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Karolina [17]
4 years ago
9

A type of transmission line for electromagnetic waves consists of two parallel conducting plates (assumed infinite in width) sep

arated by a distance aaa. Each plate carries the same uniform surface current density of 16.0 A/m, but the currents run in opposite directions. What is the magnitude of the magnetic field between the plates at a point 1.00 mm from one of the plates if aaa
Physics
1 answer:
vekshin14 years ago
8 0

Answer:

2.01 \times 10^{-5}\ T

Complete question:if a = 0.800 cm? (μ0 = 4π × 10-7 T · m/A)

Explanation:

Given:

Distance between plates = 0.8 cm

Distance from one plate = 1 mm

Current density 'J'= 16 A/m

Currents are flowing in opposite direction.

\mu _o=4\pi \times 10^{-7}

When current is flowing in opposite direction then magnetic field given as

B=\dfrac{\mu _oJ}{2}+\dfrac{\mu _oJ}{2}\\B=\mu _oJ

By putting the values we get

B=4\pi \times 10^{-7}\times 16\\B=2.01 \times 10^{-5}\ T

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The absorption spectrum would have all the wavelengths of the light source but would have black lines where the two red and one orange lines were in the emission spectrum
4 0
4 years ago
Jazz is a 172 lb athlete who exercised at 7.6 METs. At this workload, what is his energy expenditure in kcals/min.? Round to the
dezoksy [38]

Answer:

Energy expenditure in K cals/min = 10 K cals /min (approximately)

Explanation:

As we know

Energy expenditure in Kcal/min=  METs x 3.5 x Body weight (kg) / 200

Given is METs=7.6

Weight of Jazz= 172lb=78.02kg

putting the values in formula,

Energy expenditure in K cals/min=  7.6 x 3.5 x 78.02 / 200

                                                       =10.38 K cals /min

                                                       =10 K cals /min (approximately)

Therefore, Energy expenditure in K cals/min by Jazz will be approximately 10 K cals /min

8 0
4 years ago
Hi..Please answer questions???​
Nadya [2.5K]

Answer:

A

Explanation:

Becuse its complete number

3 0
3 years ago
A 15 kg bucket of water is suspended by a very light ropewrapped around a solid uniform cylinder 0.300 m in diamter withmass 12.
matrenka [14]

Answer:

Part a)

T = 42 N

Part b)

v_f = 11.8 m/s

Part c)

t = 1.7 s

Part d)

F = 159.7 N

Explanation:

Part a)

While bucket is falling downwards we have force equation of the bucket given as

mg - T = ma

for uniform cylinder we will have

TR = I\alpha

so we have

T = \frac{1}{2}MR^2(\frac{a}{R^2})

T = \frac{1}{2}Ma

now we have

mg = (\frac{M}{2} + m)a

a = \frac{mg}{(\frac{M}{2} + m)}

a = \frac{15 \times 9.81}{(6 + 15)}

a = 7 m/s^2

now we have

T = \frac{12 \times 7}{2}

T = 42 N

Part b)

speed of the bucket can be found using kinematics

so we have

v_f^2 - v_i^2 = 2 a d

v_f^2 - 0 = 2(7)(10)

v_f = 11.8 m/s

Part c)

now in order to find the time of fall we can use another equation

v_f - v_i = at

11.8 - 0 = 7 t

t = 1.7 s

Part d)

as we know that cylinder is at rest and not moving downwards

so here we can use force balance

F = T + Mg

F = 42 + (12 \times 9.81)

F = 159.7 N

5 0
3 years ago
If the same fish is attached to the end of the unstretched spring and then allowed to fall from rest, through what maximum dista
andrew11 [14]

Answer:

x=2d

Explanation:

initial stretch in the spring is d

so using Hook's law

at equilibrium position

k×d=mg

where k= spring constant

m= mass of fish

g= acceleration due to gravity.

d=mg/k ................ (1)

in second case  by energy conservation

1/2 kx^2=mgx

x=2mg/k

using equation 1

x=2d

3 0
3 years ago
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