You want to produce a magnetic field of magnitude 5.50 x 10¹ T at a distance of 0.0 6 m from a long, straight wire's center. (a) What current is required to produce this field? (b) With the current found in part (a), how strong is the magnetic field 8.00 cm from the wire's center?
Answer:
(a) I = 1650000 A
(b) 4.125 T
Explanation:
Magnetic field, B = 5.5 T
distance, r = 0.06 m
(a) Let the current is I.
The magnetic field due to a long wire is given by
[tex]B =\frac{\mu o}{4\pi }\frac{2 I}{r}\\5.5= 10^{-7}\times \frac{2\times I}{0.06}\\I =1650000 A[/tex]
(b) Let the magnetic field is B' at distance r = 0.08 m.
[tex]B =\frac{\mu o}{4\pi }\frac{2 I}{r}\\B = 10^{-7}\times \frac{2\times 1650000}{0.08}\\B'= 4.125 T[/tex]
One solenoid is centered inside another. The outer one has a length of 50.0 cm and contains 6750 coils, while the coaxial inner solenoid is 3.0 cm long and 0.120 cm in diameter and contains 15 coils. The current in the outer solenoid is changing at 49.2 A>s. (a) What is the mutual inductance of these solenoids
Answer: The mutual inductance of these solenoids is [tex]2.88 \times 10^{-7} H[/tex].
Explanation:
Given: Length = 50.0 cm (1 cm = 0.01 m) = 0.50 m
[tex]N_{1}[/tex] = 6750
[tex]N_{2}[/tex] = 15
Radius = [tex]\frac{0.120 cm}{2} = 0.6 cm = 6 \times 10^{-4} m[/tex]
As inner of a solenoid resembles the shape of a circle. So, its area is calculated as follows.
[tex]Area = \pi \times r^{2} = \pi \times (6 \times 10^{-4})^{2}[/tex]
Formula used to calculate mutual conductance of two solenoids is as follows.
[tex]M = \frac{\mu_{o} \times A \times N_{1} \times N_{2}}{l}[/tex]
where,
M = mutual conductance
A = area
[tex]\mu_{o}[/tex] = relative permeability = [tex]4 \pi \times 10^{-7} Tm/A[/tex]
[tex]N_{1}[/tex] = no. of coils in outer solenoid
[tex]N_{2}[/tex] = no. of coils in inner solenoid
l = length
Substitute the values into above formula as follows.
[tex]M = \frac{\mu_{o} \times A \times N_{1} \times N_{2}}{l}\\= \frac{4 \pi \times 10^{-7} Tm/A \times \pi (6 \times 10^{-4})^{2} \times 6750 \times 15}{0.5 m}\\= 2.88 \times 10^{-7} H[/tex]
Thus, we can conclude that the mutual inductance of these solenoids is [tex]2.88 \times 10^{-7} H[/tex].
Is a measurement is precuse it must also be accurate
As a new electrical engineer for the local power company, you are assigned the project of designing a generator of sinusoidal ac voltage with a maximum voltage of 120 V. Besides plenty of wire, you have two strong magnets that can produce a constant uniform magnetic field of 1.5 T over a square area with a length of 10.2 cm on a side when the magnets are separated by a distance of 12.8 cm . The basic design should consist of a square coil turning in the uniform magnetic field. To have an acceptable coil resistance, the coil can have at most 400 loops.
What is the minimum rotation rate of the coil so it will produce the required voltage? Express your answer using two significant figures.
Answer:
The rotation rate is 15.3 rad/s.
Explanation:
maximum voltage, V = 120 V
Magnetic field, B = 1.5 T
length, L = 10.2 cm
width, W = 12.8 cm
Number of loops, N = 400
Let the rate of rotation is w.
The maximum voltage is given by
V = N B A w
120 = 400 x 1.5 x 0.102 x 0.128 x w
w = 15.3 rad/s
What is the first quantum number of a 252 electron in phosphorus,
1322s22p3s23p3?
A. n=0
B. n= 3
O
C. n = 1
O D. n = 2
Answer:
the correct answer is B
Explanation:
The quantum numbers are the constants obtained when solving the Schrodinger equation, the first quantum number or principal quantum number (n), can take values from zero to infinity.
This quantum number is placed as a coefficient in the quantum distribution.
In this case for phosphorus, the number is n = 3
the correct answer is B
With explanation pls
Answer:
[tex](x \times 1) = (400 \times 2.5) \\ x = 1000 \: newtons[/tex]
[tex]y = 0[/tex]
Isotopes of the same element always have the same
(2 points)
O atomic mass number
O A-number
O Z-number
O neutrinos
Answer:
Z-number
Explanation:
The Z number is the number of protons in an atom, and this does not change when an isotope is created. I got it right on the test.
A 40-kg crate is being lowered with a downward acceleration is 2.0 m/s2 by means of a rope. (a) What is the magnitude of the force exerted by the rope on the crate
Answer:
F = 312 N
Explanation:
Given that,
The mass of a crate, m = 40 kg
Acceleration of the crate, a = 2 m/s²
As the carte is falling downward, the net force exerted by the rope on the carte is given by :
F = m(g-a)
Put all the values,
F = 40(9.8-2)
F = 312 N
Hence, the required force exerted by the rope on the crate is equal to 312 N.
Which image shows an example of potential energy?
Answer:
D
Explanation:
Potential energy involves the change of an object's position, which in this case a rocket is increasing its vertical displacement from the ground.
When a rocket is increasing its vertical displacement from the ground, it exhibits both potential and kinetic energy. Therefore option D is correct.
At the initial stage, when the rocket is on the ground and not moving, it possesses potential energy. This potential energy is in the form of stored energy due to its elevated position above the ground.
As the rocket launches and gains altitude, it continues to accumulate potential energy because it is moving higher against the force of gravity.
Simultaneously, as the rocket moves upward, it also gains kinetic energy. Kinetic energy is the energy associated with the rocket's motion.
The faster the rocket moves, the greater its kinetic energy becomes. As the rocket ascends, its speed increases, resulting in an increase in kinetic energy.
Therefore, in the context of a rocket increasing its vertical displacement from the ground, both potential energy (due to its height) and kinetic energy (due to its motion) are present.
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How do a parachutes work??4-5 sentences plsss help rn
Answer:
Explanation:
A parachute works by forcing air into the front of it and creating a structured 'wing' under which the canopy pilot can fly. Parachutes are controlled by pulling down on steering lines that change the shape of the wing, cause it to turn. The main forces acting on a parachute are gravity and drag. When you first release the parachute, the force of gravity pulls it downward, and the parachute speeds toward the ground. The faster the parachute falls, though, the more drag it creates.
Hey guys....
What is the advantage of a capacitor as it stores charge?
Wave 1
WWW
m
Resulting
Wave
Wave 2
Wave 2 Phase Shift
The resulting wave has the largest possible amplitude when Wave-1 and Wave-2 are exactly in step ... their peaks both happen at the same time and their troughs both happen at the same time.
This means that Wave-1 and Wave-2 have the same frequency, and the phase shift from one wave to the other is zero.
When all of that happens, the amplitude of the resulting wave is the sum of the amplitudes of Wave-1 and Wave-2. If Wave-1 and Wave-2 have the same amplitude, then the resulting wave will have double that amplitude.
Dos cargas puntuales iguales y negativas, q1=q2=-24micro C se localizan en x=0 y y=38m y x=0 y y=-7m, respectivamente. Calcula la magnitud de la fuerza electrica total en N que ejercen estas dos cargas sobre una tercera, tambien puntual, Q=26micro C en y=0 y x=16m
Answer:
F_net = 9.87 10⁻⁴ N
Explanation:
Let's use that force is a vector magnitude
∑ F = F₁₃ + F₂₃
De bold arfe vectros. The force is the electric force, we use that charges of the same sign repel and when the charges are of a different sign they attract
the charges q1 and q2 are negative and the charge q3 is positive with the positions y1 = 38 m, y2 = -7m, y3 = 16 m
∑ F = F₁₃ - F₂₃
F_net = [tex]k \frac{q_1q_3}{r_{13}^2 } - k \frac{q_2q_3}{r_{23}^2 }[/tex]
in this case q₁ = q₂ = q
F_net = k q q₃ ( )
let's look for the distance
r₂₃ = y₂ - y₃
r₂₃ = -7 -16
r₂₃ = - 23 m
r₁₃ = 38 - 16
r₁₃ = 22 m
let's calculate
F_net = 9 10⁹ 24 26 10⁻¹² ( )
F_net = 5.616 ( 1.758 10⁻⁴ )
F_net = 9.87 10⁻⁴ N
A 3.50 kg basket of cookies sits on a 2.00 m high shelf. What is the gravitational potential energy of the basket?
pls help
Answer:
68.6 J
Explanation:
Applying,
P.E = mgh............... Equation 1
Where P.E = Potential Energy of the basket, m = mass of the basket, g = acceleration due to gravity of the basket, h = height of the basket
From the question,
Given: m = 3.5 kg, h = 2.00 m
Constant: g = 9.8 m/s²
Substitute these values into equation 1
P.E = 3.5×2×9.8
P.E = 68.6 J
Hence the potential energy of the basket is 68.6 J
A parallel plate vacuum capacitor has 8.40 J of energy stored. The separation between plates is 2.30 mm. If the separation is decreased to 1.15 mm what is the energy stored if (a) the charge Q on the plates is held constant, and (b) the voltage V across the plates is held constant
(a) 4.20 J
(b) 16.74 J
Explanation:For a parallel plate vacuum capacitor with area A and whose plates are separated by by a distance of d, its capacitance C is given by;
C = A∈₀ / d --------------------(i)
Where;
∈₀ = constant called permittivity of vacuum.
The energy U stored in such capacitor is given by;
U = [tex]\frac{1}{2}[/tex]CV² ----------------------(ii)
or
U = [tex]\frac{1}{2}[/tex](Q²/C) -------------------(**)
Where;
V = potential difference or voltage across the plates.
Q = charge on the plates.
(a) If the charge is held constant
Combine equations (i) and (**) to give;
U = [tex]\frac{1}{2}[/tex]Q² / (A∈₀ / d) -----------------------(iii)
From the question;
The parallel plate capacitor has 8.40J energy stored and distance between plates is 2.30mm i.e
U = 8.40J
d = 2.30mm = 0.023m
Substitute these values into equation (iii)
8.40 = [tex]\frac{1}{2}[/tex]Q² / (A∈₀ / 0.023)
8.40 = [tex]\frac{1}{2}[/tex]Q² x (0.023 / A∈₀)
Multiply through by 2
2 x 8.40 = Q² x (0.023 / A∈₀)
16.80 = Q² x (0.023 / A∈₀)
Divide through by 0.023
16.80 / 0.023 = Q² x (0.023 / A∈₀) / 0.023
730.4 = Q² / (A∈₀)
Make Q² subject of the formula
Q² = 730.4(A∈₀)
Now, if the separation distance is decreased to 1.15mm and the voltage is held constant i.e
d = 1.15mm = 0.0115m
Q = constant [this means that Q² still remains 730.4(A∈₀) ]
The energy stored is found by substituting these values of d and Q² into equation (iii) as follows;
U = [tex]\frac{1}{2}[/tex]Q² / (A∈₀ / d)
U = [tex]\frac{1}{2}[/tex](730.4(A∈₀)) / (A∈₀ / 0.0115)
U = [tex]\frac{1}{2}[/tex](730.4(A∈₀))(0.0115 / A∈₀)
U = [tex]\frac{1}{2}[/tex](730.4)(0.0115)
U = 4.20J
Therefore, the energy stored if the charge Q on the plates is held constant is 4.20 J
(b) If the voltage is held constant
Combine equations (i) and (ii) to give;
U = [tex]\frac{1}{2}[/tex](A∈₀ / d)V² -----------------------(iv)
From the question;
The parallel plate capacitor has 8.40J energy stored and distance between plates is 2.30mm i.e
U = 8.40J
d = 2.30mm = 0.023m
Substitute these values into equation (iv)
8.40 = [tex]\frac{1}{2}[/tex](A∈₀ / 0.023)V²
Multiply through by 2 x 0.023
2 x 0.023 x 8.40 = (A∈₀)V²
2 x 0.023 x 8.40 = (A∈₀)V²
0.385 = (A∈₀)V²
Make V² subject of the formula
V² = 0.385/(A∈₀)
Now, if the separation distance is decreased to 1.15mm and the voltage is held constant i.e
d = 1.15mm = 0.0115m
V = constant [this means that V² still remains 0.385/(A∈₀) ]
The energy stored is found by substituting these values of d and V² into equation (iv) as follows;
U = [tex]\frac{1}{2}[/tex](A∈₀ / 0.0115)[0.385/(A∈₀)]
U = [tex]\frac{1}{2}[/tex](0.385/0.0115)
U = 16.74
Therefore, the energy stored if the voltage V across the plates is held constant is 16.74 J
One of the wavelengths of light emitted by hydrogen atoms under normal laboratory conditions is at ?0 = 656.3nm in the red portion of the electromagnetic spectrum. In the light emitted from a distant galaxy this same spectral line is observed to be Doppler-shifted to ? = 953.3nm , in the infrared portion of the spectrum.
How fast are the emitting atoms moving relative to the earth?
Answer:
1.07 × 10⁸ m/s
Explanation:
Using the relativistic Doppler shift formula which can be expressed as:
[tex]\lambda_o = \lambda_s \sqrt{\dfrac{c+v}{c-v}}[/tex]
here;
[tex]\lambda _o[/tex] = wavelength measured in relative motion with regard to the source at velocity v
[tex]\lambda_s =[/tex] observed wavelength from the source's frame.
Given that:
[tex]\lambda _o[/tex] = 656.3 nm
[tex]\lambda_s =[/tex] 953.3 nm
We will realize that [tex]\lambda _o[/tex] > [tex]\lambda_s[/tex]; thus, v < 0 for this to be true.
From the above equation, let's make (v/c) the subject of the formula: we have:
[tex]\dfrac{\lambda_o}{\lambda_s}=\sqrt{\dfrac{c+v}{c-v}}[/tex]
[tex]\Big(\dfrac{\lambda_o}{\lambda_s} \Big)^2=\dfrac{c+v}{c-v}[/tex]
[tex]\dfrac{v}{c} =\dfrac{\Big(\dfrac{\lambda_o}{\lambda_s} \Big)^2-1}{\Big(\dfrac{\lambda_o}{\lambda_s} \Big)^2+1}[/tex]
[tex]\dfrac{v}{c} =\dfrac{\Big(\dfrac{656.3}{953.3} \Big)^2-1}{\Big(\dfrac{656.3}{953.3} \Big)^2+1}[/tex]
[tex]\dfrac{v}{c} =0.357[/tex]
v = 0.357 c
To m/s:
1c = 299792458 m/s
∴
0.357c = (299 792 458 × 0.357) m/s
= 107025907.5 m/s
= 1.07 × 10⁸ m/s
how can you prove that acceleration is a derived unit
a = (dx / dt)²
Explanation: Unit of distance is m (metres) and unit of time is s (seconds) speed v is first derivative of distance x versus time:
v = dx / dt, unit is m/s. Acceleration is second derivative of
speed versus time a = (dx / dt)² = (dv/dt) , unit is m/s²
Answer:
Explanation:
Acceleration is derived unit because it has two fundamental units involved i.e. meter and second square.
A force of 350 newtons stretches a spring 30 centimeters. How much work is done in stretching the spring from 20 centimeters to 50 centimeters
Answer:
52.5 J
Explanation:
Applying,
Hook's law,
F = ke............... Equation 1
Where F = Force, k = spring constant, e = extension.
make k the subject of the equation
k = F/e............ Equation 2
From the question,
Given: F = 350 Newtons, e = 30 cm = 0.3 m
Substitute these values into equation 2
k = 350/0.3 N/m
Also,
W = 1/2(ke²).................. Equation 3
Where W = work done in stretching the spring.
Also given: e = (50-20) cm = 30 cm = 0.3 m, k = 350/0.3 N/m
Substitute these values into equation 3
W = 1/2(350/0.3)(0.3²)
W = 350×0.3/2
W = 52.5 J
two 100 ohm resistors are connected inparallel and one identical resister in series. The maximum power that can be delivered to any one resistor is 25W. What is the maximum voltage that can be applied between the terminals A and B ?
A. 50V
B. 75V
C. 100V
D. 125V
SOLVED DOWN BELOW
Explanation:
In series the same current goes thru both resistors, equiv resistance is 200 ohms, then using ohms law
I = 25/200
I= .125 amps or 125 ma
__________
R= r1 * r2 / r1 +r2
R= 100 * 100 / 100 + 100
R= 10000 / 200
R= 50 ohms
Please help! ❤️
I’ll make you the Brainlyest, I can’t get this one wrong.
what is the magnitude of an electric field (in 106 n/c) that balances the weight of a plastic sphere of mass 2.1 g that has been charged to 3.0 nc
Answer:
[tex]E=6.86\times 10^6\ N/C[/tex]
Explanation:
Given that,
Mass of the sphere, m = 2.1 g = 0.0021 kg
Charge, q = 3 nC
We need to find the magnitude of the electric field that balanced the weight of sphere. Let it is E. So,
qE = mg
[tex]E=\dfrac{mg}{q}[/tex]
Put all the values,
[tex]E=\dfrac{0.0021\times 9.8}{3\times 10^{-9}}\\\\E=6.86\times 10^6\ N/C[/tex]
So, the magnitude of the elecric field is [tex]6.86\times 10^6\ N/C[/tex].
What is sieving? Give an example where this method is used. (2)
Answer:
sieving is when you separate particles of different sizes.
Explanation:
separating sand mixtures
separating chaffs from local garri
Compared to its weight on Earth, a 5kg object on the moon will weigh
The same amount
Less
More
Answer:
Less
Explanation:
Weight is a force measurement. The object's mass is 5kg not its weight. To find its weight you have to take the mass of an object and multiply it by the acceleration of gravity. The acceleration of gravity is greater on earth than on the moon so therefore the object will weigh less on the moon.
magnetism/ magnetic field ana magnetic forces
Answer:
Magnetism is a physical phenomenon that manifests itself in a force acting between magnets or other magnetized or magnetisable objects, and a force acting on moving electric charges, such as in current-carrying cables. The force action takes place by means of a magnetic field, which is generated by the objects themselves or otherwise. There are natural and artificial magnets. All magnets have two poles called the north pole and the south pole. The north pole of one magnet repels the north pole of another magnet and attracts the south pole of another magnet; the same with south poles.
compare the time period of two pendulums of length 4m and 9m
area= length × length
area = 4m × 9m
ans 36
Which statement is true?
a particle of violet light has less energy than a particle of red light
a particle of violet light has more energy than a particle of red light
a particle of violet light has exactly the same energy as a particle of red light
particles of light do not have any energy, regardless of what color the light is
a particle of violet light has exactly the same energy as a particle of red light
5N
5 N
19 N
19 N
Pls help look at the pic
Answer:
b. is the correct answer ....
What do interplanetary space missions study?
the moon
stars in other galaxies
planets in the solar system
the sun
Answer:
C. Planets in the solar systemExplanation:
The one above is incorrect, and I know this is late. Even if it doesn't help you I hope it helps people in the future! YES I AM TALKING ABOUT YOU FUTURE PEOPLE!! I know this is the answer because I have taken 5.11 Quiz: Uncrewed Spacecraft in K12. There will only be the questions and correct answers below.
1. Which planetary body was Spirit designed to explore?
Mars.
2. What is the name of the most distant manmade object in space? (Credit: shathaadnan64/lak521)
Voyager 1.
3. Which group was designed to study Saturn? (Credit: Brainly User/snowballandtigoya1xa
Voyager 1, Huygens, and Cassini.
4. Why are scientists interested in exploring Mars?
Possible evidence of life.
5. What do interplanetary space missions study?
Planets in the solar system.
Have an amazing day!!
If the child has a mass of 13.9 kg, calculate the magnitude of the force in newtons the mother exerts on the child under the following conditions. (b) The elevator accelerates upward at 0.898 m/s2. 148.702 N
The elevator accelerates upward at an acceleration, then the magnitude of the force is 148.84 N.
What is Force?The force is the action of push or pull which makes an object to move or stop.
Given the mass of child m =13.9 kg, acceleration a =0.898 m/s², then the force will be given by
F = m(g-a)
F = 13.9 x (9.81 - (-0.898))
F = 148.84 N
Thus, the magnitude of the force is 148.84 N.
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If the ball is 0.60 mm from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a measure of how much the ball is speeding up. Express your answer in m/s2 and in units of g
This question is incomplete, the complete question is;
In a softball windmill pitch, the pitcher rotates her arm through just over half a circle, bringing the ball from a point above her shoulder and slightly forward to a release point below her shoulder and slightly forward. (Figure 1) shows smoothed data for the angular velocity of the upper arm of a college softball pitcher doing a windmill pitch; at time t = 0 her arm is vertical and already in motion. For the first 0.15 s there is a steady increase in speed, leading to a final push with a greater acceleration during the final 0.05 s before the release. In each part of the problem, determine the corresponding quantity during the first 0.15 s of the pitch.
Angular Velocity at time 0s = 12 rad/s
Angular Velocity at time 0.15s = 24 rad/s
a) What is the angular acceleration?
b) If the ball is 0.60 m from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a measure of how much the ball is speeding up. Express your answer in m/s2 and in units of g
Answer:
a) the angular acceleration is 80 rad/s²
b) the tangential acceleration of the ball is;
- a = 48 m/s²
- a = 4.9 g
Explanation:
Given the data in the question;
from the graph below;
Angular Velocity at time 0s [tex]w_o[/tex] = 12 rad/s
Angular Velocity at time 0.15s [tex]w_f[/tex] = 24 rad/s
a) What is the angular acceleration;
Angular acceleration ∝ = ( [tex]w_f[/tex] - [tex]w_o[/tex] ) / dt
we substitute
Angular acceleration ∝ = ( 24 - 12 ) / 0.15
Angular acceleration ∝ = 12 / 0.15
Angular acceleration ∝ = 80 rad/s²
Therefore, the angular acceleration is 80 rad/s²
b)
If the ball is 0.60 m from her shoulder, i.e s = 0.6 m
the tangential acceleration of the ball will be;
a = ∝ × s
we substitute
a = 80 × 0.6
a = 48 m/s²
a = ( 48 / 9.8 )g
a = 4.9 g
Therefore, the tangential acceleration of the ball is;
- a = 48 m/s²
- a = 4.9 g