Complete question is;
A single-phase 60-Hz overhead power line is symmetrically supported on a horizontal cross arm. Spacing between the centers of the conductors acing between the centers of the conductors (say, a and b) is 2.5 m. A telephone line is also symmetrically supported on a horizontal cross arm 1.8 m directly below the power line. Spacing between the centers of these conductors (say, c and d) is 1.0 m.
The mutual inductance per unit length between circuit a-b and circuit c-d is given as 4 x 10^(-7) ln √((D_ad × D_bc)/(D_ac × D_bd)) H/m
where, for example, D_ad denotes the distance in meters between conductors a and d.
a. Hence, compute the mutual inductance per kilometer between the power line and the telephone line.
b. Find the 60-Hz voltage per kilometer induced in the telephone line when the power line carries 150 A
Answer:
A) M = 1.01 × 10^(-4) H/km
B) v_cd = 5.712 V/km
Explanation:
A) From the distances given in the question, we can deduce that;
D_ac = √(((2.5/2) - (1/2))² + 1.8²)
D_ac = 1.95 m
Also;
D_ad = √(((2.5/2) + (1/2))² + 1.8²)
D_ad = 2.51 m
I_a and I_b are put of phase by 180°. Thus, due to a and b, the flux linkages to c and d is given as;
φ_cd = 4 x 10^(-7)I_a( ln (2.51/1.95))
Mutual inductance per km is given as;
M = φ_cd/I_a
Thus;
M = 4 x 10^(-7)( ln (2.51/1.95))
M = 1.01 × 10^(-7) H/m
Per km;
M = 1.01 × 10^(-7) × 1000
M = 1.01 × 10^(-4) H/km
B) voltage per km is gotten by;
v_cd = ωMI
Now, ω = 2πf = 2π × 60 = 377 rad/s
Thus;
v_cd = 377 × 1.01 × 10^(-4) × 150
v_cd = 5.712 V/km
PLEASE HELP!!!! ITS URGENT!!!
Answer:
dude the answer is upright
What do Ice core samples with lower ratios of O-18 to O-16 isotopes tell scientist about past climates
Answer:
Ocean-floor sediments can also be used to determine past climate. ... Ice cores contain more 16O than ocean water, so ice cores have a lower 18O/ 16O ratio than ocean water or ocean-floor sediments. Water containing the lighter isotope 16O evaporates more readily than 18O in the warmer subtropical regions
Explanation:
HELP DUE 3 MINUTESSSSD
Answer:
Break down small pebbles and sediments, like sand
Break down large rocks like mountains
Explanation:
6. As distance increases, gravitational force *
(10 Points)
increases
decreases
A 45.0 kilogram boy is riding a 15.0-kilogram bicycle with a speed of 8.00 meters per second. What is the combined kinetic energy of the boy and the bicycle? A)480.J B)240.0J C)1920J D)1440J
Answer:
1920Joules
Explanation:
The formula for calculating the kinetic energy of a body is expressed as;
KE = 1/2 mv²
m isthe mass
V is the speed
For the two masses, the combined KE is expressed as;
KE = 1/2(m1+m2)v²
KE = 1/2(45+15)(8)²
KE = 1/2 * 60 * 64
KE = 30 * 64
KE = 1920J
Hence the combined kinetic energy of the boy and the bicycle is 1920Joules
The combined kinetic energy of the boy and the bicycle is of 1920 J.
Given data:
The mass of boy is, m = 45.0 kg.
The mass of bicycle is, M = 15.0 kg.
The speed of bicycle is, v = 8.00 m/s.
The kinetic energy of an object is defined as the energy possessed by an object by virtue of motion of object. The combined kinetic energy of the boy-bicycle system is given as,
[tex]KE = \dfrac{1}{2}(m+M)v^{2}[/tex]
Solve by substituting the values as,
[tex]KE = \dfrac{1}{2}(45+15) \times 8^{2}\\\\KE = 1920 \;\rm J[/tex]
Thus, we can conclude that the combined kinetic energy of the boy and the bicycle is of 1920 J.
Learn more about the concept of kinetic energy here:
https://brainly.com/question/12669551
Part A: A group of students performed the same "Ohm's Law" experiment that you did in class. They obtained the following results:
Trial V (volts) I (mA)
1 1.00 7.2
2 2.10 14.0
3 3.10 20.7
4 4.00 27.2
5 4.90 32.2
Determine the slope and y-intercept of the graph, and report these values below. (
Part B: Your mischievous lab partner takes the resistor that you just experimented with and assembles it in a network with one other resistor and places them inside a black box. He challenges you to tell him the configuration of the resistors inside the box. Being an industrious physics student you connect the leads of the black box to your power source, voltmeter (in parallel), and ammeter (in series) and take the following simultaneous measurements. Use the measurements to find the equivalent resistance of the arrangement.
V (volts) I (mA)
2.0 5.5
Part C: Now that you've answered his challenge, your lab partner asks you to give the resistance of the resistor that he added to the one you experimented with. Using the information you obtained in parts A and B, predict this value of the resistance of the second resistor.
Answer:
Kindly check explanation
Explanation:
Given the data:
Trial V (volts) I (mA)
1 1.00 7.2
2 2.10 14.0
3 3.10 20.7
4 4.00 27.2
5 4.90 32.2
Slope = Rise / Run
Rise = y2 - y1 = 32.2 - 7.2 = 25
Run = x2 - x1 = 4.9 - 1.0 = 3.9
Slope = 25 / 3.9 = 6.410
y = mx + c
The intercept, c
Take the point ( 1; 7.2)
Put x = 0
7.2 = 6.410(1) + C
7.2 - 6.410 = C
C = 0.79
If the object on the Moon were raised to a height of 30.0 m, what would be the potential energy? PE=mgh (g on the Moon is 1.62m/s)
Se desea elevar un cuerpo de 1000 kg utilizando una elevadora hidráulica de plato grande circular de 50 cm de radio y plato pequeño circular de 8cm de radio, calcula: a) El peso del cuerpo. b) Cuanta fuerza hay que hacer en el émbolo pequeño, c) Si el émbolo pequeño desciende 60 cm, ¿qué volumen de fluido desplaza hacia el émbolo mayor?
Answer:
a) [tex]W=9810\: N[/tex]
b) [tex]F_{1}=251.14\: N[/tex]
c) [tex]V_{g}=0.012\: m^{3}[/tex]
Explanation:
a)
El peso del cuerpo es:
[tex]W=mg[/tex]
g es la gravedad (9.81 m/s²)
[tex]W=1000*9.81[/tex]
[tex]W=9810\: N[/tex]
b)
Usando el principio de Pascal tenemos:
[tex]P_{1}=P_{2}[/tex]
y la presion es la fuerza sobre el area.
[tex]\frac{F_{1}}{A_{1}}=\frac{F_{2}}{A_{2}}[/tex]
F(1) es la fuerza aplicada en el embolo pequeñoA(1) es el area del disco pequeñoF(2) es la fuerza aplicada en el embolo grandeA(2) es el area del disco grandeDespejando F(1):
[tex]F_{1}=F_{2}\frac{A_{1}}{A_{2}}[/tex]
el area del plato es: [tex]A=\pi R^{2}[/tex]
[tex]F_{1}=F_{2}\frac{\pi R_{1}^{2}}{\pi R_{2}^{2}}[/tex]
[tex]F_{1}=F_{2}\frac{R_{1}^{2}}{R_{2}^{2}}[/tex]
F(2) es el peso del cuerpo de 1000 kg (W)
[tex]F_{1}=9810\frac{8^{2}}{50^{2}}[/tex]
Por lo tanto, la fuerza que se debe hacer es:
[tex]F_{1}=251.14\: N[/tex]
c)
Como tenemos un sistema cerrado el volumen de agua que desciende por el embolo pequeño debe ser igual al que sube por el grande, por lo tanto:
[tex]V_{p}=V_{g}[/tex]
Vp es el volumen de agua en el émbolo pequeño
Vg es el volumen de agua en el émbolo grande
Como sabemos que son cilindros (V=πR²h)
[tex]\pi R^{2}h=V_{g}[/tex]
Entonces el volumen del émbolo mayor será:
[tex]V_{g}=\pi 0.08^{2}0.6[/tex]
[tex]V_{g}=0.012\: m^{3}[/tex]
Espero te haya sido de ayuda!
1. An object with a mass of 5 kg is pushed by a force of 10 N. What is the object's acceleration?
Answer:2m/s^2
Explanation:
a=f/m
There is a 247–m–high cliff at Half Dome in Yosemite National Park in California. Suppose a boulder breaks loose from the top of this cliff. What is the velocity of the boulder just before it strikes the ground?
Answer:
Vf = 69.61 m/s
Explanation:
We will use the third equation of motion to solve this problem:
[tex]2gh = V_{f}^2 - V_{i}^2\\[/tex]
where,
g = acceleration due to gravity = 9.81 m/s²
h = height of cliff = 247 m
Vf = final velocity = ?
Vi = initial velocity = 0 m/s (boulder breaks loose from rest)
Therefore,
[tex](2)(9.81\ m/s^2)(247\ m) = V_{f}^2 - (0\ m/s)^2\\V_{f} = \sqrt{4846.14\ m^2/s^2}\\[/tex]
Vf = 69.61 m/s
Are we within earths Roche limit
Answer: Closer to the Roche limit, the body is deformed by tidal forces. Within the Roche limit, the mass' own gravity can no longer withstand the tidal forces, and the body disintegrates. hope this helps can u give me brainliest
Explanation:
Find the polar angle (in radians with respect to +x-axis) of −3i + j.
Answer:
[tex]-18.43^{\circ}[/tex]
Explanation:
Let [tex]\theta[/tex] be the polar angle of −3i + j. We can find it using the formula as follows :
[tex]\tan\theta=\dfrac{y}{x}\\\\\tan\theta=\dfrac{1}{-3}\\\\\theta=\tan^{-1}(\dfrac{1}{-3})\\\\\theta=-18.43^{\circ}[/tex]
So, the required polar angle is [tex]-18.43^{\circ}[/tex].
A 4.00-m-long, 470 kg steel beam extends horizontally from the point where it has been bolted to the framework of a new building under construction. A 75.0 kg construction worker stands at the far end of the beam. For the steps and strategies involved in solving a similar problem, you may view a Video Tutor Solution. What is the magnitude of the torque about the point where the beam is bolted into place?
Answer:
12164.4 Nm
Explanation:
CHECK THE ATTACHMENT
Given values are;
m1= 470 kg
x= 4m
m2= 75kg
Cm = center of mass
g= acceleration due to gravity= 9.82 m/s^2
The distance of centre of mass is x/2
Center of mass(1) = x/2
But x= 4 m
Then substitute, we have,
Center of mass(1) = 4/2 = 2m
We can find the total torque, through the summation of moments that comes from both the man and the beam.
τ = τ(1) + τ(2)
But
τ(1)= ( Center of m1 × m1 × g)= (2× 470× 9.81)
= 9221.4Nm
τ(2)= X * m2 * g = ( 4× 75 × 9.81)= 2943Nm
τ = τ(1) + τ(2)
= 9221.4Nm + 2943Nm
= 12164.4 Nm
Hence, the magnitude of the torque about the point where the beam is bolted into place is 12164.4 Nm
An object swings in a horizontal circle, supported by a 1.8-m string. It swings at a speed of 3 m/s. What is the mass of the object given that the tension in the string is 90 N?
Answer:
Mass = 18 kg
Explanation:
Formula for force in centripetal motion is;
F = mv²/r
We have;
Mass; m.
Speed; v = 3 m/s
radius; r = 1.8 m
Force; F = 90 N
Thus;
Making m the subject;
m = Fr/v²
m = 90 × 1.8/3²
m = 18 kg
The velocity of an object is +47 m/s at 3.0 seconds and is +65 m/s at 12.0 seconds. Calculate the acceleration of the object
Answer:
[tex]\boxed{\text{\sf \Large 2.0 m/s^2 $}}[/tex]
Explanation:
Use acceleration formula
[tex]\displaystyle \text{$ \sf acceleration=\frac{change \ in\ velocity}{change \ in \ time} $}[/tex]
[tex]\displaystyle a=\frac{65-47}{12.0-3.0} =2.0[/tex]
We know that there is a relationship between work and mechanical energy change. Whenever work is done upon an object by an external force (or non-conservative force), there will be a change in the total mechanical energy of the object. If only internal forces are doing work then there is no change in the total amount of mechanical energy. The total mechanical energy is said to be conserved. Think of a real-life situation where we make use of this conservation of mechanical energy (where we can neglect external forces for the most part). Describe your example and speak to both the kinetic and potential energy of the motion.
Answer:
* roller skates and ice skates.
* roller coaster
Explanation:
One of the best examples for this situation is when we are skating, in the initial part we must create work with a force, it compensates to move, after this the external force stops working and we continue movements with kinetic energy, if there are some ramps, we can going up, where the kinetic energy is transformed into potential energy and when going down again it is transformed into kinetic energy. This is true for both roller skates and ice skates.
Another example is the roller coaster, in this case the motor creates work to increase the energy of the car by raising it, when it reaches the top the motor is disconnected, and all the movement is carried out with changes in kinetic and potential energy. In the upper part the energy is almost all potential, it only has the kinetic energy necessary to continue the movement and in the lower part it is all kinetic; At the end of the tour, the brakes are applied that bring about the non-conservative forces that decrease the mechanical energy, transforming it into heat.
a toy train is moved towards a magnet that cannot move. what happens to the potential energy in the system of magnets during the movement
Answer:
Dakota moves a magnetic toy train toward a magnet that cannot move. What happens to the potential energy in the system of magnets during the movement? The potential energy increases because the train moves against the magnetic force. The potential energy decreases because the train moves against the magnetic force.
I hope this helps you :)
When an
object is placed 15 cm from "
a concave mirror, a real image magnified
3 times is formed - Find
a) the focal length of the mirror
b) where the object must be placed to
give a virtual image 3 times the height
of the object
Answer:
Focal length(f)= -15 cm, magnification= -3 (image is real). So, -v/u=-3 ; or v=3u {v=image distance and u=object distance}. Using mirror formula,
1/f= 1/v + 1/u
-1/15 = 1/3u + 1/u
4/3u = -1/15
u= -20 cm
v =3u= 3×(-20)= -60 cm
So object is 20 cm and image is 60 cm in front of the mirror.
Explanation:
What happens to solar radiation when it is absorbed
Answer:
Absorbed sunlight is balanced by heat radiated from Earth's surface and atmosphere. ... The atmosphere radiates heat equivalent to 59 percent of incoming sunlight; the surface radiates only 12 percent. In other words, most solar heating happens at the surface, while most radiative cooling happens in the atmosphere
For these pictures is more or less friction needed?
Answer:
8: More
9: More
10: More
11: Less
12: Less
12: More
Particle A and particle B are held together with a compressed spring between them. When they are released, the spring pushes them apart, and they then fly off in opposite directions, free of the spring. The mass of A is 8.00 times the mass of B, and the energy stored in the spring was 73 J. Assume that the spring has negligible mass and that all its stored energy is transferred to the particles. (a) Once that transfer is complete, what is the kinetic energy of particle A
Answer:
K_a = 8,111 J
Explanation:
This is a collision exercise, let's define the system as formed by the two particles A and B, in this way the forces during the collision are internal and the moment is conserved
initial instant. Just before dropping the particles
p₀ = 0
final moment
p_f = m_a v_a + m_b v_b
p₀ = p_f
0 = m_a v_a + m_b v_b
tells us that
m_a = 8 m_b
0 = 8 m_b v_a + m_b v_b
v_b = - 8 v_a (1)
indicate that the transfer is complete, therefore the kinematic energy is conserved
starting point
Em₀ = K₀ = 73 J
final point. After separating the body
Em_f = K_f = ½ m_a v_a² + ½ m_b v_b²
K₀ = K_f
73 = ½ m_a (v_a² + v_b² / 8)
we substitute equation 1
73 = ½ m_a (v_a² + 8² v_a² / 8)
73 = ½ m_a (9 v_a²)
73/9 = ½ m_a (v_a²) = K_a
K_a = 8,111 J
Wind is formed when hot air rises and cool air sinks. This process is called __________
conduction
insulation
radiation
convection
plsssssssss answer this correct i put in 12 points and i will give you brainliest if you answer correct
Answer:
conduction is the correct answer
In the legend of William Tell, Tell is forced to shoot an apple from his son's head for failing to show respect to a high official. In our case, let's say Tell stands 8.7 meters from his son while shooting. The speed of the 144-g arrow just before it strikes the apple is 20.4 m/s, and at the time of impact it is traveling horizontally. If the arrow sticks in the apple and the arrow/apple combination strikes the ground 8 m behind the son's feet, how massive was the apple
Answer:
M = 0.31 kg
Explanation:
This exercise must be done in parts, let's start by finding the speed of the set arrow plus apple, for this we define a system formed by the arrow and the apple, therefore the forces during the collision are internal and the moment is conserved
let's use m for the mass of the arrow with velocity v₁ = 20.4 m / s and M for the mass of the apple
initial instant. Just before the crash
p₀ = m v₁ + M 0
instant fianl. Right after the crash
p_f = (m + M) v
p₀ = p_f
m v₁ = (m + M) v
v =[tex]\frac{m}{m+M} \ v_1[/tex] (1)
now we can work the arrow plus apple set when it leaves the child's head with horizontal speed and reaches the floor at x = 8 m. We can use kinematics to find the velocity of the set
x = v t
y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²
when it reaches the ground, its height is y = 0 and as it comes out horizontally, [tex]v_{oy} = 0[/tex]
0 = h - ½ g t²
t² = 2h / g
For the solution of the exercise, the height of the child must be known, suppose that h = 1 m
t = [tex]\sqrt{ \frac{ 2 \ 1}{9.8} }[/tex]
t = 0.452 s
let's find the initial velocity
v = v / t
v = 8 / 0.452
v = 17.7 m / s
From equation 1
v = m / (m + M) v₁
m + M = [tex]m \ \frac{v_1}{v}[/tex]
M = m + m \ \frac{v_1}{v}
we calculate
M = 0.144 + 0.144 [tex]\frac{20.4}{17.7}[/tex]
M = 0.31 kg
Which of these would have the highest temperature?
ice
· Water
water vapor
Answer:
water vapor
Explanation:
did assignment on edge
can someone please help me I am so behind I neee to catch up but I need it to be correct both of them
Answer:
1.B, 2.A
Explanation:
If the net force acting on an object is 0 N, you can be sure that the forces acting on the object are
A. balanced B.Unbalanced C. acting at the same direction
I think the answer would be A.
After all, it is 0 which is technically a dead center number meaning that the net should be balanced and still.
Hope this helps and have a nice day.
-R3TR0 Z3R0
Count how many significant figures each of the quantities below has:
a. 2.590 km
b.12.303 ml
c. 7800kg
A Sonometer wire of length
l vibrates at a
frequency of
350 Hz. when the length of the
wire is increased by 15cm, the
Wire Vibrates at 280 H₂ under
constant tension. Determine l
Answer:
-75 cm
Explanation:
At l ; F = 350 Hz
At l + 15 cm ; F = 280 Hz
I = 350
I + 15 = 280
280I = 350(I + 15)
280I = 350I + 5250
280I - 350I = 5250
-70I = 5250
I = - 75cm
The length is - 75 cm
can someone explain how to find the tangent line of something :D
Answer:
This can help you! Pictures tell more than 100s of word.
Explanation:
A 52 kg and a 95 kg skydiver jump from an airplane at an altitude of 4750 m, both falling in the pike position. Assume all values are accurate to three significant digits. (Assume that the density of air is 1.21 kg/m3 and the drag coefficient of a skydiver in a pike position is 0.7.) If each skydiver has a frontal area of 0.14 m2, calculate their terminal velocities (in m/s). 52 kg skydiver m/s 95 kg skydiver m/s How long will it take (in s) for each skydiver to reach the ground (assuming the time to reach terminal velocity is small)
Answer: 52 kg skydiver: 9.09 m/s and 522.55 s
95 kg skydiver: 12.3 m/s and 386.2 s
Explanation: Drag Force is an opposite force when an object is moving in a fluid.
For skydivers, when falling through the air, the forces acting on it are gravitational and drag forces. At a certain point, drag force equals gravitational force, which is constant on any part of the planet, producing a net force that is zero. Since there is no net force, there is no acceleration and, consequently, velocity is constant. When that happens, the person reached the Terminal Velocity.
Drag Force and Velocity are proportional to the squared speed. So, terminal velocity is given by:
[tex]F_{G}=F_{D}[/tex]
[tex]mg=\frac{1}{2}C \rho Av_{T}^{2}[/tex]
[tex]v_{T}=\sqrt{\frac{2mg}{\rho CA} }[/tex]
where
m is mass in kg
g is acceleration due to gravitational force in m/s²
ρ is density of the fluid in kg/m³
C is drag coefficient
A is area of the object in the fluid in m²
Calculating:
The 52kg skydiver has terminal velocity of:
[tex]v_{T}=\sqrt{\frac{2(52)(9.8)}{(1.21)(0.7)(0.14)} }[/tex]
[tex]v_{T}=[/tex] 9.09
The 95kg skydiver's terminal velocity is
[tex]v_{T}=\sqrt{\frac{2(95)(9.8)}{(1.21)(0.7)(0.14)} }[/tex]
[tex]v_{T}=[/tex] 12.3
The 52 kg and 95kg skydivers' terminal velocity are 9.09m/s and 12.3m/s, respectively.
The time each one will reach the floor will be:
52 kg at 9.09 m/s:
[tex]t=\frac{4750}{9.09}[/tex]
t = 522.5
95 kg at 12.3 m/s:
[tex]t=\frac{4750}{12.3}[/tex]
t = 386.2
The 52 kg and 95kg skydivers' time to reach the floor are 522.5 s and 386.2 s, respectively.