Answers
Answer:
0.000314 Am²
6.049*10^-7 T
Explanation:
A
From the definitions of magnetic dipole moment, we can establish that
= , where
= the magnetic dipole moment in itself
= Current, 100 A
= Area, πr² (r = diameter divided by 2). Converting to m², we have 0.000001 m²
On solving, we have
= ,
= 100 * 3.14 * 0.000001
= 0.000314 Am²
B
=(0)/4* 2/³, where
(0) = constant of permeability = 1.256*10^-6
z = 4.7 cm = 0.047 m
B = 1.256*10^-6 / 4*3.142 * [2 * 0.000314/0.047³]
B = 1*10^-7 * 0.000628/1.038*10^-4
B = 1*10^-7 * 6.049
B = 6.049*10^-7 T
Final answer:
The magnetic dipole moment of the superconducting ring is 0.000314 Ampere*m². The on-axis magnetic field strength 4.70 cm from the ring is 6.56 μT.
Explanation:
The magnetic dipole moment (μ) of a current (I) circulating around a loop of radius (r) is given by the formula μ = Iπr². Substituting the given values in SI units (I=100 Ampere, r=1.00 mm = 0.001 m), we get μ = 100 * π * (0.001)² = 0.000314 Ampere*m².
To find out the on-axis magnetic field strength (B) at a certain distance from the ring, we use the formula B = μ0/(4π) * (2μ/r³), where μ0 represents the permeability of free space. Plugging the values in SI units (μ0 = 4π × 10-7 T*m/A, r=4.70 cm = 0.047 m), The magnetic field is calculated to be B = (4π × 10-7 T*m/A) /(4π) * (2 * 0.000314 m² / 0.0473m³) = 6.56 × 10-6 T or 6.56 μT.
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Answers
The distance between the adjacent bright fringes is : 1.7 * 10⁻³ M
Given data :
separation between slits ( d ) = 1.5 x 10⁻³ m
wavelength of light ( λ ) = 514 * 10⁻⁹ m
Distance from narrow slit ( D ) = 5.0 m
Determine the distance between the adjacent bright fringes
we apply the formula below
w = D * λ / d ---- ( 1 )
where : w = distance between adjacent bright fringes
Back to equation ( 1 )
w = ( 5 * 514 * 10⁻⁹ ) / 1.5 x 10⁻³
= 1.7 * 10⁻³ M
Hence we can conclude that The distance between the adjacent bright fringes is : 1.7 * 10⁻³ M
Learn more about bright fringes calculations : brainly.com/question/4449144
Answer:
m
Explanation:
d = separation between the two narrow slits = 1.5 mm = 1.5 x 10⁻³ m
λ = wavelength of the light = 514 nm = 514 x 10⁻⁹ m
D = Distance of the screen from the narrow slits = 5.0 m
w = Distance between the adjacent bright fringes on the screen
Distance between the adjacent bright fringes on the screen is given as
m
b. With what speed does an electron exit the electron gun if its entry speed is close to zero? [Note: ignore relativity]
c. If the capacitance of the plates is 1 nF, how much charge is stored on each plate? How many extra electrons does the cathode have?
d. If you wanted to push an electron from the anode to the cathode, how much work would you have to do?
Answers
Answer:
A. 2.083 MV/m from anode to cathode.
B. 93648278.15 m/s
C. 2.5x10^-5 C and there are about 1.56x10^14 electrons
D. 4x10^-15 Joules
Explanation:
Voltage V across plate is 25 kV = 25x10^3 V
Distance apart x = 1.2 cm = 1.2x10^-2 m
A. Electric field strength is the potential difference per unit distance
E = V/x = 25x10^3/1.2x10^-2 = 2083333.3 V/m
= 2.083 MV/m
B. Energy of electron is electron charge times the voltage across
i.e eV
Charge on electron = 1.6x10^-19 C
Energy of electron = 1.6x10^-19 x 25x10^3 = 4x10^-15 Joules
Mass of electron m is 9.12x10^-31 kg
Kinetic energy of electron = 0.5mv^2
Where v is the speed
4x10^-15 = 0.5 x 9.12x10^-31 x v^2
v^2 = 8.77x10^15
v = 93648278.15 m/s
C. From Q = CV
Q = charge
C = capacitance = 1 nF 1x10^-9 F
V = voltage = 25x10^3 V
Q = 1x10^-9 x 25x10^3 = 2.5x10^-5 C
Total number of electrons = Q/e
= 2.5x10^-5/1.6x10^-19 = 1.56x10^14 electrons
D. To push electron from cathode to anode, I'll have to do a work of about
4x10^-15 Joules
Answers
The volume of Ampicillin IM q8h to be administered per dose is 1.5 mL when the order is to give 600 mg of it from a concentration of 400 mg per mL prepared by the dilution of 2 g in 4.8 mL of sterile water.
1. The information we know
- The order is to give 600 mg of Ampicillin.
- From the dilution of 2 g in 4.8 mL was obtained a solution of 400 mg per mL of Ampicillin.
2. We need to find:
The volume of Ampicillin in mL per dose
3. Calculation of the Ampicillin's volume to be administered
We can calculate the volume of Ampicillin as follows:
Therefore, we need to administer 1.5 mL of Ampicillin per dose.
Find more about doses here:
I hope it helps you!
Answer:
1.5 ml
Explanation:
The nurse is to administer 600 mg of Ampicillin IM q8h
the reconstitute yield 400 mg per mL
400 mg is in 1 ml
600 mg will be (600 mg × 1 ml) / 400 mg = 1.5 ml
Answers
Answer:
Monachry
Explanation:
Hope this helped!!!
Answers
Answer:
-4.40
Explanation:
explanation is in attachment
b. The tubing was unable to supply any more water to the tube for use.
c. The pressure outside the tube is higher that the water pressure inside the tube.
Answers
Answer:
a. The pressure in the tubing is equal to the barometric pressure.
Explanation:
Since in the question it is mentioned that the if you take the stoppert part of the tube than the level of warer would be fall approx 4th floor and if it is continued than it wont be continue but remains constant.
Now here first we do that the tube i.e. connected to the bucket should be taken up. In the first instance, the bucket supplies the water to the tube but it would not increased far away to the level of the barometric pressure
Hence, the correct option is a.