3 years ago

WILL GIVE BRAINLY TO CORRECT ANSWER

Physics

2 answers:

11111nata11111 [884]3 years ago

7 0

C is the answer to your question!

butalik [34]3 years ago

3 0

the second and third one

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WILL GIVE BRAINLIEST!!!!!!!!!!!!!!!

jek_recluse [69]

Answer:

Explanation:

Outer shell is complete.

8 0

3 years ago

How long does it take a person to skate the width of a hockey rink (85 feet) at a constant speed of 15 feet per second?

Zepler [3.9K]

S(travel distance)=85 ft
v (velocity)=15 ft/s
-----------------------------------
t (time)=?

Calculate the time with the formula for the velocity:
v=S/t
t=S/v
t=85 ft/(15 ft/s)
t=5.666s

3 0

3 years ago

The loop is in a magnetic field 0.20 T whose direction is perpendicular to the plane of the loop. At t = 0, the loop has area A

love history [14]

Answer:

Part a)

$EMF = 14 \times 10^{-3} V$

Part b)

$EMF = 15.67 \times 10^{-3} V$

Explanation:

As we know that magnetic flux through the loop is given as

$\phi = B.A$

now we have

$\phi = B\pi r^2$

now rate of change in flux is given as

$\frac{d\phi}{dt} = B(2\pi r)\frac{dr}{dt}$

now we know that

$A = \pi r^2$

$0.285 = \pi r^2$

$r = 0.30 m$

Now plug in all data

$EMF = (0.20)\times 2\pi\times (0.30) \times (0.037)$

$EMF = 14 \times 10^{-3} V$

Part b)

Now the radius of the loop after t = 1 s

$r_1 = r_0 + \frac{dr}{dt}$

$r_1 = 0.30 + 0.037$

$r_1 = 0.337 m$

Now plug in data in above equation

$EMF = (0.20)\times 2\pi\times (0.337) \times (0.037)$

$EMF = 15.67 \times 10^{-3} V$

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3 years ago

Please help me ,create a slogan that promotes the role of internet in education

aleksklad [387]

Answer:

¨Facts you didn´t know¨ or ¨unknown facts¨

Explanation:

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3 years ago

The Hall effect can be used to calculate the charge-carrier number density in a conductor. A conductor carrying current of 2.0 A

marysya [2.9K]

Answer:

option D

Explanation:

given,

A conductor is carrying current = 2.0 A is 0.5 mm thick

Hall voltage = 4.5 x 10-6 V

uniform magnetic field = 1.2 T

density of the charge = n =?

hall voltage = $V_h =\dfrac{i\ B}{n\ e\ L}$

$n = \dfrac{i\ B}{V\ e\ L}$

$n = \dfrac{2 \times 1.2 }{4.5 \times 10^{-6}\times 1.6 \times 10^{-19} \times 0.5 \times 10^{-3}}$

n = 6.67 × 10²⁷ charges/m

hence the correct answer is option D

7 0

3 years ago