Answers
Answer:
864 mT
Explanation:
The magnetic field due to a long straight wire B = μ₀i/2πR where μ₀ = permeability of free space = 4π × 10⁻⁷ H/m, i = current in wire, and R = distance from center of wire to point of magnetic field.
The magnitude of magnetic field due to the first wire carrying current i = 2.70 A at distance R which is mid-point between the wires is B = μ₀i/2πR.
Since the other wire also carries the same current at distance R, the magnitude of the magnetic field is B = μ₀i/2πR.
The resultant magnetic field at B is B' = B + B = 2B = 2(μ₀i/2πR) = μ₀i/πR
Now R = 2.50 cm/2 = 1.25 cm = 1.25 × 10⁻² m and i = 2.70 A.
Substituting these into B' = μ₀i/πR, we have
B' = 4π × 10⁻⁷ H/m × 2.70 A/π(1.25 × 10⁻² m)
B = 10.8/1.25 × 10⁻⁵ T
B = 8.64 × 10⁻⁵ T
B = 864 × 10⁻³ T
B = 864 mT
This question involves the concept of the magnetic field due to two current-carrying wires in the same direction, parallel to each other.
The magnitude of the magnetic field at the point P, which is equidistant from the wires is "8.64 x 10⁻⁵ T".
The following formula is used to find the magnetic field at the center distance between two parallel current-carrying wires in the same direction:
[tex]B = \frac{\mu_oI_1}{2\pi r}+\frac{\mu_oI_2}{2\pi r}\\\\But,\ I_1=I_2=I\\\\B = \frac{\mu_oI}{\pi r}[/tex]
where,
B = magnetic field at required point = ?
μ₀ = permeability of free space = 4π x 10⁻⁷ H/m
I = current = 2.7 A
r = distance from wires to the point = 2.5 cm/2 = 1.25 cm = 0.0125 m
Therefore,
[tex]B=\frac{(4\pi\ x\ 10^{-7}\ H/m)(2.7\ A)}{\pi (0.0125\ m)}[/tex]
B = 8.64 x 10⁻⁵ T
Learn more about the magnetic field here:
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Related Questions
. Assume that the batter does hit the ball. If the bat's instantaneous angular velocity is 30 rad/s at the instant of contact, and the distance from the sweet spot on the bat to the axis of rotation is 1.25 m, what is the instantaneous linear velocity of the sweet spot at the instant of ball contact
Answers
Answer:
37.5 m/s
Explanation:
Using,
Formula
v = ωr....................... Equation 1
Where ω = instantaneous angular velocity, v = instantaneous linear velocity, r = radius or distance from the sweet spot of the bat to the axis of rotation.
From the question,
Given: ω = 30 rad/s, r = 1.25 m
Substitute these values into equation 1
v = 30(1.25)
v = 37.5 m/s.
Hence the instantaneous linear velocity of the sweet spot at the instant of ball contact is 37.5 m/s
On a 10 kg cart (shown below), the cart is brought up to speed with 50N of force for 7m, horizontally. At this point (A), the cart begins to experience an average frictional force of 15N throughout the ride.
Find:
a) The total energy at (A)
b) The velocity at (B)
c) The velocity at (C)
d) Can the cart make it to Point (D)? Why or why not?
Answers
A projectile is fired with an initial velocity of 120.0 m/s at an angle, θ, above the horizontal. If the projectile’s initial horizontal speed is 55 meters per second, then angle θ measures approximately
Answers
Answer:
algm sabe tô precisando muito
The knee extensors insert on the tibia at an angle of 30 degrees (from the longitudinal axis of the tibia), at a distance of 3 cm from the axis of rotation at the knee. How much force must the knee extensors exert to produce an angular acceleration at the knee of 1 rad/s2 , given a mass of the lower leg and foot of 4.5 kg, and a radius of gyration of 23 cm
Answers
Answer:
the knee extensors must exert 15.87 N
Explanation:
Given the data in the question;
mass m = 4.5 kg
radius of gyration k = 23 cm = 0.23 m
angle ∅ = 30°
∝ = 1 rad/s²
distance of 3 cm from the axis of rotation at the knee r = 3 cm = 0.03 m
using the expression;
ζ = I∝
ζ = mk²∝
we substitute
ζ = 4.5 × (0.23)² × 1
ζ = 0.23805 N-m
so
from; ζ = rFsin∅
F = ζ / rsin∅
we substitute
F = 0.23805 / (0.03 × sin( 30 ° )
F = 0.23805 / (0.03 × 0.5)
F F = 0.23805 / 0.015
F = 15.87 N
Therefore, the knee extensors must exert 15.87 N
if the water measures -5 feet at low tide and 3ft at high tide what is the tidal range
Answers
Answer:
8 feet
................
Interactive Solution 8.29 offers a model for this problem. The drive propeller of a ship starts from rest and accelerates at 2.38 x 10-3 rad/s2 for 2.04 x 103 s. For the next 1.48 x 103 s the propeller rotates at a constant angular speed. Then it decelerates at 2.63 x 10-3 rad/s2 until it slows (without reversing direction) to an angular speed of 2.42 rad/s. Find the total angular displacement of the propeller.
Answers
Answer:
Δθ = 15747.37 rad.
Explanation:
The total angular displacement is the sum of three partial displacements: one while accelerating from rest to a certain angular speed, a second one rotating at this same angular speed, and a third one while decelerating to a final angular speed.Applying the definition of angular acceleration, we can find the final angular speed for this first part as follows:[tex]\omega_{f1} = \alpha * \Delta t = 2.38*e-3rad/s2*2.04e3s = 4.9 rad/sec (1)[/tex]
Since the angular acceleration is constant, and the propeller starts from rest, we can use the following kinematic equation in order to find the first angular displacement θ₁:[tex]\omega_{f1}^{2} = 2* \alpha *\Delta\theta (2)[/tex]
Solving for Δθ in (2):[tex]\theta_{1} = \frac{\omega_{f1}^{2}}{2*\alpha } = \frac{(4.9rad/sec)^{2}}{2*2.38*e-3rad/sec2} = 5044.12 rad (3)[/tex]
The second displacement θ₂, (since along it the propeller rotates at a constant angular speed equal to (1), can be found just applying the definition of average angular velocity, as follows:[tex]\theta_{2} =\omega_{f1} * \Delta_{t2} = 4.9 rad/s * 1.48*e3 s = 7252 rad (4)[/tex]
Finally we can find the third displacement θ₃, applying the same kinematic equation as in (2), taking into account that the angular initial speed is not zero anymore:[tex]\omega_{f2}^{2} - \omega_{o2}^{2} = 2* \alpha *\Delta\theta (5)[/tex]
Replacing by the givens (α, ωf₂) and ω₀₂ from (1) we can solve for Δθ as follows:[tex]\theta_{3} = \frac{(\omega_{f2})^{2}- (\omega_{f1}) ^{2} }{2*\alpha } = \frac{(2.42rad/s^{2}) -(4.9rad/sec)^{2}}{2*(-2.63*e-3rad/sec2)} = 3451.25 rad (6)[/tex]
The total angular displacement is just the sum of (3), (4) and (6):Δθ = θ₁ + θ₂ + θ₃ = 5044.12 rad + 7252 rad + 3451.25 rad ⇒ Δθ = 15747.37 rad.Rhodium is in period 5 of the periodic table. What does this tell you about this element
Answers
Answer:
. It is an extraordinarily rare, silvery-white, hard, corrosion-resistant, and chemically inert transition metal. It is a noble metal and a member of the platinum group.
Explanation:
Carl works hard to get a grades on his report card because his mother pays him 25 dollars for each semester he earns straight as Carl’s behavior is being influenced by
Answers
During which phase is the moon not visible?
A) Full Moon
B) First quarter
C) New moon
D) Waxing crescent
Answers
Answer:
they are right it is a new moon
Explanation:
took the test
A scientist adds different amounts of salt to 5 bottles of water. She then measures how long it takes for the water to boil. What is the responding variable in this experiment
Answers
Answer:
the responding variable is the water boiling
Explanation:
a responding variable is the same thing as a dependent variable and an independent variable you change the independent variable is the amount of salt, the control group is how long water takes to boil without adding salt, and a constant is the same amount of water
A falling 0.60 kg object experiences a frictional force due to air resistance of 1.5 N. What is the object's acceleration?
Answers
Answer:
7.5 m/s².
Explanation:
From the question given above, the following data were:
Mass (m) of object = 0.6 Kg
Force of friction (Fբ) = 1.5 N
Acceleration (a) =?
Next, we shall determine the force of gravity on the object. This can be obtained as follow:
Mass (m) of object = 0.6 Kg
Acceleration due to gravity (g) = 10 m/s²
Force of gravity (F₉) =?
F₉ = mg
F₉ = 0.6 × 10
F₉ = 6 N
Next, we shall determine the net force acting on the object. This can be obtained as follow:
Force of friction (Fբ) = 1.5 N
Force of gravity (F₉) = 6 N
Net force (Fₙ) =?
Fₙ = F₉ – Fբ
Fₙ = 6 – 1.5
Fₙ = 4.5 N
Finally, we shall determine the acceleration of the object. This can be obtained as follow:
Mass (m) of object = 0.6 Kg
Net force (Fₙ) = 4.5 N
Acceleration (a) of object =?
Fₙ = ma
4.5 = 0.6 × a
Divide both side by 0.6
a = 4.5 / 0.6
a = 7.5 m/s²
Therefore, the acceleration of the object is 7.5 m/s²
An 5 kg object moving at 10 m/s will have a momentum equaling ____________.
15 kg m/s/s
15 kg m/s
Answers
Answer:
50Kgm/s
Explanation:
Momentum=Mass*Velocity
P=mv
Given Mass=5Kg. Given Velocity=10m/s
Momentum=5*10=50Kgm/s
A class is learning about states of matter. The students set up the investigation in the diagram.
Which kinds of energy are needed in this investigation to change the state of matter of the owl made of wax?
Answers
Sorry I’m just trying to get my 2 answers down
A reaction occurs when a compound breaks down. This reaction has one reactant and two or more products. Energy, as from a battery, is usually needed to break the compound apart.
Answers
Answer:
decomposition
Explanation:
How much work is done when 100 N of force is applied to a rock to move it 20 m
Answers
Answer: 2000 J
Explanation: work W = F s
A student's backpack has a mass of 9.6 kg. The student applies a force of 94.08 N [up] while walking through 1.4 km [E] to get to school. Calculate the work done by the student on the backpack
Answers
The student does zero work on the backpack because the upward force applied by the student is acting perpendicular to the backpack's displacement parallel to the ground.
According to Newton's first law, an object at rest will _____.
never move
stay at rest forever
start moving
stay at rest unless moved by force
Answers
1. A perspex box has a 10 cm square base and contains water to a height of 10 cm. A piece of rock of mass 600g is lowered into the water and the level rises to 12 cm.
(a) What is the volume of water displaced by the rock?
(b) What is the volume of the rock?
(c) Calculate the density of the rock
Answers
Answer:
(a) The volume of water is 100 cm³
(b) The volume of the rock is 20 cm³
(c) The density of the rock is 30 g/cm³
Explanation:
The given parameters of the perspex box are;
The area of the base of the box, A = 10 cm²
The initial level of water in the box, h₁ = 10 cm
The mass of the rock placed in the box, m = 600 g
The final level of water in the box, h₂ = 12 cm
(a) The volume of water in the box, 'V', is given as follows;
V = A × h₁
∴ The volume of water in the box, V = 10 cm² × 10 cm = 100 cm³
The volume of water in the box, V = 100 cm³
(b) When the rock is placed in the box the total volume, [tex]V_T[/tex], is given by the sum of the rock, [tex]V_r[/tex], and the water, V, is given as follows;
[tex]V_T[/tex] = [tex]V_r[/tex] + V
[tex]V_T[/tex] = A × h₂
∴ [tex]V_T[/tex] = 10 cm² × 12 cm = 120 cm³
The total volume, [tex]V_T[/tex] = 120 cm³
The volume of the rock, [tex]V_r[/tex] = [tex]V_T[/tex] - V
∴ [tex]V_r[/tex] = 120 cm³ - 100 cm³ = 20 cm³
The volume of the rock, [tex]V_r[/tex] = 20 cm³
(c) The density of the rock, ρ = (Mass of the rock, m)/(The volume of the rock)
∴ The density of the rock, ρ = 600 g/(20 cm³) = 30 g/cm³
1. Alexandra and Rachel are on a train that sounds a whistle at a constant frequency as
it leaves the train station. Compared to the sound emitted by the whistle, the sound that
the passengers standing on the platform hear has a frequency that is
a. lower, because the sound-wave fronts reach the platform at a frequency
lower than the frequency at which they are produced
b. lower, because the sound waves travel more slowly in the still air above the
platform than in the rushing air near the train
c. higher, because the sound-wave fronts reach the platform at a frequency
higher than the frequency at which they are produced
d. higher, because the sound waves travel faster in the still air above the
platform than in the rushing air near the train
Answers
Answer: the answer would be C trust me i took the test if its not that its b
hope that helps
Explanation: i took the test
answer:
a) lower because the sound-wave fronts reach the platform at a frequency lower than the frequency at which they are produced
explanation :3
If the train is leaving the train station, then the people who are standing on the platform would hear a sound with a lower frequency since the train is moving further away. ^^
Kevin decides to soup up his car by replacing the car's wheels with ones that have 1.4 times the diameter of the original wheels. Note that the speedometer in a car is calibrated based on the tire's diameter and on the distance the tire covers in each revolution. (a) Will the reading of the speedometer change
Answers
Answer:
No.
Explanation:
Given that Kevin decides to soup up his car by replacing the car's wheels with ones that have 1.4 times the diameter of the original wheels. Note that the speedometer in a car is calibrated based on the tire's diameter and on the distance the tire covers in each revolution. (a) Will the reading of the speedometer change ?
Considering the formula
V = wr
Where
V = linear speed
W = angular speed
r = radius of the wheel.
But W = 2πrf
Where the the 2 and pi are constant. The radius of the first wheel will be small but counter balance with the larger frequency.
While the radius of the second wheel may be large but it will be of a small frequency.
We can therefore conclude that the reading on the speedometer will not change. Because speedometer will read the linear speed V.
which causes magnets to stick to metal
Answers
Answer:
Steel
Explanation:
Steel is a metal that magnets stick to because iron can be found inside steel
Answer:Magnets stick to any metal that contains iron, cobalt or nickel.
Explanation:Iron is found in steel, so steel attracts a magnet and sticks to it. Stainless steel, however, does not attract a magnet.
(a) What do you mean by rest?
Answers
I hope this helps.
3. Two bullets have masses of 0.003 kg and 0.006 kg, respectively. Both are fired with a speed of 40.0 m/s.
A. Which bullet has more kinetic energy?
B. When you double the mass, what happens to the kinetic energy?
Answers
Answer:
A. The bullet with 0.006kg has more energy
B. When the mass is doubled the kinetic energy increases
Explanation:
Kinetic energy increases when mass increases
kinetic energy increases when velocity increases
Highest density of electrostatic charges in a metal is found where
Answers
I don't know the answer but I just want points sorry
A cylindrical body has 6 m height and its radius is 2 metre calculate its volume. Ans :75.428m3
Answers
Answer:
75.4
Explanation:
r= 2
h= 6
v= 22/7 *r*r*h
v= 75.42
2. A uniform wire of resistance R is stretched until its length doubles. Assuming its density and resistivity remain constant, what is its new resistance
Answers
Answer:
Resistance is quadrupled.
Explanation:
Solving this requires us to use the formula of resistivity.
Resistivity is usually said to be the measure of the resistance of a particular size of any given material to the electrical conduction. It is mathematically represented as
ρ = RA/L, where
ρ = the resistivity of the given material
R = the resistance of the material
A = the area of the material
L = length of the material.
From the question, we're told that the length is doubled with the resistivity and density remaining constant. If the density is constant, this makes the volume constant as well.
Volume, V = A * L. We're then told that the length is doubled. If the length is doubled, for the volume to remain constant, then the area must be halved.
Volume, V = A/2 * 2L
Making, Resistance R, subject of the formula, we have
R = ρL/A.
Since resistivity is constant and the area is halved, we then have
R = 2L / (1/2A)
R = 4L / A
If the length is doubled, we have the resistance to be quadrupled
If an observer on Earth sees a total lunar eclipse, Group of answer choices everyone on the nighttime side of Earth is seeing it. someone elsewhere on Earth must be seeing a partial lunar eclipse. someone elsewhere on Earth must be seeing a total solar eclipse
Answers
Answer:
everyone on the nighttime side of Earth is seeing it.
Explanation:
A lunar eclipse is a phenomenon that occurs when the Earth comes between the Moon and the Sun thereby causing it to cover the Moon with its shadow.
Simply stated, lunar eclipse takes place when the Moon passes or moves through the Earth's shadow thereby blocking any ray of sunlight from reaching the Moon. Thus, the full moon appears deep red (blood moon).
Also, a lunar eclipse would occur only when the Sun, Earth, and Moon are closely aligned to form a straight line known as the syzygy.
There are three (3) types of lunar eclipse and these are;
1. Total lunar eclipse.
2. Partial lunar eclipse.
3. Penumbra lunar eclipse.
Generally, if an observer on Earth sees a total lunar eclipse, everyone on the nighttime side of Earth is seeing it because it's quite easy to see a total lunar eclipse while the full moon passes through the innermost part of the shadow of the earth.
How fast were both runners traveling after 4 seconds?
40
Distance (in yards)
30
20
10
1
2.
3
0
Time in seconds
Answers
Answer:
they were fast ⛷⛷
A girl weighing 45kg is standing on the floor, exerting a downward force of 200N on the floor. The force exerted on her by the floor is ..............
Select one:
a.
No force exerted
b.
Less than 2000N
c.
Equal to 200 N
d.
Greater than 200 N
Answers
Answer:
c.
Equal to 200 N..........
Which time interval has the greatest speed?
Answers
Answer:
es la 2
Explanation:
epero que te curva