Answer:
Not Changed
Explanation:
To know what happened with the volume you need to know the Ideal gas Law
[tex]\frac{P_{1}V_{1}}{T1} =\frac{P_{2}V_{2} }{T2}[/tex]
This law is a combination of the other four laws: Boyles's, Charles's, Avogadro's, and Guy-Lussac's.
The initial state is represented by P1, V1, T1 and the final by P2, V2, T2.
In this case:
[tex]T_{2} =3T_{1} \\P_{2} =3P_{1}[/tex]
Replacing on the equation
[tex]\frac{P_{1}V_{1}}{T1} =\frac{3P_{1}V_{2}}{3T_{1} }[/tex]
If we clear from the equation V2
[tex]\frac{P_{1}V_{1}3T_{1}}{T_{1} 3P_{1}} ={V_{2}}[/tex]
Then cancel both P1 and T1
[tex]\frac{3V_{1}}{3} =V_{2}[/tex]
You will found that
[tex]V_{1} =V_{2}[/tex]
A seismographic station receives S and P waves from an earthquake, separated in time by 17.8 s. Assume the waves have traveled over the same path at speeds of 4.50 km/s and 6.90 km/s. Find the distance from the seismograph to the focus of the quake.
Answer:
D = 230.2 Km
Explanation:
let distance between seismograph and focus of quake is D
From time distance formula we can calculate the time taken by the S wave
[tex]T_1 =\frac{D}{4.5}[/tex]
From time distance formula we can calculate the time taken by the P wave
[tex]T_2 =\frac{D}{6.90}[/tex]
It is given in equation both waves are seperated from each other by 17.8 sec
so we have
[tex]T_1 - T_2 = 17.8[/tex]sec
Putting both time value to get distance value
[tex]\frac{D}{4.5} - \frac{D}{6.90} = 17.8[/tex]
D = 230.2 Km
3. Which statement is not true of simple machines?
A They can change the direction in which you apply a force
B They can change the amount of force you exert.
CThey can change the amount of work done on an object.
D. They can change the distance over which you apply a force
Answer:
C They can change the amount of work done on an object. Controlling amount of work done would be down to user / more complex machines.
Explanation:
Simple machines help direct, increase, and affect distance force is applied.
Simple machines do not change the amount of work done but can alter the force applied and distance over which the force is exerted.
Simple machines do not change the amount of work done. Although they cannot alter the work done, they can change the amount of force applied and the distance over which the force is exerted. For instance, a simple machine like a lever can help reduce the force needed to lift an object.
An object is dropped onto the moon (gm = 5 ft/s2). How long does it take to fall from an elevation of 250 ft.?
Answer:
10 seconds
Explanation:
x = x₀ + v₀ t + ½ at²
250 = 0 + (0) t + ½ (5) t²
250 = 2.5 t²
t² = 100
t = 10
It takes 10 seconds to land from a height of 250 ft.
Answer:
10 seconds
Explanation:
It takes 10 seconds for an object to fall from an elevation of 250 ft.
250 = 0 + (0) t + ½ (5) t²
Two parallel plate capacitors are identical, except that one of them is empty and the other contains a material with a dielectric constant of 4.2 in the space between the plates. The empty capacitor is connected between the terminals of an ac generator that has a fixed frequency and rms voltage. The generator delivers a current of 0.29 A. What current does the generator deliver after the other capacitor is connected in parallel with the first one?
Answer:
current = 1.51 A
Explanation:
Initially the capacitor without any dielectric is connected across AC source
so the capacitive reactance of that capacitor is given as
[tex]x_c = \frac{1}{\omega c}[/tex]
now we have
[tex]i = \frac{V_{rms}}{x_c}[/tex]
here we know that
[tex]i = 0.29 A[/tex]
now other capacitor with dielectric of 4.2 is connected in parallel with the first capacitor
so here net capacitance is given as
[tex]c_{eq} = 4.2c + c = 5.2c[/tex]
now the equivalent capacitive reactance is given as
[tex]x_c' = \frac{1}{\omega(5.2c)}[/tex]
[tex]x_c' = \frac{x_c}{5.2}[/tex]
so here we have new current in that circuit is given as
[tex]i' = \frac{V_{rms}}{x_c'}[/tex]
[tex]i' = 5.2 (i) = 5.2(0.29)[/tex]
[tex]i' = 1.51 A[/tex]
A car drives over the top of a hill that has a radius of 40 m. ? Part A What maximum speed can the car have without flying off the road at the top of the hill?
Answer:
Maximum speed = 19.81 m/s
Explanation:
Maximum speed can the car have without flying off the road at the top of the hill.
For this condition to occur we have
Centripetal force ≥ Weight of car.
[tex]\frac{mv^2}{r}\geq mg[/tex]
For maximum speed without flying we have
[tex]\frac{mv^2}{r}=mg\\\\\frac{v^2}{r}=g\\\\v=\sqrt{rg}=\sqrt{40\times 9.81}=19.81m/s[/tex]
Maximum speed = 19.81 m/s
The maximum speed of the car on top of hill is 19.8 m/s.
The given parameters;
radius of the hill, r = 40 m
The maximum speed of the car on top of hill is calculated as follows;
the centripetal force must be equal or greater than weight of the car.
[tex]F_c = mg\\\\\frac{mv^2}{r} = mg\\\\\frac{v^2}{r} = g\\\\v^2 = rg\\\\v = \sqrt{rg} \\\\v = \sqrt{40 \times 9.8} \\\\v = 19.80 \ m/s[/tex]
Thus, the maximum speed of the car on top of hill is 19.8 m/s.
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An aquarium open at the top has 30-cm-deep water in it. You shine a laser pointer into the top opening so it is incident on the air-water interface at a 45∘ angle relative to the vertical. You see a bright spot where the beam hits the bottom of the aquarium. The index of refraction of water is 1.33. How much water (in terms of height) should you add to the tank so the bright spot on the bottom moves 5.0 cm?
Final answer:
To find out how much water should be added to move the bright spot by 5.0 cm, apply Snell's Law to find the angle of refraction and use trigonometry to determine the change in horizontal distance of the spot due to the change in water depth.
Explanation:
The student is asking about the behavior of light as it passes through different media, specifically how the depth of the water in an aquarium would need to change to move the bright spot created by a laser. The scenario of shining a laser at an angle into water involves Snell's Law, which describes how light bends at the interface between two media with different indices of refraction. To calculate the necessary increase in water height to move the bright spot by 5.0 cm, we need to apply the principles of refraction and trigonometry.
First, we determine the initial angle of refraction using Snell's Law:
nairsin(θair) = nwatersin(θwater)1.00sin(45°) = 1.33sin(θwater)
Upon finding the angle of refraction θwater, we can use trigonometric functions to calculate the initial horizontal distance (Xinitial) from the point of incidence to the bright spot. To find the new distance (Xfinal) after adding water, we repeat the process with the new depth. The difference in the horizontal distances gives the amount the bright spot moved, 5.0 cm. Adjusting the depth to match this movement provides the answer to the student's question.
A vibratory system in a vehicle is to be designed with the following parameters: k-295 N/m, C-2N-s/m, m-13 kg. Calculate the natural frequency of damped vibration
Answer:
[tex]w_{damped}= 4.76[/tex] s^-1
Explanation:
The mathematical relationship is
[tex]w_{damped}=w_{undamped} *\sqrt{1-(\frac{c}{2\sqrt{km}})^{2}}[/tex]
where:
c is the damper constant
k is the spring constant
m is the mass
ω_undamped is the natural frequency
ω_damped is the damped frequency
[tex]w_{undamped} =\sqrt{\frac{k}{m}}=4.79[/tex] s^-1
[tex]w_{damped}= 4.79 *\sqrt{1-(\frac{2}{2\sqrt{295*13}})^{2}}[/tex]
[tex]w_{damped}= 4.76[/tex] s^-1
A superball and a clay ball are dropped from a height of 10cm above a tabletop. They have the same mass 0.05kg and the same size. The superball bounces off the table and rises back to the same height. The clay ball sticks to the table. For the superball in the previous question, if it was in contact with the table for 34.3ms, calculate the average force exerted on the ball by the table. Hint: First calculate the momentum before and after hitting the table. Don't forget the gravitational force.
Answer:4.08 N
Explanation:
Given data
superball dropped from a height of 10 cm
Mass of ball[tex]\left ( m\right )[/tex]=0.05kg
time of contact[tex]\left ( t\right )[/tex]=34.3[tex]\times 10^{-3}[/tex] s
Now we know impulse =[tex]Force\times time\ of\ contact[/tex]=Change in momentum
[tex]F_{average}\times t[/tex]=[tex]m\left ( v-(-v)\right )[/tex]
and velocity at the bottom is given by
v=[tex]\sqrt{2gh}[/tex]
[tex]F_{average}\times 34.3\times 10^{-3}[/tex]=[tex]0.05\left ( 1.4-(-1.4)\right )[/tex]
[tex]F_{average}[/tex]=4.081N
To find the average force exerted on the superball by the table, we calculate its change in momentum during the bounce and divide by the contact time. The result is an approximate average force of 4.08 N upward.
The question involves a superball and a clay ball, both with the same mass of 0.05 kg, dropped from a height of 10 cm. The superball bounces back to its original height, while the clay ball sticks to the table. We need to calculate the average force exerted on the superball by the table.
Step-by-Step Solution
Calculate the velocity of the superball just before hitting the table: Using the equation for free fall, v = √(2gh), where g is 9.8 m/s² (acceleration due to gravity) and h is 0.10 m.
v = √(2 * 9.8 * 0.10)
v = √(1.96)
v ≈ 1.40 m/s
Determine the change in momentum: Before impact, the momentum is [tex]P_{before[/tex] = m * v. Since the superball bounces back with the same speed, the momentum after impact is [tex]P_{after[/tex] = -m * v because the direction changes.
[tex]P_{before[/tex] = 0.05 kg * 1.40 m/s = 0.07 kg·m/s
[tex]P_{after[/tex] = 0.05 kg * (-1.40 m/s) = -0.07 kg·m/s
Calculate the impulse: Impulse is the change in momentum, so Impulse = [tex]P_{after[/tex] - [tex]P_{before[/tex].
Impulse = -0.07 kg·m/s - 0.07 kg·m/s
Impulse = -0.14 kg·m/s
Calculate the average force: The impulse-momentum theorem states that Impulse = [tex]F_{avg[/tex] * Δt, where [tex]F_{avg[/tex] is the average force and Δt is the time of contact (34.3 ms = 0.0343 s). Solving for [tex]F_{avg[/tex] ,
[tex]F_{avg[/tex] = Impulse / Δt
[tex]F_{avg[/tex] = -0.14 kg·m/s / 0.0343 s
[tex]F_{avg[/tex] ≈ -4.08 N
The magnitude of the average force exerted on the ball by the table is approximately 4.08 N in the upward direction.
An uncharged 5.0-µF capacitor and a resistor are connected in series to a 12-V battery and an open switch to form a simple RC circuit. The switch is closed at t = 0 s. The time constant of the circuit is 4.0 s. What is the charge on either plate after one time constant has elapsed?
For a direct current resistor-capacitor circuit where the capacitor is initially uncharged, the charge stored on one of the capacitor's plates is given by:
Q(t) = Cℰ(1-e^{-t/(RC)})
Q(t) is the charge, t is time, ℰ is the battery's terminal voltage, R is the resistor's resistance, and C is the capacitor's capacitance.
The time constant of the circuit τ is the product of the resistance and capacitance:
τ = RC
Q(t) can be rewritten as:
Q(t) = Cℰ(1-e^{-t/τ})
We want to know how much charge is stored when one time constant has elapsed, i.e. what Q(t) is when t = τ. Let us plug in this time value:
Q(τ) = Cℰ(1-e^{-τ/τ})
Q(τ) = Cℰ(1-1/e)
Q(τ) = Cℰ(0.63)
Given values:
C = 5.0×10⁻⁶F
ℰ = 12V
Plug in these values and solve for Q(τ):
Q(τ) = (5.0×10⁻⁶)(12)(0.63)
Q(τ) = 3.8×10⁻⁵C
The charge on either plate of given capacitor after one time constant has elapsed is 3.8×10⁻⁵C.
The charge stored on one of the capacitor's plates can be calculated by,
[tex]\bold{Q(t) = C\epsilon(1-e^{-t/(RC)})}[/tex]
Where,
Q(t) - the charge,
t - time,
ℰ- the battery's terminal voltage,
R - the resistor's resistance,
C - the capacitor's capacitance.
The time constant of the circuit τ is equal to the product of the resistance R and capacitance C :
τ = RC
The amount of charge Q(t) when t = τ. put the values,
Q(τ) = Cℰ(1-e^{-τ/τ})
Q(τ) = Cℰ(1-1/e)
Q(τ) = Cℰ(0.63)
Given values:
C = 5.0×10⁻⁶F
ℰ = 12V
Put the values in the formula and solve for Q(τ):
Q(τ) = (5.0×10⁻⁶)(12)(0.63)
Q(τ) = 3.8×10⁻⁵C
Therefore, the charge on either plate of given capacitor after one time constant has elapsed is 3.8×10⁻⁵C.
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An electric current heats a 221 g (0.221 kg) copper wire from 20.0 °C to 38.0 °C. How much heat was generated by the current? (Ccopper = 0.093 kcal/kgC°)
Answer:
Heat generated by the current = 1547.89 J
Explanation:
We have equation for heat energy H = mCΔT
Mass of copper = 0.221 kg
Specific heat of copper = 0.093 kcal/kgC° = 389.112 J/kgC°
ΔT = 38 - 20 = 18°C
Substituting in H = mCΔT
H = 0.221 x 389.112 x 18 = 1547.89 J
Heat generated by the current = 1547.89 J
The heat generated by the electric current that heated a 221 g copper wire from 20.0 °C to 38.0 °C is calculated to be 1.542 kJ, using the specific heat capacity of copper and the formula Q = mcΔT.
An electric current heats a 221 g (0.221 kg) copper wire from 20.0 °C to 38.0 °C. To calculate the heat generated by the current, we use the formula for heat energy, Q = mcΔT, where m is the mass in kg, c is the specific heat capacity in kcal/kg°C, and ΔT is the change in temperature in °C.
Given:
m = 0.221 kg
c = 0.093 kcal/kg°C
ΔT = (38.0 - 20.0) °C = 18.0 °C
Substituting the values into the formula:
Q = 0.221 kg * 0.093 kcal/kg°C * 18.0 °C = 0.368658 kcal
To convert kcal to joules (since 1 kcal = 4.184 kJ),
Q = 0.368658 kcal * 4.184 kJ/kcal = 1.542 kJ
Therefore, the heat generated by the electric current is 1.542 kJ.
If a 0.9 kg ball is dropped from a height of 7 m, what is its kinetic energy when it hits the ground?
Answer:
Kinetic energy, [tex]E_k=61.74\ J[/tex]
Explanation:
It is given that,
Mass of ball, m = 0.9 kg
It is dropped form a height of 7 m, h = 7 m
When the ball is at certain height it will have only potential energy. As it hits the ground, its potential energy gets converted to kinetic energy as per the law of conservation of energy.
So, [tex]E_k=PE=mgh[/tex]
[tex]E_k=0.9\ kg\times 9.8\ m/s^2\times 7\ m[/tex]
[tex]E_k=61.74\ J[/tex]
So, the kinetic energy when it hits the ground is 61.74 Joules.
The kinetic energy of a 0.9 kg ball dropped from a height of 7 m is calculated using the conservation of mechanical energy, resulting in 61.74 Joules when it hits the ground.
The kinetic energy of a 0.9 kg ball when it hits the ground after being dropped from a height of 7 m. The principle used to determine this is the conservation of mechanical energy, which states that the sum of the potential and kinetic energies in a system remains constant if only conservative forces are doing work.
When the ball is at the height of 7 m, it posses gravitational potential energy (GPE) which can be calculated using the formula GPE = mgh, where m is the mass of the ball, g is the acceleration due to gravity (9.8 m/s²), and h is the height from which the ball is dropped.
Since the ball is dropped from rest, its initial kinetic energy is 0. When the ball reaches the ground, its potential energy is converted into kinetic energy.
Therefore, the kinetic energy (KE) of the ball when it hits the ground can be calculated using the potential energy formula, assuming no energy is lost:
KE = mgh = 0.9 kg x 9.8 m/s² x 7 m = 61.74 Joules.
This means that the ball will have a kinetic energy of 61.74 Joules when it impacts the ground.
A 300.0 W immersion heater is used to heat 0.250 kg of water from 10.0 °C to 70.0 °C. About how many seconds did this take? (There are 4186 Joules per kcal.)
Answer:
209.3 seconds
Explanation:
P = 300 W, m = 0.250 kg, T1 = 10 degree C, T2 = 70 degree C
c = 4186 J / kg C
Heat given to water = mass x specific heat of water x rise in temperature
H = 0.250 x 4186 x (70 - 10)
H = 62790 J
Power = Heat / Time
Time, t = heat / Power
t = 62790 / 300 = 209.3 seconds
It took approximately [tex]\( 209.3 \)[/tex] seconds to heat the water from [tex]\( 10.0^\circ \text{C} \)[/tex] to [tex]\( 70.0^\circ \text{C} \)[/tex].
To determine how many seconds it took to heat the water using the immersion heater, we need to calculate the amount of energy required and then use the power of the heater to find the time.
First, calculate the change in temperature of the water:
[tex]\[ \Delta T = 70.0^\circ \text{C} - 10.0^\circ \text{C} = 60.0^\circ \text{C} \][/tex]
Next, calculate the energy [tex]\( Q \)[/tex] required to heat the water using the specific heat capacity of water [tex]\( C = 4186 \text{ J/kg}^\circ \text{C} \)[/tex]:
[tex]\[ Q = mc\Delta T \][/tex]
where [tex]\( m \)[/tex] is the mass of water and [tex]\( c \)[/tex] is the specific heat capacity of water.
Given:
[tex]\[ m = 0.250 \text{ kg} \][/tex]
[tex]\[ c = 4186 \text{ J/kg}^\circ \text{C} \][/tex]
[tex]\[ \Delta T = 60.0^\circ \text{C} \][/tex]
[tex]\[ Q = 0.250 \times 4186 \times 60.0 \][/tex]
[tex]\[ Q = 62790 \text{ Joules} \][/tex]
Now, calculate the time t required using the power P of the heater:
[tex]\[ P = 300.0 \text{ W} \][/tex]
The time [tex]\( t \)[/tex] is given by:
[tex]\[ t = \frac{Q}{P} \][/tex]
[tex]\[ t = \frac{62790}{300.0} \][/tex]
[tex]\[ t = 209.3 \text{ seconds} \][/tex]
A camera is equipped with a lens with a focal length of 34 cm. When an object 2.4 m (240 cm) away is being photographed, what is the magnification?
Answer:
The magnification is -6.05.
Explanation:
Given that,
Focal length = 34 cm
Distance of the image =2.4 m = 240 cm
We need to calculate the distance of the object
[tex]\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}[/tex]
Where, u = distance of the object
v = distance of the image
f = focal length
Put the value into the formula
[tex]\dfrac{1}{u}=\dfrac{1}{34}-\dfrac{1}{240}[/tex]
[tex]\dfrac{1}{u}=\dfrac{103}{4080}[/tex]
[tex]u =\dfrac{4080}{103}[/tex]
The magnification is
[tex]m = \dfrac{-v}{u}[/tex]
[tex]m=\dfrac{-240\times103}{4080}[/tex]
[tex]m = -6.05[/tex]
Hence, The magnification is -6.05.
If a sound is 30 dB and its absolute pressure was 66 x 10-9 Pa, what must have been the reference pressure?
Given:
I = 30dB
P = 66 × [tex]10^{-9}[/tex] Pa
Solution:
Formula used:
I = [tex]20\log_{10}(\frac{P}{P_{o}})[/tex] (1)
where,
I = intensity of sound
P = absolute pressure
[tex]P_{o}[/tex] = reference pressure
Using Eqn (1), we get:
[tex]30 = 20\log _{10}\frac{66\times 10^{-9}}{P^{o}}[/tex]
[tex]P_{o}[/tex] = [tex]\frac{66\times 10^{-9}}{10^{1.5}}[/tex]
[tex]P_{o}[/tex] = 2.08 × [tex]10^{-9}[/tex] Pa
The reference pressure for a sound intensity level of 0 dB is always 20 micropascals, or 2 x 10^-5 Pa, regardless of the absolute pressure of the sound.
If a sound is 30 dB and its absolute pressure was 66 x 10-9 Pa, we need to find the reference pressure. The reference pressure is known as the threshold of hearing and corresponds to a sound intensity level of 0 dB. In acoustics, 0 dB is quantified relative to a reference which has been set at a sound pressure level of 20 micropascals, equivalent to 2 x 10-5 Pa. The question of what is the reference pressure can be answered easily: the reference pressure is always 20 micropascals or 2 x 10-5 Pa, because the decibel scale is logarithmic and based on this fixed reference.
A satellite orbits at a distance from the Earth's center of about 2.60 Earth radii and takes 5.89 hours to go around once. What distance (in meters) does the satellite travel in one day?
Answer:
424088766.068 m
Explanation:
Radius of the circular orbit that the satellite is 2.6 Earth radii (r) = 2.6 R
R = Radius of earth = 6371000 m (mean radius)
In order to find the distance that the satellite travels in 5.89 hours to complete one complete revolution is the circumference of the circular orbit
Circumference of a circle = 2×π×r
⇒Distance travelled in 5.89 hours = 2×π×2.6 R
⇒Distance travelled in 5.89 hours = 2×π×2.6×6371000
⇒Distance travelled in 5.89 hours = 104078451.3393m
Distance travelled in 1 hour = 104078451.3393/5.89 = 17670365.252 m
∴ Distance travelled in 24 hours = 17670365.252×24 = 424088766.068 m
Final answer:
The satellite orbits Earth at approximately 2.60 Earth radii and travels a total distance of about 4.23×10⁸ meters in one day.
Explanation:
The question pertains to calculating the total distance traveled by a satellite orbiting Earth at a given distance over the duration of one day. To solve this, we need to determine the circumference of the satellite's circular orbit and then calculate how many orbits it completes in a day.
To determine the satellite's orbit circumference, we use the formula for the circumference of a circle, C = 2πr, where r represents the orbit radius. Given that the satellite orbits at a distance of about 2.60 Earth radii, we can express this distance in meters by multiplying the Earth's radius, 6.37×10⁶ meters, by 2.60, yielding a radius of approximately 1.6562×10⁷ meters. Therefore, the satellite's orbit circumference is approximately 1.04×10⁸ meters.
Since the satellite completes one orbit in 5.89 hours, we can now calculate how many orbits it completes in a day (24 hours) by dividing 24 by 5.89, which is approximately 4.07 orbits per day. The total distance traveled in one day is the circumference multiplied by the number of orbits, resulting in a distance of approximately 4.23×10⁸ meters per day.
A student performs this experiment and measures the bar to have a mass of 150g and length of 36cm. What is the moment of inertia of the bar?
Answer:
The moment of inertia of the bar is [tex]45\times10^{-4}\ kg-m^2[/tex]
Explanation:
Given that,
mass of bar = 150 g
Length l = 36 cm
We need to calculate the moment of inertia of the bar
Using formula of moment inertia
[tex]I=\dfrac{1}{12}Ml^2[/tex]
Where,
M = mass of the bar
L = length of the bar
Put the value into the formula
[tex]I=\dfrac{1}{12}\times150\times10^-3\times36\times10^{-2}[/tex]
[tex]I=45\times10^{-4}\ kg-m^2[/tex]
Hence, The moment of inertia of the bar is [tex]45\times10^{-4}\ kg-m^2[/tex]
The moment of inertia of a bar with a mass of 150g and length of 36cm, rotating about its center, is approximately 0.00162 kg·m². The calculation uses the formula I = (1/12) * M * L². First, convert the mass and length to SI units and then substitute them into the formula.
To find the moment of inertia of a uniform bar with a mass of 150g and a length of 36cm, we can use the formula for a rod rotating about its center:
I = (1/12) * M * L²
Where I is the moment of inertia, M is the mass of the bar, and L is the length of the bar. Let's convert the mass to kilograms and the length to meters:Mass, M = 150g = 0.15kgLength, L = 36cm = 0.36mNow substitute these values into the formula:I = (1/12) * 0.15kg * (0.36m)²
I = (1/12) * 0.15kg * 0.1296m²
I ≈ 0.00162 kg·m²
Therefore, the moment of inertia of the bar rotating about its center is approximately 0.00162 kg·m².
What is the beat frequency heard when two organ pipes, each open at both ends, are sounded together at their fundamental frequencies if one pipe is 60 cm long and the other is 68 cm long?(The speed of sound is 340 m/s)
Answer:
The beat frequency is 33.33 Hz.
Explanation:
Given that,
Length of first pipe =60 cm
Length of other pipe = 68 cm
Speed of sound = 340 m/s
We need to calculate the frequency
We know that,
When they operate at fundamental frequency then the length is given by,
[tex]L=\dfrac{\lambda}{2}[/tex]
The wavelength is given by
[tex]\lambda=2L[/tex]
For first organ pipe,
Using formula of frequency
[tex]f=\dfrac{v}{\lambda}[/tex]
[tex]f_{1}=\dfrac{v}{2L_{1}}[/tex]...(I)
Put the value into the formula
[tex]f_{1}=\dfrac{340}{2\times60\times10^{-2}}[/tex]
[tex]f_{1}=283.33\ Hz[/tex]
For second organ pipe,
[tex]f_{2}=\dfrac{v}{2L_{2}}[/tex]...(II)
Put the value in the equation (II)
[tex]f_{2}=\dfrac{340}{2\times68\times10^{-2}}[/tex]
[tex]f_{2}=250\ Hz[/tex]
Therefore the beat frequency
[tex]\Delta f=f_{1}-f_{2}[/tex]
[tex]\Delta f=283.33-250[/tex]
[tex]\Delta f=33.33\ Hz[/tex]
Hence, The beat frequency is 33.33 Hz.
The beat frequency heard when two organ pipes are sounded together at their fundamental frequencies, where one pipe is 60 cm long and the other is 68 cm long, is 33.33 Hz.
Explanation:To find the beat frequency heard when two organ pipes are sounded together at their fundamental frequencies, we need to calculate the frequencies of each pipe. The formula for frequency is f = v/λ, where v is the speed of sound and λ is the wavelength. Since the pipes are open at both ends, the wavelength is twice the length of the pipe. So, the frequency of the 60 cm pipe would be f₁ = v/(2L₁), and the frequency of the 68 cm pipe would be f₂ = v/(2L₂). Now, to calculate the beat frequency, we subtract the frequencies: beat frequency = |f₁ - f₂|.
Using the given speed of sound (340 m/s), we can substitute the lengths of the pipes into the frequency formula to find the frequencies:
f₁ = 340/(2 x 0.6) Hz = 283.33 Hz
f₂ = 340/(2 x 0.68) Hz = 250 Hz
Now, we can calculate the beat frequency:
beat frequency = |283.33 - 250| Hz = 33.33 Hz
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A 0.50-kg mass attached to the end of a string swings in a vertical circle (radius 2.0 m). When the mass is at the highest point on the circle, the speed of the mass is 12 m/s. What is the magnitude of the force of the string on the mass at this position?
Answer:
31.1 N
Explanation:
m = mass attached to string = 0.50 kg
r = radius of the vertical circle = 2.0 m
v = speed of the mass at the highest point = 12 m/s
T = force of the string on the mass attached.
At the highest point, force equation is given as
[tex]T + mg =\frac{mv^{2}}{r}[/tex]
Inserting the values
[tex]T + (0.50)(9.8) =\frac{(0.50)(12)^{2}}{2}[/tex]
T = 31.1 N
Answer: 41N
Explanation :
T= mv^2/R + mgcos θ
At the highest point on the circle θ=0
Cos 0 = 1
T= mv^2/R + mg
m = 0.5kg
Velocity at the highest point (amplitude)= 12m/s
T = 0.5× 12^2/2 + 0.5×10
0.5×144/2 +5
T = 0.5×72 + 5
T = 36+5
T = 41N
A garden hose with an inside diameter of 0.75 inches is used to fill a round swimming pool 3.0 m in diameter. How many hours will it take to fill the pool to a depth of 1.0 m if water flows from the hose at a speed of 0.30 m/s? Enter your answer without units.
Answer:
23
Explanation:
First, we need to convert the hose diameter from inches to meters.
0.75 in × (2.54 cm / in) × (1 m / 100 cm) = 0.0191 m
Calculate the flow rate given the velocity and hose diameter:
Q = vA
Q = v (¼ π d²)
Q = (0.30 m/s) (¼ π (0.0191 m)²)
Q = 8.55×10⁻⁵ m³/s
Find the volume of the pool:
V = π r² h
V = π (1.5 m)² (1.0 m)
V = 7.07 m³
Find the time:
t = V / Q
t = (7.07 m³) / (8.55×10⁻⁵ m³/s)
t = 82700 s
t = 23 hr
The magnetic flux that passes through one turn of a 18-turn coil of wire changes to 2.67 from 8.19 Wb in a time of 0.0386 s. The average induced current in the coil is 275 A. What is the resistance of the wire?
Two ships of equal mass are 110 m apart. What is the acceleration of either ship due to the gravitational attraction of the other? Treat the ships as particles and assume each has a mass of 39,000 metric tons. (Give the magnitude of your answer in m/s2.)
Answer:
Acceleration of the ship, [tex]a=2.14\times 10^{-7}\ m/s^2[/tex]
Explanation:
It is given that,
Mass of both ships, [tex]m=39000\ metric\ tons=39\times 10^6\ kg[/tex]
Distance between two ships, d = 110 m
The gravitational force between two ships is given by :
[tex]F=G\dfrac{m^2}{d^2}[/tex]
[tex]F=6.67\times 10^{-11}\ Nm^2/kg^2\times \dfrac{(39\times 10^6\ kg)^2}{(110\ m)^2}[/tex]
F = 8.38 N
Let a is the acceleration. Now, using second law of motion as :
[tex]a=\dfrac{F}{m}[/tex]
[tex]a=\dfrac{8.38\ N}{39\times 10^6\ kg}[/tex]
[tex]a=2.14\times 10^{-7}\ m/s^2[/tex]
So, the acceleration of either ship due to the gravitational attraction of the other is [tex]2.14\times 10^{-7}\ m/s^2[/tex]. Hence, this is the required solution.
The gravitational attraction between the two ships can be determined by using Newton's law of gravitation. Afterwards, the acceleration of one ship due to this force is found by applying Newton's second law of motion.
Explanation:In this scenario, we're dealing with the concept of gravitational attraction between two massive bodies. We can find the gravitational force between the ships using Newton's law of gravitation:
F = G * (m1 * m2) / r²,
where F is the gravitational force, G is the gravitational constant (6.674 x 10^-11 N(m²/kg²)), m1 and m2 are the two masses, and r is the distance between them.
So, let's plug in the values. Each ship mass (m1 = m2) is 39,000 metric tons, equivalent to 3.9 x 10^7 kg. The distance between them (r) is 110 m, which we need to square. Calculating for F, we get an incredibly small value, which evidences the weak nature of gravitational force.
To find the acceleration of either ship (let's say m1), due to the gravitational force of the other, we will apply Newton's second law (F = m*a). Here, F is the gravitational force we found, m is the mass of the ship, and a is the acceleration we're looking for:
a = F / m,
Plugging the values found into this equation will result in the desired acceleration in m/s².
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A silicon wafer is used to attenuate the intensity from a laser emitting at a wavelength of 0.8 µm. If the laser output power is 100 mW what is the wafer thickness required to attenuate the power to 1 mW?
Answer:
[tex]t =4.605 *10^{-3}[/tex]
Explanation:
given data:
wavelength of emission =[tex]0.8 \mu m[/tex]
output power = 100 mW
We can deduce value of obsorption coefficient from the graph obsorption coefficient vs wavelength
for wavelength [tex]0.8 \mu m[/tex] the obsorption coefficient value is 10^{3}
intensity can be expressed as a function of thickness as following:
[tex]I(t) = I_{O} *e^{-\lambda *t}[/tex]
putting all value to get thickness
[tex]1*10^{-3} =100*10^{-3}e^{-10^{3}*t}[/tex]
[tex]0.01 = e^{10^{3}t}[/tex]
[tex]t =4.605 *10^{-3}[/tex]
A police car on the side of the road (at rest) uses a radar gun to catch speeders as they approach it. The frequency the radar gun emits is 8 x 109 Hz and the speed limit is 65 mi/hr. What is the difference in frequency of the emitted and returned radar wave
Final answer:
Using the Doppler Effect formulae for electromagnetic waves, the difference in frequency (Dopler shift) experienced by radar signals upon hitting and returning from a moving vehicle allows a radar gun to calculate the vehicle's speed. The principle involves the use of shift in frequency and the speed of light to measure the speed at which the vehicle is moving.
Explanation:
The question involves the Doppler Effect in physics, specifically its application in a radar speed trap. To find the frequency shift, we must use the Doppler Effect formulae for electromagnetic waves. However, it seems there might have been a mix-up with the frequencies provided in several example problems. Since those seem to be examples rather than the actual frequencies we are working with, let's focus on finding the principle behind calculating the speed of the vehicle based on a known frequency shift of a radar emission and its return signal.
The frequency shift (Δf) in the Doppler Effect for electromagnetic waves such as radar can be calculated by the formula: Δf = (2 * f * v) / c, where f is the original frequency emitted by the radar gun, v is the speed of the vehicle, and c is the speed of light. The factor of 2 is because the radar signal experiences a frequency shift once when it hits the moving vehicle and another shift when the echo returns. The radar gun's internal processors calculate the difference in frequency (the Dopler shift) to find the speed of the vehicle. For accurate measurement, the radar unit must be able to discern even small frequency shifts to effectively differentiate speeds with fine resolution.
The electric field at a point 2.8 cm from a small object points toward the object with a strength of 180,000 N/C. What is the object's charge q? ( k = 1/4πε 0 = 8.99 × 10 9 N ∙ m 2/C 2)
Answer:
Charge, [tex]Q=1.56\times 10^{-8}\ C[/tex]
Explanation:
It is given that,
Electric field strength, E = 180000 N/C
Distance from a small object, r = 2.8 cm = 0.028 m
Electric field at a point is given by :
[tex]E=\dfrac{kQ}{r^2}[/tex]
Q is the charge on an object
[tex]Q=\dfrac{Er^2}{k}[/tex]
[tex]Q=\dfrac{180000\ N/C\times (0.028\ m)^2}{9\times 10^9\ Nm^2/C^2}[/tex]
[tex]Q=1.56\times 10^{-8}\ C[/tex]
So, the charge on the object is [tex]1.56\times 10^{-8}\ C[/tex]. Hence, this is the required solution.
The object's charge q is calculated to be approximately 1.57 × 10⁻⁵ C.
To find the object's charge q, we use the formula for the electric field due to a point charge:
E = kQ/r²
Here, E is the electric field, k is the Coulomb's constant (8.99 × 10⁹ N·m²/C²), Q is the charge, and r is the distance from the charge.
We are provided with:
Electric field, E = 180,000 N/C
Distance, r = 2.8 cm = 0.028 m
Rearranging the formula to solve for Q gives:
Q = Er²/k
Substituting in the given values:
Q = (180,000 N/C) × (0.028 m)² / (8.99 × 10⁹ N·m²/C²)
Q = (180,000) × (0.000784) / (8.99 × 10⁹)
Q ≈ 1.57 × 10⁻⁵ C
The object's charge q is approximately 1.57 × 10⁻⁵ C.
The rate constant for this first‑order reaction is 0.150 s−1 at 400 ∘C. A⟶products How long, in seconds, would it take for the concentration of A to decrease from 0.860 M to 0.260 M?
Answer : The time taken for the concentration will be, 7.98 seconds
Explanation :
First order reaction : A reaction is said to be of first order if the rate is depend on the concentration of the reactants, that means the rate depends linearly on one reactant concentration.
Expression for rate law for first order kinetics is given by :
[tex]k=\frac{2.303}{t}\log\frac{[A]_o}{[A]}[/tex]
where,
k = rate constant = [tex]0.150s^{-1}[/tex]
t = time taken for the process = ?
[tex][A]_o[/tex] = initial concentration = 0.860 M
[tex][A][/tex] = concentration after time 't' = 0.260 M
Now put all the given values in above equation, we get:
[tex]0.150s^{-1}=\frac{2.303}{t}\log\frac{0.860}{0.260}[/tex]
[tex]t=7.98s[/tex]
Therefore, the time taken for the concentration will be, 7.98 seconds
To determine the time for the concentration of substance A to decrease in a first-order reaction, the formula t = (1/k) * ln([A]0/[A]t) is used with the known rate constant and initial and final concentrations.
Explanation:The question asks how long it would take for the concentration of substance A to decrease from 0.860 M to 0.260 M given a first-order reaction with a rate constant of 0.150 s−¹ at 400 °C.
For a first-order reaction, the time (t) it takes for the concentration of a reactant to change can be found using the formula:
t = (1/k) * ln([A]0/[A]t)
Where:
t = Timek = Rate constant[A]0 = Initial concentration of A[A]t = Final concentration of AIn this case, we can substitute the given values into the equation:
t = (1/0.150 s−¹) * ln(0.860 M / 0.260 M) = (1/0.150 s−¹) * ln(3.3077)
The time can be calculated by completing the computation for ln(3.3077) and dividing by the rate constant.
A 200-loop coil of cross sectional area 8.5 cm2 lies in the plane of the paper. Directed out of the plane of the paper is a magnetic field of 0.06 T. The field out of the paper decreases to 0.02 T in 12 milliseconds. What is the direction of the current induced?
Explanation:
It is given that,
Number of turns, N = 200
Area of cross section, A = 8.5 cm²
Magnetic field is directed out of the paper and is, B = 0.06 T
The magnetic field is out of the paper decreases to 0.02 T in 12 milliseconds. We need to find the direction of current induced. The induced emf is given by :
[tex]\epsilon=-N\dfrac{d\phi}{dt}[/tex]
Since, [tex]\epsilon=IR[/tex]
I is the induced current
[tex]I=-\dfrac{N}{R}\dfrac{d\phi}{dt}[/tex]
According to Lenz's law, the direction of induced current is such that it always opposes the change in current that causes it.
Here, the field is directed out of the plane of the paper, this gives the induced current in counterclockwise direction.
The sides of a square increase in length at a rate of 3 m/sec. a. At what rate is the area of the square changing when the sides are 14 m long? b. At what rate is the area of the square changing when the sides are 25 m long?
The area of a square is given by:
A = s²
A is the square's area
s is the length of one of the square's sides
Let us take the derivative of both sides of the equation with respect to time t in order to determine a formula for finding the rate of change of the square's area over time:
d[A]/dt = d[s²]/dt
The chain rule says to take the derivative of s² with respect to s then multiply the result by ds/dt
dA/dt = 2s(ds/dt)
A) Given values:
s = 14m
ds/dt = 3m/s
Plug in these values and solve for dA/dt:
dA/dt = 2(14)(3)
dA/dt = 84m²/s
B) Given values:
s = 25m
ds/dt = 3m/s
Plug in these values and solve for dA/dt:
dA/dt = 2(25)(3)
dA/dt = 150m²/s
When the side of the square is 14 m, the rate at which the area is changing is 84 m²/s.
When the side of the square is 25 m, the rate at which the area is changing is 150 m²/s.
The given parameters;
rate at which the side of the square is increasing, [tex]\frac{dl}{dt}[/tex] = 3 m/sThe area of the square is calculated as;
A = L²
The change in the area is calculated as;
[tex]\frac{dA}{dt} = 2l\frac{dl}{dt}[/tex]
When the side of the square is 14 m, the rate at which the area is changing is calculated as;
[tex]\frac{dA}{dt} = 2l \frac{dl}{dt} \\\\\frac{dA}{dt} = 2 \times l \times \frac{dl}{dt}\\\\\frac{dA}{dt} = 2 \times 14 \times 3\\\\\frac{dA}{dt} = 84 \ m^2/s[/tex]
When the side of the square is 25 m, the rate at which the area is changing is calculated as;
[tex]\frac{dA}{dt} = 2l \frac{dl}{dt} \\\\\frac{dA}{dt} = 2 \times l \times \frac{dl}{dt}\\\\\frac{dA}{dt} = 2 \times 25 \times 3\\\\\frac{dA}{dt} = 150 \ m^2/s[/tex]
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g 'A fire hose ejects a stream of water at an angle of 31.6o above the horizontal. The water leaves the nozzle with a speed of 29.3 m/s. Assuming that the water behaves like a projectile, how far from a building should the fire hose be located to hit the highest possible fire?
Answer:
39.08 m
Explanation:
θ = 31.6 degree
u = 29.3 m/s
Let it hits at the maximum height h. It takes half of the time of flight to cover the distance.
T = 2 u Sinθ / g
t = T / 2 = u Sinθ / g = (29.3 x Sin 31.6) / 9.8 = 1.566 s
The horizontal distance covered in this time is
d = u Cosθ x t = 29.3 x Cos 31.6 x 1.566 = 39.08 m
Thus, the fire hose be located at a distance of 39.08 m from the building.
A uniform electric field, with a magnitude of 370 N/C, is directed parallel to the positive x-axis. If the electric potential at x = 2.00 m is 1 000 V, what is the change in potential energy of a particle with a charge of + 2.80 x 10-3 C as it moves from x = 1.9 m to x = 2.1 m?
Answer:
[tex]\Delta U = 0.2072 J[/tex]
Explanation:
Potential difference between two points in constant electric field is given by the formula
[tex]\Delta V = E.\Delta x[/tex]
here we know that
[tex]E = 370 N/C[/tex]
also we know that
[tex]\Delta x = 2.1 - 1.9 = 0.2 m[/tex]
now we have
[tex]\Delta V = 370 (0.2) = 74 V[/tex]
now change in potential energy is given as
[tex]\Delta U = Q\Delta V[/tex]
[tex]\Delta U = (2.80 \times 10^{-3})(74)[/tex]
[tex]\Delta U = 0.2072 J[/tex]
The capacitance of a fully-charged capacitor is 11 F. Determine the capacitor's capacitance when it is half charged.
Answer:
The capacitance is 11 F for half and fully charged capacitor.
Explanation:
Capacitance of capacitor is given by the expression
[tex]C=\frac{\epsilon A}{d}[/tex]
Where ε is the vacuum permittivity, A is the area of plates and d is the separation between plates.
So capacitance does not depend upon charge and potential. So capacitance fully and half charged capacitors are same.
Here the capacitance is 11 F for half and fully charged capacitor.
Hi there!
[tex]\boxed{C = 11F}[/tex]
Even though C = Q/V, the capacitance is NEVER changed by the charge or voltage.
The only factors that change the capacitance of a capacitor are those related to its geometry or if a dielectric is inserted. We can look at some examples:
For a parallel plate capacitor:
[tex]C = \frac{\epsilon_0A}{d}[/tex]
C = Capacitance (F)
A = Area of plates (m²)
d = distance between plates (m)
For a spherical capacitor:
[tex]C = 4\pi \epsilon_0 (\frac{r_{outer}r_{inner}}{r_{outer} - r_{inner}}})[/tex]
Notice how the capacitance is strictly determined by its geometric properties. Therefore, changing the charge or voltage has no effect, so the capacitance will remain 11 F.