Answer:
r² / 110² − 3z² / 1375² = 1
Explanation:
The equation of a hyperboloid (which is a hyperbola rotated about the z axis or conjugate axis) that is centered at the origin is:
x² / a² + y² / b² − z² / c² = 1
If the cross sections are circular rather than elliptical, then a = b.
(x² + y²) / a² − z² / c² = 1
Or, if you prefer cylindrical coordinates:
r² / a² − z² / c² = 1
We know that at z = 0, r = 110. And at z = -500, r = 130.
110² / a² − 0 = 1
130² / a² − (-500)² / c² = 1
Solving:
a² = 110²
c² = 1375² / 3
Plugging in:
r² / 110² − 3z² / 1375² = 1
Compute the voltage drop along a 24-m length of household no. 14 copper wire (used in 15-A circuits). The wire has diameter 1.628 mm and carries a 13-A current. The resistivity of copper is 1.68×10−8Ω⋅m.
Answer:
Explanation:
We know that the formula for resistance in terms of length and area is given by
R = p l/ a
Where p be the resistivity and a be the area of crossection
a = pi × d^2 / 4
Where d be the diameter
a = 3.14 × (1.628 × 10^-3)^2 / 4
a = 2 × 10^-6 m^2
R = 1.68 × 10^-8 × 24 / (2 × 10^-6)
R = 20.16 × 10^-2 ohm
By use of Ohm's law
V = IR
V = 13 × 20.16 × 10^-2
V = 2.62 Volt
Romeo (75.1 kg) entertains Juliet (58.0 kg) by playing his guitar from the rear of their boat at rest in still water, 2.70 m away from Juliet, who is in the front of the boat. After the serenade, Juliet carefully moves to the rear of the boat (away from shore) to plant a kiss on Romeo's cheek. How far does the 76.5-kg boat move toward the shore it is facing?
Answer: I think 8.7 m
Explanation:
initially, a particle is moving at 5.33 m/s at an angle of 37.9° above the horizontal. Two seconds later, its velocity is 6.11 m/s at an angle of 54.2° below the horizontal. What was the particle's average acceleration during these 2.00 seconds in the x-direction (enter first) and the y-direction?
Explanation:
Average acceleration is the change in velocity over the change in time:
a = (v − v₀) / t
In the x direction:
aₓ = (6.11 cos (-54.2°) − 5.33 cos (37.9°)) / 2.00
aₓ = -0.316 m/s²
In the y direction:
aᵧ = (6.11 sin (-54.2°) − 5.33 sin (37.9°)) / 2.00
aᵧ = -4.11 m/s²
To find the average acceleration, decompose the given velocities into their x and y components. Then calculate the change in velocities in both directions divided by the time interval.
Explanation:The average acceleration of a particle can be found using the change in velocity and the time interval. The given velocities are in magnitude and direction, so we need to decompose these into their x and y components. The initial and final x-components are Vix = 5.33 m/s cos(37.9°) and Vfx = 6.11 m/s cos(54.2°), respectively. The average x-direction acceleration (ax) can be calculated as ax = (Vfx - Vix)/time. Similarly, the initial and final y-components are Viy = 5.33 m/s sin(37.9°) and Vfy = 6.11 m/s sin(54.2°), respectively. The y-direction acceleration (ay) is ay = (Vfy - Viy)/time.
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A train slows down as it rounds a sharp horizontal turn, going from 94.0 km/h to 46.0 km/h in the 17.0 s that it takes to round the bend. The radius of the curve is 140 m. Compute the acceleration at the moment the train speed reaches 46.0 km/h. Assume the train continues to slow down at this time at the same rate.
Answer:
1.41 m/s^2
Explanation:
First of all, let's convert the two speeds from km/h to m/s:
[tex]u = 94.0 km/h \cdot \frac{1000 m/km}{3600 s/h} = 26.1 m/s[/tex]
[tex]v=46.0 km/h \cdot \frac{1000 m/km}{3600 s/h}=12.8 m/s[/tex]
Now we find the centripetal acceleration which is given by
[tex]a_c=\frac{v^2}{r}[/tex]
where
v = 12.8 m/s is the speed
r = 140 m is the radius of the curve
Substituting values, we find
[tex]a_c=\frac{(12.8 m/s)^2}{140 m}=1.17 m/s^2[/tex]
we also have a tangential acceleration, which is given by
[tex]a_t = \frac{v-u}{t}[/tex]
where
t = 17.0 s
Substituting values,
[tex]a_t=\frac{12.8 m/s-26.1 m/s}{17.0 s}=-0.78 m/s^2[/tex]
The two components of the acceleration are perpendicular to each other, so we can find the resultant acceleration by using Pythagorean theorem:
[tex]a=\sqrt{a_c^2+a_t^2}=\sqrt{(1.17 m/s^2)+(-0.78 m/s^2)}=1.41 m/s^2[/tex]
Answer:
a = 1.406 m/s²
Explanation:
We are told that, the speed of the train decreases from 94.0 km/h to 46.0 km/h
Let's convert both to m/s.
Thus,
v1 =94 km/h =(94 x 10)/36 =26.11 m/s
v2=46 km/h =(46 x 10)/36=12.78 m/s
The formula to calculate the tangential acceleration is given by;
a_t = -dv/dt
Where;
dv is change in velocity
dt is time difference
dv is calculated as; dv = v1 - v2
Thus, dv = 26.11 - 12.78 = 13.33 m/s
We are given that, t = 17 seconds
Thus;
a_t = -13.33/17 = -0.784 m/s²
The negative sign implies that the acceleration is inwards.
Now, let's calculate the radial acceleration;
a_r = v²/r
Where;
r is the radius of the path = 140m
v is the velocity at the instant given
a_r is radial acceleration.
Thus,
a_r = 12.78²/140 = 1.167 m/s²
Now, the tangential and radial components of acceleration are perpendicular to each other. Thus, we can use using Pythagoreas theorem to find the resultant acceleration;
Thus;
a² = (a_t)² + (a_r)²
Plugging in the relevant values, we have;
a² = (-0.784)² + (1.167)²
a² = 1.976545
a = √1.976545
a = 1.406 m/s²
For a rigid body in rotational motion, what can be stated about the angular velocity of all of the particles? The linear velocity?
Answer:
Explanation:
In case of rotational motion, every particle is rotated with a same angular velocity .
The relation between linear velocity and angular velocity is
V = r w
Where v is linear velocity, r is the radius of circular path and w be the angular velocity.
Here w is same for all the particles but every particle has different radius of circular path I which they are rotating, so linear velocity is differnet for all.
A force of 34N stretches a very light ideal spring 0.73 m from equilibrium, What is the force constant (spring constant) of the spring? (A) 47N/m (B) 38N/m (C) 53N/m (D) 25N/m
Answer:
Spring constant, k = 47 N/m
Explanation:
It is given that,
Force applied to a spring, F = 34 N
A very light ideal spring moves 0.73 m from equilibrium position i.e. x = 0.73 m
We have to find the force constant or spring constant of the spring. It can be calculated using Hooke's law. According to him, the force acting on the spring when it compresses or stretches is given by :
[tex]F=-kx[/tex] (-ve sign shows opposite direction)
[tex]k=\dfrac{F}{x}[/tex]
[tex]k=\dfrac{34\ N}{0.73\ m}[/tex]
k = 46.5 N/m
or
k = 47 N/m
Hence, the spring constant of the spring is 47 N/m.
Utilizing Hooke's law, which posits that the force needed to compress or extend a spring is proportional to the distance, the spring constant can be calculated to be approximately 46.6 N/m by dividing the force (34N) by the displacement (0.73m). So, the closest option amongst the provided ones is (A) 47N/m.
Explanation:In the situation described where a force of 34N stretches a very light ideal spring 0.73 m from equilibrium, you would want to find the spring's force constant, often denoted as k. This involves a principle in physics known as Hooke's law, which states that the force needed to extend or compress a spring by some distance is proportional to that distance. The equation for Hooke's law is F=kx.
To find the spring constant (k), you rearrange the equation for Hooke's Law to give you: k = F/x. Substituting the force (F = 34N) and displacement from equilibrium (x = 0.73m) into the equation gives: k = 34N / 0.73m. Calculating this gives a spring constant of approximately 46.6 N/m. So, none of the given options (A) 47N/m (B) 38N/m (C) 53N/m (D) 25N/m are exactly correct, but Option A is the closest.
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A parallel-plate capacitor has a plate area of 0.3 m2 and a plate separation of 0.1 mm. If the charge on each plate has a magnitude of 5 × 10–6 C then the force exerted by one plate on the other has a magnitude of about:
Answer:
4.72 N
Explanation:
The charge density across each plate is given by:
[tex]\sigma = \frac{Q}{A}[/tex]
where
[tex]Q=5\cdot 10^{-6}C[/tex] is the charge on each plate
[tex]A=0.3 m^2[/tex] is the area of each plate
Solving,
[tex]sigma = \frac{5\cdot 10^{-6}C}{0.3 m^2}=1.67\cdot 10^{-5} C/m^2[/tex]
The force exerted by one plate on the other is given by:
[tex]F=\frac{Q\sigma}{2\epsilon_0}[/tex]
where
[tex]Q=5\cdot 10^{-6}C[/tex] is the charge on each plate
[tex]\sigma=1.67\cdot 10^{-5} C/m^2[/tex] is the surface charge density
[tex]\epsilon_0[/tex] is the vacuum permittivity
Substituting,
[tex]F=\frac{(5\cdot 10^{-6} C)(1.67\cdot 10^{-5} C/m^2)}{2(8.85\cdot 10^{-12}F/m)}=4.72 N[/tex]
We can find the force exerted by one plate of a parallel-plate capacitor on the other by utilizing the known quantities and the formula F = Q² / (A * ε), which arises from force definition and electric field for a parallel-plate capacitor.
Explanation:The magnitude of the force between two charges is given by Coulomb's law: F = k * (Q1 * Q2) / d², where k is the Coulomb's constant, Q1 and Q2 are the charges, and d is the distance. The electric field for a parallel-plate capacitor is E = Q / (A * ε), where ε is the permittivity of free space.
The force exerted on each plate of the capacitor is F = QE which, once substituted, becomes F = Q² / (A * ε) when charge, plate area, and permittivity are known. Given the values of the problem, by substituting into the equation, you can calculate the force between the plates.
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At what height above the ocean the acceleration of gravity of the earth will have a value of 9.3m /s^2 ?
Answer:
160 km
Explanation:
g = 9.8 m/s^2, g' = 9.3 m/s^2
Let h be the height from ocean level.
Use the formula for acceleration due to gravity at height.
g' = g (1 - 2h/R)
The radius of earth is 6400 km
9.3 = 9.8 ( 1 - 2 H / R)
0.05 = 2 h/R
h = 0.05 × 6400 / 2 = 160 km
How long does it take a wheel that is rotating at 33.3 rpm to speed up to 78.0 rpm if it has an angular acceleration of 2.15 rad/s^2?
Answer:
2.17 s
Explanation:
Here, w0 = 33.3 rpm = 33.3 /60 rps
= 2 × pi × 33.3 / 60 rad/ s
= 3.4854 rad / s
w = 78 rpm = 78 / 60 rps
= 2 × pi × 78 / 60 = 8.164 rad / s
a = 2.15 rad/s^2
Use first equation of motion for rotational motion
w = w0 + a t
8.164 - 3.485 = 2.15 × t
t = 2.17 s
A solenoid 0.425 m long has 950 turns of wire. What is the magnetic field in the center of the solenoid when it carries a current of 2.75 A? (JC 19.57)
Answer:
The magnetic field in the center of the solenoid is [tex]7.8\times10^{-3}\ T[/tex].
Explanation:
Given that,
Length of solenoid = 0.425 m
Number of turns N = 950
Current I = 2.75 A
The magnetic field in the center of the solenoid is the product of the current , number of turns per unit length and permeability.
In mathematical form,
[tex]B = \mu_{0}nI[/tex]
Where, [tex]n = \dfrac{N}{l}[/tex]
N = number of turns
L = length
I = current
Now, The magnetic field
[tex]B = \dfrac{\mu_{0}NI}{l}[/tex]
Put the value into the formula
[tex]B=\dfrac{4\pi\times10^{-7}\times950\times2.75}{0.425}[/tex]
[tex]B=\dfrac{4\times3.14\times10^{-7}\times950\times2.75}{0.425}[/tex]
[tex]B=7.8\times10^{-3}\ T[/tex]
Hence, The magnetic field in the center of the solenoid is [tex]7.8\times10^{-3}\ T[/tex].
A water main is to be constructed with a 12.512.5% grade in the north direction and a 2020% grade in the east direction. Determine the angle thetaθ required in the water main for the turn from north to east.
Final answer:
The angle theta required for the water main to turn from north to east can be determined by treating the grades as the rise over run and using vector addition with trigonometry to find the resultant direction and its angle east of north.
Explanation:
To determine the angle theta (θ) required for the water main to make the turn from north to east with a 12.5% grade in the north direction and a 20% grade in the east direction, we need to use vector addition and trigonometry. The water main's path can be viewed as containing two components: one in the north direction and one in the east direction. We will represent these as vectors and find their resultant to determine the required angle between them.
The grades can be understood as the rise over run, which means the north vector has a rise of 12.5 units for every 100 units of run, and the east vector has a rise of 20 units for every 100 units of run. To find the resultant vector, we can use the arctangent function:
θ = arctan(opposite/adjacent) = arctan(east grade / north grade) = arctan(0.20 / 0.125) = arctan(1.6)
Once we calculate θ, this will give us the angle between the north direction and the resultant direction of the water main. Keep in mind that this angle should be measured east of north to reflect the correct orientation of the water main's turn.
What will happen to the volume of a gas if its absolute temperature triples (increases by a factor of three) as the amount of gas and the pressure are held constant? Select the correct answer below: Question 3 options: The volume will decrease by a factor of three. The volume will triple (increase by a factor of three). The volume will double (increase by a factor of two).
Explanation:
The expression for an Ideal Gas is:
[tex]P.V=n.R.T[/tex] (1)
Where:
[tex]P[/tex] is the pressure of the gas
[tex]V[/tex] is the volume of the gas
[tex]n[/tex] the number of moles of gas
[tex]R[/tex] is the gas constant
[tex]T[/tex] is the absolute temperature of the gas
Finding [tex]V[/tex]:
[tex]V=\frac{n.R.T}{P}[/tex] (2)
If we are told the the amount of gas [tex]n[/tex] and pressure [tex]P[/tex] remain constant, but we increase the temperature [tex]T[/tex] by a factor of three; we will have to rewrite (2) with the new temperature [tex]T_{N}[/tex]:
[tex]T_{N}=3T[/tex]
[tex]V=\frac{n.R.3T}{P}[/tex] (3)
[tex]V=3\frac{n.R.T}{P}[/tex] (4)
Now, if we compare (2) with (4), it is clearly noticeable the volume of the gas has increased by a factor of 3.
The volume will triple ( increase by a factor of three )
[tex]\texttt{ }[/tex]
Further explanationThe basic formula of pressure that needs to be recalled is:
Pressure = Force / Cross-sectional Area
or symbolized:
[tex]\large {\boxed {P = F \div A} }[/tex]
P = Pressure (Pa)
F = Force (N)
A = Cross-sectional Area (m²)
Let us now tackle the problem !
[tex]\texttt{ }[/tex]
In this problem , we will use Ideal Gas Law as follows:
Given:
Initial Temperature of The Gas = T₁
Final Temperature of The Gas = T₂ = 3T₁
Asked:
Final Volume of The Gas = V₂ = ?
Solution:
[tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}[/tex]
[tex]\frac{PV_1}{T_1} = \frac{PV_2}{3T_1}[/tex]
[tex]\frac{V_1}{1} = \frac{V_2}{3}[/tex]
[tex]3(V_1) = 1(V_2)[/tex]
[tex]V_2 = 3V_1[/tex]
[tex]\texttt{ }[/tex]
Conclusion:The volume will triple ( increase by a factor of three )
[tex]\texttt{ }[/tex]
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Answer detailsGrade: High School
Subject: Physics
Chapter: Pressure
[tex]\texttt{ }[/tex]
Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant , Liquid , Pressure, Volume , Ideal , Gas , Law
Suppose you dissolve 4.75 g of another solid (not urea) in 50.0 mL of water. You note that the temperature changes from 25.0 °C to 28.0 °C. Part 1: What is the mass of the solution?
Answer:
Mass of the solution = 54.75 g
Explanation:
Mass of solid dissolved = 4.75 g
Mass of the solution = Mass of solid dissolved + Mass of water.
Mass of water = Volume x Density
Volume = 50 mL = 50 cm³
Density = 1 g/cm³
Mass of water = 50 x 1 = 50 g
Mass of the solution = 4.75 + 50 = 54.75 g
Two parallel-plate capacitors have the same plate area, but the plate gap in capacitor 1 is twice as big as capacitor 2. If capacitance of the first capacitor is C, then capacitance of the second one is:
Answer:
Capacitance of the second capacitor = 2C
Explanation:
[tex]\texttt{Capacitance, C}=\frac{\varepsilon_0A}{d}[/tex]
Where A is the area, d is the gap between plates and ε₀ is the dielectric constant.
Let C₁ be the capacitance of first capacitor with area A₁ and gap between plates d₁.
We have
[tex]\texttt{Capacitance, C}_1=\frac{\varepsilon_0A_1}{d_1}=C[/tex]
Similarly for capacitor 2
[tex]\texttt{Capacitance, C}_2=\frac{\varepsilon_0A_2}{d_2}=\frac{\varepsilon_0A_1}{\frac{d_1}{2}}=2\times \frac{\varepsilon_0A_1}{d_1}=2C[/tex]
Capacitance of the second capacitor = 2C
A(n) 1100 kg car is parked on a 4◦ incline. The acceleration of gravity is 9.8 m/s 2 . Find the force of friction keeping the car from sliding down the incline.
Answer:
Force of friction, f = 751.97 N
Explanation:
it is given that,
Mass of the car, m = 1100 kg
It is parked on a 4° incline. We need to find the force of friction keeping the car from sliding down the incline.
From the attached figure, it is clear that the normal and its weight is acting on the car. f is the force of friction such that it balances the x component of its weight i.e.
[tex]f=mg\ sin\theta[/tex]
[tex]f=1100\ kg\times 9.8\ m/s^2\ sin(4)[/tex]
f = 751.97 N
So, the force of friction on the car is 751.97 N. Hence, this is the required solution.
In essence, the force of friction that prevents the car from sliding down the incline is equal to the component of the weight of the car down the incline. After performing the necessary calculations, this is found to be roughly 766.9 Newtons.
Explanation:In response to your question, the force of friction keeping the car from sliding down the incline can be calculated using basic physics principles. Firstly, we need to calculate the component of the gravitational force parallel to the incline, which is given by F_gravity = m * g * sin(theta), where m is the mass of the car, g is the acceleration due to gravity, and theta is the angle of incline.
Substituting the given values, we get F_gravity = 1100 kg * 9.8 m/s² * sin(4°), which approximates to 766.9 N.
Now, since the car is stationary and not sliding down, this suggests that the force of friction equals the component of the weight down the incline. Therefore, the force of friction keeping the car from sliding down is approximately 766.9 Newtons.
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A circular coil of radius 10 cm and a separate square coil of side 20 cm are both rotated in a magnetic field of 1.5 T. If the circular coil is rotated at a frequency of 60 Hz, then at what frequency must the square coil be rotated in order for both coils to have the same maximum induced voltage? A) 47 Hz
B) 60 Hz
C) 76 Hz
D) 19 Hz
Answer:
The frequency of square coil is 47 Hz.
(A) is correct option.
Explanation:
Given that,
Radius =10 cm
Side = 20 cm
Magnetic field = 1.5 T
Frequency = 60 Hz
We need to calculate the the maximum induced voltage
[tex]V_{m}=BAN\omega[/tex]
Where, B = magnetic field
A = area of cross section
N = number of turns
[tex]\omega=2\pi f[/tex]= angular frequency
Put the value into the formula
[tex]V_{m}=1.5\times3.14\times(10\times10^{-2})^2\times1\times2\times3.14\times60[/tex]
[tex]V_{m}=17.75\ V[/tex]
If the square coil have the same induced voltage.
Area of square of coil [tex]A =(20\times10^{-2})^2[/tex]
[tex]A=0.04\ m^2[/tex]
Now, The angular velocity of square coil
[tex]\omega=\dfrac{V_{m}}{AB}[/tex]
[tex]\omega=\dfrac{17.75}{0.04\times1.5}[/tex]
[tex]\omega=295.8\ \dfrac{rad}{s}[/tex]
Now, frequency of rotation
[tex]f = \dfrac{\omega}{2\pi}[/tex]
Put the value into the formula of frequency
[tex]f=\dfrac{295.8}{2\times3.14}[/tex]
[tex]f=47\ Hz[/tex]
Hence, The frequency of square coil is 47 Hz.
As a woman walks, her entire weight is momentarily placed on one heel of her high-heeled shoes. Calculate the pressure exerted on the floor by the heel if it has an area of 1.43 cm2 and the woman's mass is 60.5 kg. Express the pressure in pascals. (In the early days of commercial flight, women were not allowed to wear high-heeled shoes because aircraft floors were too thin to withstand such large pressures.)
Answer:
4.15 x 10^6 N
Explanation:
Area, A = 1.43 cm^2 = 1.43 x 10^-4 m^2
mass, m = 60.5 kg
Weight, F = m g = 60.5 x 9.8 = 592.9 N
Pressure = Force / Area
P = Weight / Area
P = 592.9 / (1.43 x 10^-4)
P = 4.15 x 10^6 N
An ideal transformer has 60 turns in its primary coil and 360 turns in its secondary coil. If the input rms voltage for the 60-turn coil is 120 V, what is the output rms voltage of the secondary coil?
Answer:
Voltage of the secondary coil is 720 volts.
Explanation:
Number of turns in primary coil, N₁ = 60
Number of turns in secondary coil, N₂ = 360
Input rms voltage of primary coil, V₁ = 120 V
We have to find the output rms voltage of the secondary coil. The relationship between the number of coil and voltage is given by :
[tex]\dfrac{N_1}{N_2}=\dfrac{V_1}{V_2}[/tex]
[tex]\dfrac{60}{360}=\dfrac{120}{V_2}[/tex]
V₂ = output rms voltage of the secondary coil.
After solving above equation, we get :
V₂ = 720 V
Hence, the output rms voltage of the secondary coil is 720 volts.
In an amusement park water slide, people slide down an essentially frictionless tube. The top of the slide is 3.2 m above the bottom where they exit the slide, moving horizontally, 1.2 m above a swimming pool. Does the mass of the person make any difference?
Answer:
No
Explanation:
When the person slides down, the change in gravitational potential energy is converted into kinetic energy, according to
[tex]\Delta U = \Delta K\\mg\Delta h = \frac{1}{2}mv^2[/tex]
where
m is the mass of the person
g is the acceleration of gravity
v is the final speed
[tex]\Delta h[/tex] is the change in heigth of the person
Here we have assumed that the initial speed is zero.
Re-arranging the equation,
[tex]v = \sqrt{2g \Delta h}[/tex]
and we see that this quantity does not depend on the mass of the person, so every person will have the same speed at the bottom of the slide, equal to:
[tex]v=\sqrt{2(9.8 m/s^2)(3.2 m-1.2 m)}=6.3 m/s[/tex]
At a distance of 20 m from a source of sound, the sound level is 40 dB. If the observer backs up to a distance of 40 m from the source, what will the sound level be? O A. 34 dB O B. 28 dB ° C. 20 dB O D. 10 dB
Answer:
The sound level is 34 dB.
(A) is correct option.
Explanation:
Given that,
Distance from a source of sound [tex]r_{1}=20 m[/tex]
Sound level [tex]L_{1}= 40\ dB[/tex]
Distance from a source of sound[tex]r_{2}=40\ m[/tex]
We need to calculate the sound level
Using equation
[tex]L_{2}=L_{1}-|20 log(\dfrac{r_{1}}{r_{2}})|[/tex]
[tex]L_{2}=40-|20 log(\dfrac{20}{40})|[/tex]
[tex]L_{2}=40-|20 log(0.5)|[/tex]
[tex]L_{2}=33.9\ dB[/tex]
[tex]L_{2}=34\ dB[/tex]
Hence, The sound level is 34 dB.
The sound level decreases by 20 dB when the distance from the source is doubled because the intensity of sound decreases with the square of the distance. Thus, the sound level at 40 m would be 20 dB. The answer is option C.
When the observer moves from a distance of 20 m to 40 m away from the source of sound, the sound intensity decreases. This is due to the inverse square law, which states that the intensity of sound is inversely proportional to the square of the distance from the source. Since the observer doubles the distance from the source, the intensity of the sound will be reduced to one-fourth. Because each 10-fold decrease in intensity corresponds to a reduction of 10 dB in sound level, halving the distance corresponds to an intensity ratio of 1/4, which is two factors of 10 (or 102 times less intense). Thus, the sound level decreases by 20 dB (2 factors of 10).
Therefore, at a distance of 40 m, the sound level would be 20 dB less than the original 40 dB, leading to a new sound level of 20 dB. The answer is option C: 20 dB.
A voltage source Vs = 10 V is in series with a resistor of 10 kOhm. If the source transformation theorem is applied, what will the value of the current source be in mA ?
Answer:
The current source is 1 mA.
Explanation:
Given that,
Voltage = 10 V
Resistor = 10 k ohm
We need to calculate the current
Using formula of transformation theorem
The source transformation must be constrained is given by
[tex]v_{s}=iR_{s}[/tex]
Where, V = voltage
I = current
R = resistor
Put the value into the formula
[tex]10=i\times10\times10^{3}[/tex]
[tex]i =\dfrac{10}{10\times10^{3}}[/tex]
[tex]i = 0.001\ A[/tex]
[tex]i=1\ mA[/tex]
Hence, The current source is 1 mA.
By using Ohm's Law (I = V/R), it can be determined that for a voltage source Vs = 10 V in series with a resistor of 10 kOhm, the current source would be 1mA after source transformation.
Explanation:
The question asks for the current value in a circuit when the source transformation theorem is applied. This involves using Ohm's law, which in basic form states that the current through a resistor equals the voltage across the resistor divided by the resistance. It's represented by the equation I = V/R.
In our case, the voltage (V) is 10 volts, and the resistance (R) is 10 kilo-Ohms or 10,000 Ohms. Applying Ohm's law gives I = V/R, so I = 10V/10,000Ω = 0.001 A. However, because the question asks for the current in milliamperes (mA), we need to convert from amperes to milliamperes, knowing that 1A = 1000mA. Therefore, 0.001A = 1mA.
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A bullet with a mass ????=12.5 g and speed ????=86.4 m/s is fired into a wooden block with ????=113 g which is initially at rest on a horizontal surface. The bullet is embedded into the block. The block-bullet combination slides across the surface for a distance ???? before stopping due to friction between the block and surface. The coefficients of friction are ????????=0.753 and ????????=0.659. (a) What is the speed (m/s) of the block-bullet combination immediately after the collision? (b) What is the distance d (m)?
Answer:
a) 8.61 m/s, b) 5.73 m
Explanation:
a) During the collision, momentum is conserved.
mv = (m + M) V
(12.5 g) (86.4 m/s) = (12.5 g + 113 g) V
V = 8.61 m/s
b) After the collision, energy is conserved.
Kinetic energy = Work done by friction
1/2 (m + M) V² = F d
1/2 (m + M) V² = N μk d
1/2 (m + M) V² = (m + M) g μk d
1/2 V² = g μk d
d = V² / (2g μk)
d = (8.61 m/s)² / (2 × 9.8 m/s² × 0.659)
d = 5.73 m
Notice we used the kinetic coefficient of friction. That's the friction when an object is moving. The static coefficient of friction is the friction on a stationary object. Since the bullet/block combination is sliding across the surface, we use the kinetic coefficient.
Suppose that an electromagnetic wave which is linearly polarized along the x−axis is propagating in vacuum along the z−axis. The wave is incident on a conductor which is placed at z > 0 region of the space. The conductor has conductivity σ, magnetic permeability µ and electric permittivity ε.
(a) Find the characteristic time for the free charge density which dissipates at the conductor.
(b) Write the Maxwell equations and derive the wave equation for a plane wave propagating in a conductor.
(c) Find the attenuation distance at which the incident amplitude reduces to e ^−1 of its initial value.
(d) Find the electric and magnetic fields inside the conductor. 8 (e) Find the power loss per area of the incident electromagnetic wave at the surface of conductor.
Suppose the earth is shaped as a sphere with radius 4,0004,000 miles and suppose it rotates once every 24 hours. How many miles along the equator does it rotate each hour? (approximation is acceptable)
Answer:
1047 miles
Explanation:
The radius of the Earth is
[tex]r = 4000[/tex] (miles)
So its circumference, which is the total length of the equator, is given by
[tex]L=2\pi r= 2\pi(4000)=25133 mi[/tex]
Now we know that the Earth rotates once every 24 hours. So the distance through which the equator moves in one hour is equal to its total length divided by the number of hours, 24:
[tex]L' = \frac{25133 mi}{24h}=1047 mi[/tex]
A highway curve with radius 80 m is to be banked so that a car traveling 21 m/s will not skid sideways even in the absence of friction. At what angle should the curve be banked?
Answer:
Angle, θ = 29.35
Explanation:
It is given that,
Radius of the curve, r = 80 m
Velocity of the car, v = 21 m/s
The highway curve is to be banked so that the car will not skid sideways in the presence of friction. We need to find the angle should the curve be banked. Let the angle is θ. It is given by :
[tex]tan{\theta}=\dfrac{v^2}{rg}[/tex]
Where
g = acceleration due to gravity
[tex]tan\ {\theta}=\dfrac{(21\ m/s)^2}{80\ m\times 9.8\ m/s^2}[/tex]
[tex]\theta=tan^{-1}(0.5625)[/tex]
θ = 29.35°
Hence, this is the required solution.
A particle leaves the origin with an initial velocity of 3.00 m/s in the x direction, and moves with constant acceleration ax = -2.70 m/s2 and ay = 3.90 m/s2. How far does the particle move in the x direction before turning around?
Answer:
1.67 m
Explanation:
Ux = 3 m/s, ax = -2.7 m/s^2, vx =0
Let the distance travelled before stopping along x axis is x.
Use third equation of motion along x axis
Vx^2 = ux^2 + 2 a x
0 = 9 - 2 × 2.7 × x
X = 1.67 m
Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.57 A out of the junction.What is the magnitude of the current in wire 3?
I₃ = 0.17 A into the junction.
The key to solve this problem is using Kirchhoff's Current Law which statements that the algebraic sum of the currents that enter and leave a particular junction must be 0 (I₁+I₂+...+In = 0). Be careful, the currents that leaves a point is considered positive current, and the one that enters a point is considered negative.
In other words, the sum of the currents that enter to the joint is equal to the sum of the currents that come out of the joint.
Three wire meet at junction. Wire 1 has a current of 0.40 A into the junction, the current of the wire 2 is 0.57 A out of the junction. What is the magnitude of the current in wire 3?
To calculate the current in the wire 3:
First, let's name the currents in the circuit. So:
The current in the wire 1 is I₁ = 0.40 A, the current in the wire 2 is I₂ = 0.57 A, and the current of wire 3 is I₃ = ?.
Second, we make a scheme of the circuit, with the current I₁ of wire 1 into the junction, the current I₂ of wire 2 out of the junction, and let's suppose that the current I₃ of the wire 3 into the junction. (See the image attached)
Using Kirchhoff's Current Law, the sum of the currents I₁ and I₃ into the junction is equal to the current I₂ out of junction.
I₁ + I₃ = I₂
0.40 A + I₃ = 0.57 A
I₃ = 0.57 A - 0.40A
I₃ = 0.17 A
Checking the Kirchhoff's Current Law, the sum of all currents is equal to 0 :
I₁ + I₃ = I₂
I₁ - I₂ + I₃ = 0
0.40 A - 0.57 A + 0.17 A = 0
Let's suppose that the current I₃ of the wire 3 out the junction.
I₁ = I₂ + I₃
0.40 A = 0.57 A + I₃
I₃ = 0.40 A - 0.57 A
I₃ = -0.17 A which means that the current flow in the opposite direction that we selected.
The magnitude of the current in wire 3 entering the junction is 0.17A. This result is based on the principle of conservation of electric current which states that the total current entering the junction must equal the total current exiting the junction.
Explanation:This question involves the principle of the conservation of electric current, also known as Kirchhoff's Current Law. According to this concept, at any junction point the total current entering the junction is equal to the total current leaving the junction. In this case, if wire 1 has a current of 0.4A entering the junction and wire 2 has a current of 0.57A leaving the junction, the current in wire 3 would be the difference between these two values.
Since current can't be negative, if the current in wire 1 (entering the junction) is less than the current in wire 2 (exiting), the current in wire 3 must be entering the junction to balance the current flow.
Therefore, the magnitude of the current in wire 3 would be 0.57A - 0.4A = 0.17A. This means wire 3 has a current of 0.17A entering the junction.
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As mentioned before, our asteroid is in the shape of a sphere and has a mass of 1000 kilograms. Determine the density (in grams per cubic centimeter) of this asteroid if its diameter is known to be 1.2 meters. Useful information: 1 kg = 1000 g, 1 m = 100 cm, volume of sphere = 4/3 ? r3. Remember that the radius of a sphere is equal to half its diameter. Show all of your work. (20 points)
Answer: [tex]1.1052g/cm^{3}[/tex]
Explanation:
Density [tex]D[/tex] is a characteristic property of a material and is defined as the relationship between the mass [tex]m[/tex] and volume [tex]V[/tex] of a specific substance or material. So, the density of the asteroid is given by the following equation:
[tex]D=\frac{m}{V}[/tex] (1)
On the other hand, we know the asteroid has a mass [tex]m=1000kg[/tex] and is spherical. This means its volume is given by the following formula:
[tex]V=\frac{4}{3}}\pi r^{3}[/tex] (2)
Where [tex]r=\frac{d}{2}=\frac{1.2m}{2}=0.6m[/tex] is the radius of the sphere and is half its diameter [tex]d[/tex].
Knowing this, we can calculate the volume:
[tex]V=\frac{4}{3}}\pi (0.6m)^{3}[/tex] (3)
[tex]V=0.904m^{3}[/tex] (4)
Substituting (4) in (1):
[tex]D=\frac{1000kg}{0.904m^{3}}=1105.242\frac{kg}{m^{3}}[/tex] (5) This is the density of the asteroid, but we were asked to find it in [tex]\frac{g}{cm^{3}}[/tex]. This means we have to make the conversion:
[tex]D=1105.242\frac{kg}{m^{3}}.\frac{1000g}{1kg}.\frac{1m^{3}}{(100cm)^{3}}[/tex]
Finally:
[tex]D=1.1052\frac{g}{cm^{3}}[/tex]
A particle moves along a circle with radius R, so that the tangential component of its acceleration is constant. At t = 0 the velocity of the particle was equal to zero. Find
(a) the normal component an of the acceleration as function of time.
(b) the magnitude of the acceleration vector a as well as the angle the vector a forms with the position vector r as functions of time.
To find the normal component of acceleration, differentiate the velocity and acceleration functions.
Explanation:In order to find the normal component of the acceleration, we need to first find the velocity and acceleration vectors as functions of time. Using the given position function, we can differentiate it to find the velocity function:
v(t) = dr/dt = -3(4sin(3t)i + 4cos(3t)ĵ)
Next, we can differentiate the velocity function to find the acceleration function:
a(t) = dv/dt = -3(3cos(3t)i - 3sin(3t)ĵ)
Since the tangential component of the acceleration is constant, we can set the coefficient of the tangential component equal to a constant value:
-3(3sin(3t)) = Constant
Solving this equation will give us the value of the constant. Once we know the value of the constant, we can substitute it back into the acceleration function to find the normal component of the acceleration as a function of time.
True or False? The surfaces of equal potential around an infinite charged plane are parallel to the plane (i.e. the surfaces never intersect the charged plane). False True [5 points] Suppose I center a conducting sphere of charge (radius R, total charge +Q) at the origin. How much energy would it take to move a point charge (total charge +q) from the surface of the sphere (r = R) to the center of the sphere (r = 0)? E = kq / R
Answer:
true
Explanation:
the sun hits it ig idek
The surfaces of equal potential around an infinite charged plane are parallel to the plane: True.
The surfaces around an infinite charged plane.According to the law of electrostatic forces, the surfaces of equal potential around an infinite charged plane are always parallel to the plane because they cannot intersect the charged plane.
Since this conducting sphere of charge is centered at the origin (0), the quantity of energy that would be required to move a point charge from the surface of this sphere (r = R) to the center of the sphere is given by:
E = qV
E = q × 0
E = 0 Joules.
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