The magnitude J(r) of the current density in a certain cylindrical wire is given as a function of radial distance from the center of the wire's cross section as J(r) = Br, where r is in meters, J is in amperes per square meter, and B = 2.35 ✕ 105 A/m3. This function applies out to the wire's radius of 2.00 mm. How much current is contained within the width of a thin ring concentric with the wire if the ring has a radial width of 11.5 μm and is at a radial distance of 1.20 mm?
Answers
The current contained within the width of a thin ring concentric is 18.1 x 10⁻⁶A
What is Current?
This is defined as electric charges moving through an electric conductor or space.
Parameters
Current density of J(r) = Br, where B = 2.35 x 10⁵ A/m³.
I = Jₓ A
where I is current, A is area and J is current density
A= 2rΔr
where 2r = circumference, Δr = width,
Substitute the values into the equation.
I=J(2rΔr)
I=2Br^2Δr
I= 2(2.35 x 10⁵)(1.2 x 10⁻³)^2(11.5 x 10⁻⁶)
= 18.1 x 10^-6 A
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Answer:
18.1 x 10^-6 A
Explanation:
A cylindrical wire carries a current density of J(r) = Br, where B = 2.35 x 10^5 A/m^3, to find the current within a certain area we multiply the current density with the are of this area:
I = J*A
for a ring with r distance from the center and width Δr, where Δr<< r, the area is:
A= 2rΔr
where 2r is the circumference and Δr is the width, substitute to get:
I=J(2rΔr)
I=2Br^2Δr
substitute with the given values to get:
I= 2(2.35 x 10^5)(1.2 x 10^-3)^2(11.5 x 10^-6)
= 18.1 x 10^-6 A
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Temperature°F = (9/5 * °C) + 32°°C = 5/9 * (°F - 32°) 1 pt each. Using the table above as a guide, complete the following conversions. Be sure to show your work to the side:1. 5 cm = ________ mm2. 83 cm = ________ m3. 459 L = _______ ml4. .378 Kg = ______ g5. 45°F = ________ °C6. 80°C = _________ °F
An electric field of 710,000 N/C points due west at a certain spot. What is the magnitude of the force that acts on a charge of -6.00 C at this spot? (14C - 10 6C) Give your answer in Si unit rounded to two decimal places
Answers
Answer:
kilogram
Explanation:
Answer:
SI base units of mass=KG
B. What is the change in the total momentum of the pair?
C. What is the magnitude of the change in the momentum Δp2, of mass M2?
Answers
Answer:
a). ΔP1=-2.4
b). Pp=0 F=0
c). ΔP2=2.4
Explanation:
Initial momentum
Final momentum
The change of momentum m1 is:
a).
ΔP1=
ΔP1=
ΔP1=
ΔP1=
ΔP1=
b).
The law of conservation of energy in this case there is not external forces so the momentum of the pair change is equal to zero
P=0
Fx=0
c).
ΔP1+ΔP2=0
ΔP2=-ΔP1
ΔP2=-
ΔP2=
Final answer:
The magnitude of the change in momentum of mass M1 is 2400 Daltons*m/s. The change in the total momentum of the pair is 2000 Daltons*m/s. The magnitude of the change in momentum of mass M2 is -400 Daltons*m/s.
Explanation:
A. To find the magnitude of the change in momentum of mass M1, we use the formula Δp1 = m1 * Δv1, where m1 is the mass of M1 and Δv1 is the change in velocity of M1. Since M1 simply changes direction, its change in velocity is equal to 2 times its original velocity. Therefore, Δp1 = m1 * (2v1) = 6 * (2 * 200) = 2400 Daltons*m/s.
B. The change in the total momentum of the pair is equal to the sum of the changes in momentum of M1 and M2. Since M2 also changes direction, its change in velocity is equal to 2 times its original velocity. Therefore, the change in the total momentum is Δp1 + Δp2 = 2400 Daltons*m/s + (-400 Daltons*m/s) = 2000 Daltons*m/s.
C. To find the magnitude of the change in momentum of mass M2, we use the same formula as in part A, but with the values for M2. Δp2 = m2 * Δv2 = 1 * (2 * (-200)) = -400 Daltons*m/s.
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C) The distance between the minima stays the same.
D) The distance between the minima increases.
E) The distance between the maxima increases.
Answers
Answer:
The correct statements are D and E.
Explanation:
The fringe width is given by the following formula as :
Here,
is wavelength of light
D is distance between slit and the screen
d is slit width.
If the between these slits is decreased, the fringe width increases. As a result, the distance between the minima increases and also the distance between the maxima increases.
m
(b) If each holds one end of a rope, and the man pulls on the rope so that he moves 1.3 m, how far from the woman will he be now?
m
(c) How far will the man have moved when he collides with the woman?
m
Answers
Answer:
Given that
m₁ = 50 kg
m₂=80 kg
d= 12 m
a)
We know that center of mass given as
X = (x₁m₁+x₂m₂)/(m₁+m₂)
Lets take distance of CM from woman is X
So now by putting the value
X = (0 x 50+12 x 80)/(50+80)
x=7.38 m
b)
There is no any external force so the CM will not move.
So we can say that
x₁m₁+x₂m₂ = 0
50(x) - 80(1.3)=0
x=2.08
So the distance move by woman d=12-2.08-1.3=8.62 m
d=8.62 m
c) lets take distance move by man is x
50 (x) - 80 (12-x) =0
x=7.38
So the distance move by woman d=12-7.38
d=4.62 m
Answers
Answer:
π*R²*E
Explanation:
According to the definition of electric flux, it can be calculated integrating the product E*dA, across the surface.
As the electric field E is uniform and parallel to the hemisphere axis, and no charge is enclosed within it, the net flux will be zero, so, in magnitude, the flux across the opening defining the hemisphere, must be equal to the one across the surface.
The flux across the open surface can be expressed as follows:
As E is constant, and parallel to the surface vector dA at any point, can be taken out of the integral, which is just the area of the surface, π*R².
⇒Flux = E*π*R²
x = vi(cos )t
x = ayt
x = vxt (RIGHT ANSWER)
Answers
The formula for calculating the horizontal displacement of a horizontally launched projectile is
A projectile launched horizontally with a velocity v, at a height y ,travels a horizontal distance x, while falling through a distance y. The horizontal velocity of a projectile remains constant throughout its motion, in the absence of air resistance. The vertical component of the velocity is under the action of the gravitational force and hence it increases in magnitude as it falls through the height.
The horizontal displacement of the projectile is a uniform motion and it occurs at a constant speed v.
Thus, the horizontal displacement of the projectile is given by the expression.