Answer:
326,000
Explanation:
One acre-foot equals about 326,000 gallons, or enough water to cover an acre of land, about the size of a football field, one foot deep. An average California household uses between one-half and one acre-foot of water per year for indoor and outdoor use.
On a car trip you drive for 2 hours and 41 minutes on a highway at a speed of 107.0 km/h. Then you stop at a gas station to fill up your tank. You also eat a quick lunch. The whole break lasts 23 minutes. After the break you start your engine up and you switch to a state road. You drive for another 3 hours and 31 minutes at a speed of 67.0 km/h before you arrive to your destination. What was your average speed for the whole trip with the lunchbreak included
Answer:
v = 79.3 km/h
Explanation:
By definition, the average speed, is the quotient between the total distance traveled and the time needed to travel that distance.The total time, is the sum of three times: one while driving before stopping at the gas station (t₁), the time spent there (t₂) and the time since leaving the gas station until reaching the final destination (t₃) .Let's convert these times to seconds first:[tex]t_{1} = 161 min* \frac{60s}{1min} = 9660 s (1)[/tex]
[tex]t_{2} = 23 min* \frac{60s}{1min} = 1380 s (2)[/tex]
[tex]t_{3} = 211 min* \frac{60s}{1min} = 12660 s (3)[/tex]
[tex]t_{tot} =t_{1} +t_{2} +t_{3} = 9660s + 1380s + 12660s = 23700s (4)[/tex]
In order to find the total distance traveled, we need to add the distance traveled before stopping at the gas station (x₁) and the distance traveled after leaving it (x₂).Applying the definition of average speed, we can find these distances as follows:[tex]x_{1} = v_{1} * t_{1} (5)[/tex]
[tex]x_{2} = v_{2} * t_{3} (6)[/tex]
where v₁ = 107.0 km/h, and v₂= 67.0 km/hAs we did with time, let's convert v₁ and v₂ to m/s:[tex]v_{1} = 107.0 km/h*\frac{1000m}{1km}*\frac{1h}{3600s} = 29.7 m/s (7)[/tex]
[tex]v_{2} = 67.0 km/h*\frac{1000m}{1km}*\frac{1h}{3600s} = 18.6 m/s (8)[/tex]
Replacing (7) and (1) in (5) we get x₁, as follows (in meters):[tex]x_{1} = v_{1} * t_{1} = 29.7 m/s * 9660 s = 286902 m (9)[/tex]
Doing the same for x₂ with (3) and (8):[tex]x_{2} = v_{2} * t_{3} = 18.6 m/s * 12660 s = 235476 m (10)[/tex]
Total distance traveled is just the sum of (9) and (10):[tex]x_{tot} = x_{1} +x_{2} = 286902 m + 235476 m = 522378 m (11)[/tex]
As we have already said, the average speed is just the quotient between (11) and (4), as follows:[tex]v_{avg} =\frac{\Delta x}{\Delta t} = \frac{522378m}{23700s} = 22.0 m/s (12)[/tex]
Converted back to km/h:[tex]v_{avg} = 22.0 m/s*\frac{1km}{1000m}*\frac{3600s}{1h} = 79.3 km/h (13)[/tex]
The new springs will be identical to the original springs, except the force constant will be 5655.00 N/m smaller. When James removes the original springs, he discovers that the length of each spring expands from 8.55 cm (its length when installed) to 12.00 cm (its length with no load placed on it). If the mass of the car body is 1355.00 kg, by how much will the body be lowered with the new springs installed, compared to its original height
Answer:
Explanation:
For original spring , compression in spring due to a load of 1355 kg is
x = 12 - 8.55 = 3.45 cm = .0345 m
spring constant = W / x
= 1355 x 9.8 / .0345
= 384898.55 N /m
Spring constant of new spring
k = 384898.55 - 5655 = 379243.55 N /m
New compression for new spring
= W / k
= 1355 x 9.8 / 379243.55
= .035 m
= 3.50 cm
Difference of compression = 3.50 - 3.45
= .05 cm .
In later case , car will be more lowered by .05 cm .
who has brown hair and brown eyes but is a boy
Answer:
I have strawberry blonde/brown hair blue eyes and a girl lol
Explanation:
A loaded wagon of mass 10,000 kg moving with a speed of 15 m/s strikes a stationary wagon of the same mass making a perfect inelastic collision. What will be the speed of coupled wagons after collision?
Answer:
7.5 m/s
Explanation:
Unfortunately, I don't have an explanation but I guessed the correct answer.
While talking to a friend, a construction worker momentarily set her cell phone down on one end of an iron rail of length 7.50 m. At that moment, a second worker dropped a wrench so that it hit the other end of the rail. The person on the phone detected two pulses of sound, one that traveled through the air and a longitudinal wave that traveled through the rail. (Assume the speed of sound in iron is 5,950 m/s and the speed of sound in air is 343 m/s).
A) Which pulse reaches the cell phone first?
B) Find the separation in time (in s) between the arrivals of the two pulses.
Answer:
A)
The impulse that reaches the cell phone first is the Longitudinal wave
B) 0.0206 seconds
Explanation:
length of Iron rail = 7.5 m
speed of sound in Iron = 5950 m/s
speed of sound in Air = 343 m/s
A) Determine which pulse reaches the cell phone first
The impulse that reaches the cell phone first is the Longitudinal wave
Time for longitudinal pulse to be detected = 7.5 / 5950 = 0.00126 s
Time for pulse through air to be detected = 7.5 / 343 = 0.02186 s
B) separation in time between the arrivals of the two pulses
ΔT = 0.02186 - 0.00126 = 0.0206 seconds
In a certain region of space the electric potential increases uniformly from east to west and does not vary in any other direction. The electric field:Group of answer choicespoints east and varies with positionpoints east and does not vary with positionpoints west and varies with positionpoints west and does not vary with positionpoints north and does not vary with position
Answer:
Explanation:
The relation between electric field and potential difference is as follows
E = - dV / dr
That means if dV is positive , E is negative . In other words , if potential increases , E is negative or in opposite direction in which potential increases .
Here the electric potential increases uniformly from east to west , that means electric field is from west to east . Since potential is uniformly increasing that means
dV / dr = constant
E = constant
Electric field is constant .
So the option which is correct is
" points east and does not vary with position " .
In which situation are waves transmitted?
O A. A patient wears a lead apron at the dentist's office when getting
teeth X-rays.
O B. A light in a swimming pool comes on after dark to prevent
accidents in the water.
O C. A person wears earplugs to prevent hearing damage when fueling
a jet plane at the airport.
O D. A reflective screen is put on a parked car's dashboard to keep the
car from heating up in sunlight.
Answer: B. A light in a swimming pool comes on after dark to prevent
accidents in the water.
To fully describe velocity you must have a _____
A. Magnitude and unit
B. Speed and unit
C. Average speed and position
D. Magnitude and direction
A toy car can go 5 mph. How long would it take to go 12 miles?
How do dog whistles work?
The sound it emits comes from what is known as the ultrasonic range, a pitch that is so high humans can't hear it. Dogs can hear these sounds, however, as can cats and other animals. Because of this, the dog whistle is a favored training tool, though it may not be for every dog parent.
Which option correctly matches the chemical formula of a compound with its name?
A. N2O3, dinitrogen trioxide
B. N2O, trinitrogen dioxide
C. N2O, dinitrogen trioxide
D. N2O3, trinitrogen dioxide
Answer:
A is indeed correct
Explanation:
just did the question
The option that correctly matches the chemical formula of a compound with its name is N₂O₃ dinitrogen trioxide.
What is dinitrogen trioxide?
Dinitrogen trioxide is one of the simple nitrogen oxide. The chemical compound formula of Dinitrogen trioxide N₂O₃.
That is, it consists of 2 molecules of nitrogen, hence the prefix "Di" and 3 molecules of oxygen given the prefix tri.
Thus, the option that correctly matches the chemical formula of a compound with its name is N₂O₃ dinitrogen trioxide.
Learn more about Dinitrogen trioxide here: https://brainly.com/question/21392313
#SPJ2
Which of the following is the BEST explanation for why oceans have two different types of currents?
Answer:
sddww
Explanation:
szsswa
The power lines are at a high potential relative to the ground, so there is an electric field between the power lines and the ground. To maximize the potential difference between one end of the fluorescent tube and the other, how should the tube be held? Select the best answer from the choices provided. View Available Hint(s) Select the best answer from the choices provided. The potential difference between the ends of the tube does not depend on the tube's orientation. The tube should be held horizontally, parallel to the ground. The tube should be held vertically, perpendicular to the ground.
Answer:
The tube should be held vertically and perpendicular to the ground.
Explanation:
Answer: The tube should be held vertically and perpendicular to the ground. The reason is as follows:
Reasoning:
The power lines are parallel to the ground hence, their electric field will be perpendicular to the ground and equipotential surface will be cylindrical.
Hence, if you will put fluorescent tube parallel to the ground then both the ends of the tube will lie on the same equipotential surface and the potential difference will be zero.
So, to maximize the potential the ends of the tube must be on different equipotential surfaces. The surface which is near to the power line has high potential value and the surface which is farther from the line has lower potential value.
hence, to maximize the potential difference, the tube must be placed perpendicular to the ground.
a girl whose mass is 40kg walk up a flight of 20steps each 15mm hight in 10seconds.find power developed by the girl showing the solution
Answer: Approximately 11.76 joules per second
=========================================================
Work Shown:
Mass = 40 kg
Force pulling down = (mass)*(gravity) = 40*9.8 = 392 newtons
Roughly 392 newtons of force are pulling down on her.
To climb the steps, she must apply 392 newtons of force upward.
---------------
Displacement = 20*(15 mm) = 300 mm = 0.3 m
Work = Force*Displacement
Work = 392*0.3
Work = 117.6 joules of energy
---------------
Power = (Work)/(Time)
Power = (117.6 joules)/(10 seconds)
Power = (117.6/10) joules per second
Power = 11.76 joules per second, which is approximate
Explain the difference in the function of plant and animal cell organelles, including cell membrane, cell wall, nucleus, cytoplasm, mitochondria, chloroplast, and vacuole
Answer:
Plant cell Animal cell
2. Have a cell membrane. 2. Have no chloroplasts.
3. Have cytoplasm. 3. Have only small vacuoles.
4. Have a nucleus. 4. Often irregular in shape.
5. Often have chloroplasts
containing chlorophyll. 5. Do not contain plastids.
Suppose one Sherpa uses a force of 980 N to move a load of equipment to a height of 20 meters in 25 seconds. How much power is used?
F = 980 N
h = 20 m
t = 25 s
P=? (power)
W=F*h (work)
P=W*t
P=F*h*t
P=980*20*25 =490000 W = 490 kW = 0.49 MW
As mentioned in the text, the tangent line to a smooth curve r(t) = ƒ(t)i + g(t)j + h(t)k at t = t0 is the line that passes through the point (ƒ(t0), g(t0), h(t0)) parallel to v(t0), the curve’s velocity vector at t0. In Exercises 23–26, find parametric equations for the line that is tangent to the given curve at the given parameter value t = t0.
Answer:
[tex]x = t[/tex]
[tex]y = \frac{1}{3}t[/tex]
[tex]z =t[/tex]
Explanation:
Given
[tex]r(t) = f(t)i + g(t)j + h(t)k[/tex] at [tex]t = 0[/tex]
Point: [tex](f(t0), g(t0), h(t0))[/tex]
[tex]r(t) = ln\ t_i + \frac{t-1}{t+2}j + t\ ln\ tk[/tex], [tex]t0 = 1[/tex] -- Missing Information
Required
Determine the parametric equations
[tex]r(t) = ln\ ti + \frac{t-1}{t+2}j + t\ ln\ tk[/tex]
Differentiate with respect to t
[tex]r'(t) = \frac{1}{t}i +\frac{3}{(t+2)^2}j + (ln\ t + 1)k[/tex]
Let t = 1 (i.e [tex]t0 = 1[/tex])
[tex]r'(1) = \frac{1}{1}i +\frac{3}{(1+2)^2}j + (ln\ 1 + 1)k[/tex]
[tex]r'(1) = i +\frac{3}{3^2}j + (0 + 1)k[/tex]
[tex]r'(1) = i +\frac{3}{9}j + (1)k[/tex]
[tex]r'(1) = i +\frac{1}{3}j + (1)k[/tex]
[tex]r'(1) = i +\frac{1}{3}j + k[/tex]
To solve for x, y and z, we make use of:
[tex]r(t) = f(t)i + g(t)j + h(t)k[/tex]
This implies that:
[tex]r'(1)t = xi + yj + zk[/tex]
So, we have:
[tex]xi + yj + zk = (i +\frac{1}{3}j + k)t[/tex]
[tex]xi + yj + zk = it +\frac{1}{3}jt + kt[/tex]
By comparison:
[tex]xi = it[/tex]
Divide by i
[tex]x = t[/tex]
[tex]yj = \frac{1}{3}jt[/tex]
Divide by j
[tex]y = \frac{1}{3}t[/tex]
[tex]zk = kt[/tex]
Divide by k
[tex]z = t[/tex]
Hence, the parametric equations are:
[tex]x = t[/tex]
[tex]y = \frac{1}{3}t[/tex]
[tex]z =t[/tex]
What are regular and irregular reflection of light? plz help its
urgent..
Explanation:
Regular reflection: It is the reflection from a smooth surface such that the light rays are evenly parallel to each other and an image is formed. ... Irregular reflection: It is the diffused reflection from uneven surface such that the light rays are not parallel to each other and do not form an image.
what happens when a wave passes through a medium ?
Answer:
When waves travel from one medium to another the frequency never changes. As waves travel into the denser medium, they slow down and wavelength decreases. Part of the wave travels faster for longer causing the wave to turn. The wave is slower but the wavelength is shorter meaning frequency remains the same.
Explanation:
A 1 m3tank containing air at 10oC and 350 kPa is connected through a valve to another tank containing 3 kg of air at 35oC and 150 kPa. Now the valve is opened, and the entire system is allowed to reach thermal equilibrium with the surroundings, which are at 20oC. Determine the volume of the second tank and the final equilibrium pressure of air.
Answer:
- the volume of the second tank is 1.77 m³
- the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa
Explanation:
Given that;
[tex]V_{A}[/tex] = 1 m³
[tex]T_{A}[/tex] = 10°C = 283 K
[tex]P_{A}[/tex] = 350 kPa
[tex]m_{B}[/tex] = 3 kg
[tex]T_{B}[/tex] = 35°C = 308 K
[tex]P_{B}[/tex] = 150 kPa
Now, lets apply the ideal gas equation;
[tex]P_{B}[/tex] [tex]V_{B}[/tex] = [tex]m_{B}[/tex]R[tex]T_{B}[/tex]
[tex]V_{B}[/tex] = [tex]m_{B}[/tex]R[tex]T_{B}[/tex] / [tex]P_{B}[/tex]
The gas constant of air R = 0.287 kPa⋅m³/kg⋅K
we substitute
[tex]V_{B}[/tex] = ( 3 × 0.287 × 308) / 150
[tex]V_{B}[/tex] = 265.188 / 150
[tex]V_{B}[/tex] = 1.77 m³
Therefore, the volume of the second tank is 1.77 m³
Also, [tex]m_{A}[/tex] = [tex]P_{A}[/tex][tex]V_{A}[/tex] / R[tex]T_{A}[/tex] = (350 × 1)/(0.287 × 283) = 350 / 81.221
[tex]m_{A}[/tex] = 4.309 kg
Total mass, [tex]m_{f}[/tex] = [tex]m_{A}[/tex] + [tex]m_{B}[/tex] = 4.309 + 3 = 7.309 kg
Total volume [tex]V_{f}[/tex] = [tex]V_{A}[/tex] + [tex]V_{B}[/tex] = 1 + 1.77 = 2.77 m³
Now, from ideal gas equation;
[tex]P_{f}[/tex] = [tex]m_{f}[/tex]R[tex]T_{f}[/tex] / [tex]V_{f}[/tex]
given that; final temperature [tex]T_{f}[/tex] = 20°C = 293 K
we substitute
[tex]P_{f}[/tex] = ( 7.309 × 0.287 × 293) / 2.77
[tex]P_{f}[/tex] = 614.6211119 / 2.77
[tex]P_{f}[/tex] = 221.88 kPa ≈ 222 kPa
Therefore, the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa
Two students are on a balcony a distance h above the street. One student throws a ball vertically downward at a speed vi; at the same time, the other student throws a ball vertically upward at the same speed. Answer the following symbolically in terms of vi, g, h, and t. (Take upward to be the positive direction.)
(a) What is the time interval between when the first ball strikes the ground and the second ball strikes the ground?
?t = ______
(b) Find the velocity of each ball as it strikes the ground.
For the ball thrown upward vf = ______
For the ball thrown downward vf = ______
(c) How far apart are the balls at a time t after they are thrown and before they strike the ground?
d = _______
Answer:
Explanation:
a )
Time for first ball to reach top position
v = u - gt
0 = vi - gt
t = vi / g
Time to reach balcony while going downwards
= vi /g
Total time = 2 vi / g
Time to go down further to the ground = t₁
Total time = 2 vi / g + t₁
Time for the other ball to go to the ground = t₁
Time difference = ( 2 vi / g + t₁ ) - t₁
= 2vi / g .
( b )
v² = u² + 2gh
For both the throw ,
final displacement = h , initial velocity downwards = vi
( For the first ball also , when it go down while passing the balcony , it acquires the same velocity vi but its direction is downwards.)
vf² = vi² + 2gh
vf = √ ( vi² + 2gh )
(c )
displacement of first ball after time t
s₁ = - vi t + 1/2 g t² [ As initial velocity is upwards , vi is negative ]
displacement of second ball after time t
s₂ = vi t + 1/2 g t²
Difference = d = s₂ - s₁
= vi t + 1/2 g t² - ( - vi t + 1/2 g t² )
d = 2 vi t .
An aluminum wire having a cross-sectional area equal to 2.20 10-6 m2 carries a current of 4.50 A. The density of aluminum is 2.70 g/cm3. Assume each aluminum atom supplies one conduction electron per atom. Find the drift speed of the electrons in the wire.
Answer:
The drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.
Explanation:
We can find the drift speed by using the following equation:
[tex] v = \frac{I}{nqA} [/tex]
Where:
I: is the current = 4.50 A
n: is the number of electrons
q: is the modulus of the electron's charge = 1.6x10⁻¹⁹ C
A: is the cross-sectional area = 2.20x10⁻⁶ m²
We need to find the number of electrons:
[tex] n = \frac{6.022\cdot 10^{23} atoms}{1 mol}*\frac{1 mol}{26.982 g}*\frac{2.70 g}{1 cm^{3}}*\frac{(100 cm)^{3}}{1 m^{3}} = 6.03 \cdot 10^{28} atom/m^{3} [/tex]
Now, we can find the drift speed:
[tex]v = \frac{I}{nqA} = \frac{4.50 A}{6.03 \cdot 10^{28} atom/m^{3}*1.6 \cdot 10^{-19} C*2.20 \cdot 10^{-6} m^{2}} = 2.12 \cdot 10^{-4} m/s[/tex]
Therefore, the drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.
I hope it helps you!
What is the correct coefficient for 2H2 + O2 →2H2O
Explanation:
2forH2,1for02,and2forH20
A point charge, Q1 = -4.2 μC, is located at the origin. A rod of length L = 0.35 m is located along the x-axis with the near side a distance d = 0.45 m from the origin. A charge Q2 = 10.4 μC is uniformly spread over the length of the rod.Part (a) Consider a thin slice of the rod, of thickness dx, located a distance x away from the origin. What is the direction of the force on the charge located at the origin due to the charge on this thin slice of the rod? Part (b) Write an expression for the magnitude of the force on the point charge, |dF|, due to the thin slice of the rod. Give your answer in terms of the variables Q1, Q2, L, x, dx, and the Coulomb constant, k. Part (c) Integrate the force from each slice over the length of the rod, and write an expression for the magnitude of the electric force on the charge at the origin. Part (d) Calculate the magnitude of the force |F|, in newtons, that the rod exerts on the point charge at the origin.
Answer:
a) attractiva, b) dF = [tex]k \frac{Q_1 \ dQ_2}{dx}[/tex], c) F = [tex]k Q_1 \frac{Q_2}{d \ (d+L)}[/tex], d) F = -1.09 N
Explanation:
a) q1 is negative and the charge of the bar is positive therefore the force is attractive
b) For this exercise we use Coulomb's law, where we assume a card dQ₂ at a distance x
dF = [tex]k \frac{Q_1 \ dQ_2}{dx}[/tex]
where k is a constant, Q₁ the charge at the origin, x the distance
c) To find the total force we must integrate from the beginning of the bar at x = d to the end point of the bar x = d + L
∫ dF = [tex]k \ Q_1 \int\limits^{d+L}_d {\frac{1}{x^2} } \, dQ_2[/tex]
as they indicate that the load on the bar is uniformly distributed, we use the concept of linear density
λ = dQ₂ / dx
DQ₂ = λ dx
we substitute
F = [tex]k \ Q_1 \lambda \int\limits^{d+L}_d \, \frac{dx}{x^2}[/tex]
F = k Q1 λ ([tex]-\frac{1}{x}[/tex])
we evaluate the integral
F = k Q₁ λ [tex](- \frac{1}{d+L} + \frac{1}{d} )[/tex]
F = k Q₁ λ [tex]( \frac{L}{d \ (d+L)})[/tex]
we change the linear density by its value
λ = Q2 / L
F = [tex]k Q_1 \frac{Q_2}{d \ (d+L)}[/tex]
d) we calculate the magnitude of F
F =9 10⁹ (-4.2 10⁻⁶) [tex]\frac{10.4 10x^{-6} }{0.45 ( 0.45 +0.35)}[/tex]
F = -1.09 N
the sign indicates that the force is attractive
Answer:
a)Toward the rod
b)|dF| = k|Q1|Q2(dx/L)/x^2
c)|F| = k|Q1|Q2/(d(d+L))
d)Plug in for answer c and solve
Explanation:
A)
Q1 is negative and Q2 is positive so it is an attractive force to where the rod is located.
B)
The formula for Force due to electric charges is F=kQ1Q2/r^2
In this case, Q2 is distrusted through the length of the rod as opposed to a single point charge. As such Q2 is actually Q2*dx/L as dx is a small portion of the full length, L.
The radius between Q1 and Q2 depends on the section of the rod taken so r will be the variable x distance from Q1.
The force is only from a small portion of the rod so more accurately, we are finding |dF| as opposed to the full force, F, caused by the whole rod.
The final formula is |dF| = k|Q1|Q2(dx/L)/x^2
C)
Integrating with respect to the only changing variable, x, which spans the length of the rod, from radius = d to d+L we get this:
F = integral from d to d+L of k|Q1|Q2(dx/L)/x^2
factor out constants
F = kQ1Q2/L * integral d to d+L(1/x^2)dx
F = kQ1Q2/L * (-1/x)| from d to d+L
F = kQ1Q2/L * (-1/d+L - -1/d)
F = kQ1Q2/L * (-d/(d(d+L)) + (d+L)/(d(d+L))
F = kQ1Q2/L * (L)/(d(d+L))
F = kQ1Q2/(d(d+L))
D)
Plug in the given values into c and you have your answer.
Find the GCF of each set of numbers.
12, 21, 30
Math
Answer:
3 is the GCF for all these numbers if thats what you're asking
20. For each improvement in glider design, engineers follow
O A. the written instructions that are provided in the hang glider build kit.
O B. an iterative process of testing, modifying, retesting, and modifying again.
O C. a complicated process of checks and balances while obtaining financing.
O D. a mathematical process, rejecting designs that don't follow blueprint dimensions.
Turn In
This is the build up of substance such as pesticides in an organism and occurs when an organism absorb a substance at a rate faster than that at which the substance is lost
Answer:
which the substance is lost by catabolism and excretion.
Explanation:
A cyclist cover 6km in 20minutes. His speed is
Answer:
The speed of a cyclist is 0.3 km/min.
Explanation:
Given
The distance d = 6km Time t = 20 minutesTo determine
We need to determine the speed of a cyclist.
In order to determine the speed of a cyclist, all we need to do is to divide the distance covered by a cyclist by the time taken to cover the distance.
We know the formula involving speed, time, and distance
[tex]s=\frac{d}{t}[/tex]
where
s = speedd = distance coveredt = time takensubstitute d = 6, and t = 20 in the formula
[tex]s=\frac{d}{t}[/tex]
[tex]s=\frac{6}{20}[/tex]
Cancel the common factor 2
[tex]s=\frac{3}{10}[/tex]
[tex]s=0.3[/tex] km/min
Thus, the speed of a cyclist is 0.3 km/min.
Earth's magnetic field is approximately 1/2 gauss, that is 50 micro-tesla because the SI field unit of a tesla is 10,000 gauss. Earth's north geographic pole is close to its south magnetic pole, and magnetic field is directed from the north to the south poles of a magnetic dipole so it goes from Earth's south geographic pole towards its north. Suppose you have wire carrying a large DC current from the south wall of a building to its north wall and that it is horizontal, on the floor. If Earth's field is parallel to the ground and does not dip, what force if any would the wire experience
Answer:
F = 0
Explanation:
The magnetic force is described by two expressions
for a moving charge
F = q v x B
for a wire with a current
F = I L xB
bold indicates vectors
let's write this equation in module form
F = I L B sin θ
where the angle is between the direction of the current and the direction of the magnetic field
In this case they indicate that the cable goes from the South wall to the North wall, so this is the direction of the current
The magnetic field of the Earth goes from the south to the north and in this part it is horizontal
Therefore the current and the magnetic field are parallel, the angle between them is zero
sin 0 = 0
consequently the magnetic force is zero
F = 0
convert 0.0345mW
to MW
Answer:
3.45e-11MV
that is ur answer