = Velocity of object B.
First, we will convert 45.0 km/h to m/s, and we get;45.0 km/h = 12.5 m/s
Secondly, we will convert 35.0 km/h to m/s, and we get;35.0 km/h = 9.72 m/s.
Using the formula for relative velocity, we get;V= Va - VbV= 12.5 m/s - 9.72 m/sV= 2.78 m/s
We can now use the formula below to find the time (t) it will take for the two cars to become even with each other:
V= d / twhere;
V = 2.78 m/s (Relative Velocity)and
d = 75m (Initial distance)
Solving for t we get;
t = d / V = 75m / 2.78m/s
= 27s.
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A tennis ball of mass 57 g trivels with velocity <55,0,0>m/s toward a wall After bouncing off the wall, the tennis ball is observed to be traveling with velocty <. 49,0,0>m (a) Draw a diagram showing the intial and final momentum of the tennis ball. This will help you answer the following questions. (b) What is the change in the momentum of the tennis ball? Δ
r
=kg⋅m/s (c) What is the maditude of the change of momentum of the tennis baili? ∣A
P
∣
2
∣kgm/s (d) What is the chance in the magnitude of the tennis bali's momentum? A∣
P
∣= kg⋅m/s Note that the magnitude of the change of the vector momentum is large, while the change in the magnitude of the momentum is small.
A tennis ball of mass 57 g travels towards a wall with velocity <55,0,0>m/s. After bouncing off the wall, it has velocity <49,0,0>m/s. The change in momentum is -0.008 kg*m/s, and the magnitude of the change is 0.008 kg*m/s.
(a) In the diagram, the initial momentum of the tennis ball is represented by an arrow pointing to the right with a magnitude of 0.057 kg*m/s. The final momentum is represented by an arrow pointing to the left with a magnitude of 0.049 kg*m/s.
(b) The change in momentum of the tennis ball is calculated by subtracting the initial momentum from the final momentum: ΔP = 0.049 kg*m/s - 0.057 kg*m/s = -0.008 kg*m/s.
(c) The magnitude of the change in momentum is the absolute value of the change in momentum: |ΔP| = |-0.008 kg*m/s| = 0.008 kg*m/s.
(d) The change in the magnitude of the tennis ball's momentum is the difference between the magnitudes of the initial and final momenta: |Δ|P| = |0.057 kg*m/s| - |0.049 kg*m/s| = 0.008 kg*m/s.
Note that the magnitude of the change of the vector momentum is large (0.008 kg*m/s), while the change in the magnitude of the momentum is small (0.008 kg*m/s).
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According to Newton's law of gravity, where would you feel the strongest gravitational pull?
On Earth.
On a planet with twice the Earth's mass and half the Earth's radius.
On a planet with half the Earth's mass and twice the Earth's radius.
You would feel the same gravitational pull at all of these locations.
You would feel the strongest gravitational pull on a planet with twice the Earth's mass and half the Earth's radius.
The correct option to the given question is option b.
According to Newton's law of gravity, the strength of the gravitational pull is determined by two factors: the mass of the object and the distance between them. Since the law of gravity states that the gravitational force is directly proportional to the mass of the object, we can see that the gravitational pull is directly proportional to the object's mass. Therefore, the stronger the mass of the object, the stronger the gravitational force will be.
Additionally, the force of gravity is inversely proportional to the square of the distance between the two objects. Therefore, as the distance between two objects increases, the force of gravity decreases and vice versa. Given these two factors, we can calculate the gravitational pull at various locations on different objects.
Based on this information, it can be said that you would feel the strongest gravitational pull on a planet with twice the Earth's mass and half the Earth's radius. This is because the mass of the planet is much larger than that of Earth, and therefore the gravitational pull is much stronger. Additionally, the radius of the planet is much smaller, which means that the distance between the center of mass and the surface of the planet is much smaller, further increasing the strength of the gravitational pull.
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You may want to review Part A Find the equivalent resistance between points A and B shown in the figure( R
2
=3.1Ω,R
3
=3.1Ω,R
1
=3.1Ω,R
5
=6.6Ω, and R
6
=6.6Ω. . Consider R
1
=3.1Ω, Express your answer using two significant figures. X Incorrect; Try Again; 4 attempts remaining
The equivalent resistance between points A and B in the given circuit is approximately 4.68Ω.
To find the equivalent resistance between points A and B in the given figure, we need to analyze the series and parallel combinations of resistors.
Looking at the diagram, we can see that R2 and R3 are in series, as well as R5 and R6. Let's calculate the equivalent resistances for these series combinations:
R2 and R3 in series:
Rs1 = R2 + R3 = 3.1Ω + 3.1Ω = 6.2Ω.
R5 and R6 in series:
Rs2 = R5 + R6 = 6.6Ω + 6.6Ω = 13.2Ω.
Now, we have two parallel combinations: Rs1 and R1, and Rs2 and R4.
Rs1 and R1 in parallel:
Rp1 = (Rs1 * R1) / (Rs1 + R1) = (6.2Ω * 3.1Ω) / (6.2Ω + 3.1Ω) ≈ 2.07Ω.
Rs2 and R4 in parallel:
Rp2 = (Rs2 * R4) / (Rs2 + R4) = (13.2Ω * 3.1Ω) / (13.2Ω + 3.1Ω) ≈ 2.61Ω.
Finally, we have Rp1 and Rp2 in series:
Req = Rp1 + Rp2 ≈ 2.07Ω + 2.61Ω ≈ 4.68Ω.
Therefore, the equivalent resistance between points A and B is approximately 4.68Ω.
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A battery (AA) with the capacity of 2.0Ah (amp hr) is used to power a small 1.5 V DC motor. The DC motor power for optimal operation is 1Watt. a. Calculate the current needed to power the motor. b. Describe how electrical charge/energy is contained in a battery and how an electrical charge is released. c. How long can a 2.0 Ah battery power the 1.5V DC motor? d. How many electrons would have passed through the motor until the battery is fully depleted assuming that all electrons in the battery can be fully utilized?
Calculating the current needed to power the motor The power of the motor is given as 1WattVoltage of the motor is given as 1.5V Current needed to power the motor can be calculated as shown below: P=VI or V = P / I or I = P / V Putting values, I = 1Watt/1.5V = 0.67Amps
Therefore, the current needed to power the motor is 0.67 Amps.
How electrical charge/energy is contained in a battery and how an electrical charge is released Electrical charge is contained in a battery in the form of chemical energy. This chemical energy is stored in the battery’s cells. When an electric load is connected to the battery, the chemical energy is converted into electrical energy and then it flows in the circuit. A battery is designed in such a way that the chemical reaction is slowed down until the time it is connected to an electric load and the electrical energy is needed. In this way, the battery’s energy does not get wasted.
How long can a 2.0 Ah battery power the 1.5V DC motor? Given that the capacity of the battery is 2Ah and the voltage required by the motor is 1.5V. The energy stored in the battery is given as,
E = V * Q where, E = Energy stored V = Voltage Q = Capacity of the battery
Therefore,
E = 1.5V * 2Ah
= 3 Wh (Watt-hour)
Thus, a 2Ah battery can power a 1.5V motor for 2 hours, as
energy = Power * Time or
Time = Energy / Power
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A uniform electric field of magnitude 250 V/m is directed in the positive x-direction. A 12.0-mC charge moves from the origin to the point (x,y) = (20.0 cm, 50.0 cm).
a. What was the change in the potential energy of this charge?
b. Through what potential difference did the charge move?
A. The change in potential energy of the charge is 1.617 Joules , b. The charge moved through a potential difference of 134.75 Volts.
Calculate the change in potential energy (ΔPE) of a charge moving in an electric field, we can use the formula:
ΔPE = q * ΔV,
where q is the charge and ΔV is the potential difference.
Magnitude of the electric field (E) = 250 V/m
Charge (q) = 12.0 mC = 12.0 × 10^(-3) C
a. the change in potential energy, we need to determine the potential difference (ΔV) experienced by the charge. The potential difference can be calculated using the formula:
ΔV = E * d,
where E is the electric field magnitude and d is the displacement.
the charge moves from the origin (x = 0, y = 0) to the point (x,y) = (20.0 cm, 50.0 cm). The displacement can be calculated as:
d =[tex]\sqrt{ ((\triangle x)^2 + (\triangle y)^2),[/tex]
where Δx is the change in x-coordinate and Δy is the change in y-coordinate.
Δx = 20.0 cm = 20.0 ×[tex]10^{(-2)[/tex]m,
Δy = 50.0 cm = 50.0 × [tex]10^{(-2)[/tex] m.
Substituting the values into the formula, we have:
d = [tex]\sqrt {((20.0 * 10^{(-2))^2} + (50.0 * 10^{(-2))^2})[/tex]
=[tex]\sqrt {(0.04 + 0.25)[/tex]
=[tex]\sqrt (0.29)[/tex]
≈ 0.539 m.
we can calculate the potential difference:
ΔV = (250 V/m) * (0.539 m)
= 134.75 V.
b. The change in potential energy is given by:
ΔPE = q * ΔV
= (12.0 × [tex]10^{(-3)[/tex] C) * (134.75 V)
≈ 1.617 J.
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Calculate the shielding effectiveness of an enclosure when 10
∗
12 holes are created (grouped close to each other) with radius 7.4 mm. The frequency of the radiation is 100MHz and the material of the shield is Cu with thickness of 0.2 mm
Calculate the shielding effectiveness of an enclosure when 10∗12 holes are created (grouped close to each other) with radius 7.4 mm. The frequency of the radiation is 100MHz and the material of the shield is Cu with thickness of 0.2 mm. The shielding effectiveness of the enclosure is approximately 288.4 dB.
The shielding effectiveness of an enclosure can be calculated using the formula:
SE = 20 * log10(1 + (λ * N * d / A))
Where:
SE = Shielding Effectiveness
λ = Wavelength of radiation in meters
N = Number of holes
d = Distance between the holes in meters
A = Area of the enclosure in square meters
To calculate the shielding effectiveness, we need to determine the values of λ, N, d, and A.
First, let's calculate the wavelength (λ) of the radiation:
λ = c / f
where c is the speed of light in a vacuum (3 x 10^8 m/s) and f is the frequency of the radiation (100 MHz).
Substituting the values, we have:
λ = (3 x 10^8 m/s) / (100 x 10^6 Hz)
λ = 3 meters
Next, let's calculate the area (A) of the enclosure:
A = π * r^2
where r is the radius of the holes (7.4 mm).
Converting the radius to meters:
r = 7.4 mm / 1000
r = 0.0074 meters
Substituting the value of r, we have:
A = π * (0.0074 meters)^2
A = 0.000171 square meters
Now, let's calculate the distance (d) between the holes:
Since the holes are grouped close to each other, let's assume the distance between the centers of adjacent holes is equal to the diameter of one hole. Therefore, d = 2r.
Substituting the value of r, we have:
d = 2 * 0.0074 meters
d = 0.0148 meters
Finally, let's substitute the values of λ, N, d, and A into the formula for shielding effectiveness (SE):
SE = 20 * log10(1 + (λ * N * d / A))
SE = 20 * log10(1 + (3 meters * 10^12 holes * 0.0148 meters / 0.000171 square meters))
Calculating the expression inside the logarithm:
(3 meters * 10^12 holes * 0.0148 meters / 0.000171 square meters) = 2.61 x 10^14
Substituting this value into the formula:
SE = 20 * log10(1 + 2.61 x 10^14)
SE ≈ 20 * log10(2.61 x 10^14)
Using a calculator or logarithm tables, we can find the logarithm of 2.61 x 10^14 to be approximately 14.42.
Substituting this value into the formula:
SE ≈ 20 * 14.42
SE ≈ 288.4
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Refer to your text book or a web resource to compare its explanation of the double-slit
interference with your observations. Determine how the superposition of waves can give
rise to either constructive or destructive interference. Show that the small angle
approximation allows you to substitute y/D for sin θ in your calculation of fringe spacing.
What variables affect the spacing of the bright fringes in the double-slit pattern? Write
theoretical expressions used to determine the separation between dark and bright
interferences of the double-slit experiment.
The double-slit interference is a phenomenon that occurs when a coherent light source passes through two closely spaced slits, creating an interference pattern on a screen or detector placed behind the slits. The explanation of this phenomenon involves the superposition of waves.
When light passes through the slits, it diffracts and creates a pattern of overlapping wavefronts. These wavefronts interfere with each other, resulting in regions of constructive and destructive interference. Constructive interference occurs when the peaks of the waves align, leading to bright fringes or maximum intensity. Destructive interference occurs when the peaks and troughs of the waves cancel each other out, resulting in dark fringes or minimum intensity.
The small angle approximation allows us to substitute y/D for sin θ, where y is the fringe spacing, D is the distance between the slits and the screen, and θ is the angle of the fringe from the central maximum. This approximation is valid when the angle θ is small, and it simplifies the calculation of fringe spacing.
The variables that affect the spacing of the bright fringes in the double-slit pattern include the wavelength of light, the distance between the slits, and the distance between the slits and the screen.
The theoretical expression used to determine the separation between dark and bright interferences in the double-slit experiment is given by y = (λD) / d, where λ is the wavelength of light, D is the distance between the slits and the screen, and d is the distance between the two slits. This expression relates the fringe spacing (y) to the other variables involved in the experiment.
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A roller-coaster car with mass of 760 kgkg is launched from a giant spring of constant 25 kN/mkN/m into a frictionless loop-the-loop track of radius 7.3 mm, as shown in the figure.What is the minimum amount that the spring must be compressed if the car is to stay on the track?
The minimum amount that the spring must be compressed for the car to stay on the track is approximately 1.35 meters.
Mass of the car (m) = 760 kg
Spring constant (k) = 25 kN/m (Note: 1 kN/m = 1000 N/m)
Radius of the loop-the-loop track (r) = 7.3 m
Acceleration due to gravity (g) = 9.8 m/s²
To find the minimum compression of the spring (x), we need to determine the point where the normal force (N) is equal to the centripetal force.
At the top of the loop, the normal force (N) and the gravitational force (mg) add up to provide the centripetal force.
The centripetal force (F_centripetal) is given by:
F_centripetal = (m * v²) / r
The normal force (N) is given by:
N = mg + F_centripetal
Substituting the expressions for F_centripetal and N:
N = mg + (m * v²) / r
N = mg + (m * (kx² / m)) / r
Simplifying the equation:
N = mg + (kx²) / r
Now, let's calculate the minimum compression of the spring (x) by finding the point where the normal force is equal to the centripetal force:
N = F_centripetal
mg + (kx²) / r = (m * v²) / r
mg + (kx²) / r = (m * (kx²) / m) / r
Simplifying further:
mg = (kx²) / r
Now, let's solve for x:
x² = (mgr) / k
x = sqrt((mgr) / k)
Substituting the given values into the equation:
x = sqrt((760 kg * 9.8 m/s² * 7.3 m) / (25,000 N/m))
Calculating the value:
x ≈ 1.35 meters
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If
d
1
=3.11
−1.92
+3.97
κ
and
d
2
=−4.81
+2.17
−
k
, then what is (
d
1
+
d
2
) ) (
d
1
×4
d
2
)? Tries 0/99
We are given two different terms d1 and d2.
the answer is (−255λ + 37.5λμ + 443.5k − 32.5μk − 52.92λμk + 28.16μ²k − 42.3λk² + 44.52k³).
We need to find the value of (d1 + d2) × (d1 × 4d2).
We have, d1 = 3.11λ − 1.92μ + 3.97k and
d2 = −4.81λ + 2.17μ − k
Then, (d1 + d2) = (3.11λ − 1.92μ + 3.97k) + (−4.81λ + 2.17μ − k)= −1.7λ + 0.25μ + 2.97k
Now, d1 × 4d2 = (3.11λ − 1.92μ + 3.97k) × 4(−4.81λ + 2.17μ − k)= 150 − 13.28λμ + 6.36μk − 15.88λk + 16.76k²
Thus, (d1 + d2) × (d1 × 4d2) = (−1.7λ + 0.25μ + 2.97k) × (150 − 13.28λμ + 6.36μk − 15.88λk + 16.76k²)= −255λ + 37.5λμ + 443.5k − 32.5μk − 52.92λμk + 28.16μ²k − 42.3λk² + 44.52k³
So, (d1 + d2) × (d1 × 4d2) is −255λ + 37.5λμ + 443.5k − 32.5μk − 52.92λμk + 28.16μ²k − 42.3λk² + 44.52k³.
Hence, the answer is (−255λ + 37.5λμ + 443.5k − 32.5μk − 52.92λμk + 28.16μ²k − 42.3λk² + 44.52k³).
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A train is traveling up a 2.68
∘
incline at a speed of 4.51 m/s when the Part A last car breaks tree and begins to coast without friction. How long does it take for the last car to come to rest momentanly? Express your answer using two signiffcant figures. Part B How far did the last car travel before momentarily coming to rest? Express your answer using two significant figureu.
The distance traveled by the last car before coming to rest momentarily is 227.3 m.
Given data Train is traveling up a 2.68∘ incline at a speed of 4.51 m/sPart A: Calculate time taken by last car to come to restmomentarilyInitial velocity (u) of the last car = 4.51 m/sFinal velocity (v) of the last car = 0 m/sAcceleration (a) of the last car can be calculated by resolving the gravitational force into the parallel and perpendicular components to the incline:
a = g sin θ
= 9.8 m/s² × sin(2.68°)
= 0.448 m/s².
Using the equation of motion:
v = u + at0
= 4.51 + (-0.448)t Solving for t, we get
t = 10.06 s Hence, the time taken by the last car to come to rest momentarily is 10.06 s Part B: Calculate the distance traveled by the last car before coming to rest momentarily Using the equation of motion:
s = ut + 1/2 at²
s = 4.51 × 10.06 + 0.5 × 0.448 × (10.06)²
s = 227.3 m Hence, the distance traveled by the last car before coming to rest momentarily is 227.3 m.
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Suppose the average current passing through a device in 0.5 h is 336 mA. Determine the total charge. 1.87×10
−4
C
3.63×10
4
C
0.168C
605C
168C 10.1C
336C
5.35×10
3
C
The total charge passing through the device in 0.5 hours is determined to be 3.63×10^4 C.
To calculate the total charge, we can use the equation Q = I × t, where Q is the charge, I is the current, and t is the time.
Given that the average current passing through the device is 336 mA (milliamperes) and the time is 0.5 hours, we need to convert the current to amperes and the time to seconds to ensure consistent units.
Converting the current from milliamperes to amperes, we divide by 1000, resulting in 0.336 A.
Converting the time from hours to seconds, we multiply by 3600, as there are 3600 seconds in an hour, giving us 1800 seconds.
Substituting the values into the equation Q = I × t, we can calculate the total charge:
Q = 0.336 A × 1800 s = 604.8 C.
Therefore, the total charge passing through the device in 0.5 hours is approximately 604.8 C, which can be rounded to 3.63×10^4 C in scientific notation.
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A 12.0-V battery is connected in series with a capacitor. If the capacitance of the capacitor is 0.43 µF, how much charge is on each of its plates?
The charge on each of the plates of the capacitor is 5.16 x 10^-6 C.
A capacitor is a passive electronic component with two terminals. It stores electrical energy in an electric field by virtue of accumulating electric charges on two close surfaces insulated from each other. The effect of a capacitor is known as capacitance.
Given a 12.0-V battery connected in series with a capacitor having a capacitance of 0.43 µF
Step-by-step explanation :
The capacitance of the capacitor is given to be 0.43 µF.
The battery voltage is given to be 12.0 V.
Let V be the voltage across the capacitor.
Then, charge (Q) on each of the plates of the capacitor is given by the equation Q = CV
where, C is the capacitance of the capacitor and V is the voltage across the capacitor.
Substituting the given values, we have :
Q = (0.43 x 10^-6 F) × (12.0 V)
Q = 5.16 x 10^-6 C
Therefore, the charge on each of the plate = 5.16 x 10^-6 C.
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Explain relativistic effects experienced when travelling at speeds greater than 0.10c with respect to: a) Time dilation b) Length Contraction c) Momentum
The Theory of Relativity explains relativistic effects experienced at high velocities, including time dilation, length contraction, and increased momentum, with time passing more slowly, objects appearing shorter, and momentum increasing as speeds approach the speed of light.
The Theory of Relativity is a physical law that explains the world around us. Its implications, when applied to situations of high velocities, cause some surprising outcomes. When travelling at speeds greater than 0.10c, these outcomes are called relativistic effects. Three relativistic effects are time dilation, length contraction, and momentum.
Time dilation refers to the notion that time is relative and not absolute. Time dilation occurs when one object is moving relative to another object. When travelling at high speeds, time seems to pass more slowly. This phenomenon can be observed in the twin paradox. Consider two twins. One twin remains on Earth while the other twin is sent on a spaceship travelling near the speed of light. When the travelling twin returns, they will have aged less than their Earthbound sibling.
Length contraction refers to the fact that an object's length is not absolute. The length of an object can change based on the speed at which it is travelling. At speeds approaching the speed of light, an object will appear to be shorter. This phenomenon can be observed when examining muons, which are subatomic particles.
Momentum refers to the idea that an object in motion has a certain amount of force behind it. When an object is travelling at high speeds, its momentum increases. The faster an object moves, the more mass it has. This means that its momentum also increases. At speeds approaching the speed of light, the momentum of an object can be incredibly high.These three relativistic effects — time dilation, length contraction, and momentum — all occur when an object is travelling at speeds greater than 0.10c. They are fundamental concepts in the Theory of Relativity.
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Q3. Given the vectors A=2.00
∧
i+6.00
∧
j and B=3.00
∧
i−2.00
∧
j, Find (a) C=A+B (b) the vector difference D=A−B
The Vectors are: A) C = 5.00i + 4.00j and B) D = -1.00i + 8.00j.
A vector is a mathematical quantity that represents both magnitude (size) and direction. It is typically represented by an arrow, where the length of the arrow represents the magnitude and the direction of the arrow represents the direction.
In physics, vectors are used to describe quantities such as displacement, velocity, acceleration, force, and more. Vectors have both magnitude and direction, which means they convey not only how much of a quantity is present but also the direction in which it acts.
(a) To find C = A + B, we simply add the corresponding components of vectors A and B:
C = (2.00i + 6.00j) + (3.00i - 2.00j)
= (2.00 + 3.00)i + (6.00 - 2.00)j
= 5.00i + 4.00j
Therefore, C = 5.00i + 4.00j.
(b) To find D = A - B, we subtract the corresponding components of vectors A and B:
D = (2.00i + 6.00j) - (3.00i - 2.00j)
= (2.00 - 3.00)i + (6.00 + 2.00)j
= -1.00i + 8.00j
Therefore, D = -1.00i + 8.00j.
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Determine the magnitude of this force. 6. A point charge of +3.00μC is 12.0 cm distant from a second point charge of −1.50×10−6C. Calculate the magnitude of the force on each charge.
The magnitude of the force exerted by Q2 on Q1 is given by Coulomb's Law as follows:F21= kq1q2r2
where k is Coulomb's constant = 9 × 10^9 N m^2 C^-2
The distance between point charges, r is given in cm. Convert it to m as follows:r = 12.0 cm = 12.0 × 10^-2 mThen the magnitude of the force between the two charges is:F21= kq1q2r2
= 9 × 10^9 N m^2 C^-2 × +3.00 μC × -1.50 × 10^-6C (12.0 × 10^-2 m)^2
= -405 N
The negative sign indicates that the force between the two charges is attractive rather than repulsive. Since the force on Q2 is equal and opposite to that on Q1, the magnitude of the force on Q2 is also 405 N.The magnitude of the force on Q1 is 405 N and the magnitude of the force on Q2 is also 405 N.Therefore, magnitude of the force on Q1 is 405 N and the magnitude of the force on Q2 is also 405 N.
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An aircraft has a wing area of 20 m² and whose wings resemble the NACA 23012 with [13] no flaps and is flying horizontally (0° angle of attack) at a constant speed of 250 km/h. To gain height the pilot adjusts the controls so that the angle of attack becomes 10°. Take the density of the air as 1.23 kg/m³. Determine the total power required to execute this action at the same constant speed.
The total power needed for the aircraft to execute this action at the same constant speed is 80.91 KW.
Aircraft:
The lift produced by an airplane is an essential factor in determining the aircraft's takeoff and landing capabilities and is proportional to the wing area. The wing area of the aircraft is 20 m², and its wings are similar to the NACA 23012 [13]. Therefore, the aircraft is flying at a constant speed of 250 km/h, and to gain altitude, the pilot adjusts the controls to an angle of attack of 10°.
Calculate the aircraft's total power requirements at the same constant speed if the density of the air is 1.23 kg/m³.
Solution:
Total power is equal to the sum of the induced power and the parasitic power.
PT = PI + PP
Step 1: Find the induced power, PI
Power can be calculated using the following equation:
PI = CL × ½ρV³ × S
where,
CL = lift coefficient of the wing
ρ = density of air
V = velocity of the airplane
S = surface area of the wing
Therefore,
PI = (1.5) × ½ (1.23 kg/m³) × (69.44 m/s)³ × 20 m²
PI = 47,810 Watt or 47.81 KW
Step 2: Find the parasitic power, PP
The parasitic power can be computed using the following formula:
PP = Cd × ½ρV³ × S
where,
Cd = coefficient of drag of the wing
ρ = density of air
V = velocity of the airplane
S = surface area of the wing
Therefore,
PP = 0.024 × ½ (1.23 kg/m³) × (69.44 m/s)³ × 20 m²
PP = 33,102 Watt or 33.10 KW
Step 3: Total power, PT
Power required for the airplane to increase its altitude to a 10° angle of attack at the same constant speed is given by,
PT = PI + PP
Total power, PT = 47.81 KW + 33.10 KW
PT = 80.91 KW
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A batter hit a baseball so that it leaves the bat at speed v
o
=40.0 m/s at an angle α
0
=30.0
∘
. (a) Find the position of the ball at t=3.00 s. (b) Find the time when the ball reaches the highest point of its flight, and its height H at this time. (c) Find the horizontal distance R from the starting point to where the ball hits the ground and the ball's velocity just before it hits.
(a) The position of the ball at time, t = 3 s is y = -24.1 m and x = 103.92 m.
(b) The time when the ball reaches highest point is 2.04 s. And the maximum height reached by the ball is 20.4 m
(c) The horizontal distance of the ball is 141.4 m.
What is the position of the ball?(a) The position of the ball at time, t = 3 s is calculated as follows;
vertical position of the ball;
y = (40 m/s x sin30) x (3 s) - ¹/₂(9.8 m/s²) x ( 3 s)²
y = -24.1 m
horizontal position of the ball;
x = ( 40 m/s x cos 30) x ( 3 s)
x = 103.92 m
(b) The time when the ball reaches highest point is calculated as;
v = u - gt
where;
v is the final velocity = 0, at highest pointu is the initial velocity0 = u - gt
gt = u
t = u/g
t = (40 x sin 30)/9.8
t = 2.04 s
The maximum height reached by the ball is calculated as;
H = (40 m/s x sin30) x (2.04 s) - ¹/₂(9.8 m/s²) x ( 2.04 s)²
H = 20.4 m
(c) The horizontal distance of the ball is calculated as follows;
R = u²sin(2θ) / g
R = (40² x sin(2 x 30) ) / 9.8
R = 141.4 m
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An electron with an initial speed of 282,000 m/s is brought to rest by an electric field. What was the potential difference that stopped the electron? (Express your answer in volts to three decimal places) An electron with an initial speed of 282,000 m/s is brought to rest by an electric field. What was the potential difference that stopped the electron? (Express your answer in volts to three decimal places)
Therefore, the potential difference that stopped the electron is 24131.25 V.
Given that, the initial speed of the electron (u) = 282000 m/s
To find the potential difference (V) that stopped the electron, we need to use the formula for work done by electric field:
W = qV
where q is the charge of the electron.
W = K.E. of the electron (initial) - K.E. of the electron (final)
K.E. of the electron (initial) = 1/2
mu² = 1/2(9.11 x 10⁻³¹ kg) (282000 m/s)²
K.E. of the electron (initial) = 3.861 J
Since the electron is brought to rest, the final kinetic energy is zero.
K.E. of the electron (final) = 0J
Therefore, the work done by the electric field (W) is equal to the initial kinetic energy of the electron.
W = 3.861 J
Let the potential difference be V volts. q = charge of the electron = -1.6 x 10⁻¹⁹ C.W = qV
Therefore, V = W/q
Substituting the values, we get
V = 3.861 J / (-1.6 x 10⁻¹⁹ C)V = -24131.25 V
Since potential difference cannot be negative, we take the absolute value of the result:
The potential difference is 24131.25 V.
The potential difference that stopped the electron is 24131.25 V.
An electron with an initial speed of 282,000 m/s is stopped by a potential difference of 24131.25 V.
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Two balls collide, one of mass m
1
moving at a velocity of 10 m s
−1
and the other, of mass m
2
=3m
1
, at rest. Calculate the velocity of m
2
after collision if the collision is perfectly elastic.
The final velocity of object 2 is v2 = 150/4 = 37.5 m/s.
Given: Mass of object m1= m2 and its velocity u1= 10m/s,
Second object m2= 3m1 and its initial velocity u2= 0
To find : Final velocity of object 2 after perfectly elastic collision.
In an elastic collision, both momentum and kinetic energy are conserved.
Thus, according to law of conservation of momentum,
we have,m1u1 + m2u2 = m1v1 + m2v2 ........
(1)where, m1, u1 and v1 are mass, initial velocity and final velocity of object 1 while m2, u2 and v2 are mass, initial velocity and final velocity of object 2 respectively.
Now substituting the given values in the above equation,
we get,
m1u1 + m2u2 = m1v1 + m2v2 => m1(10) + m2(0) = m1(v1) + m2(v2) => 10m1 = m1v1 + m2v2 => 10m1 = m1v1 + 3m1v2 => 10 = v1 + 3v2 => 10 = v2 (3/4)
(since v1 = 10 - 3v2 and m1 = m2/3)Thus,
the final velocity of object 2 is
v2 = 150/4
= 37.5 m/s.
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