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mestny [16]
3 years ago
5

A magnet falls through a loop of wire with the south pole entering first. After it has fallen all the way through the wire loop

so that the north pole end is moving away from the loop, what is the direction of the induced current in the loop (as viewed from above, looking down on the magnet)
Physics
1 answer:
Alisiya [41]3 years ago
5 0

Answer:

The induced current direction as viewed is clockwise

Explanation:

Lenz's Law states that the induced e. m. f. causes current to be driven in the loop of wire in such a way as to generate magnetic field that are oppose the magnetic flux change which is the source of the induced current

Therefore, as the magnet approaches the coil with the south pole,  the coil produces current equivalent to the upward movement of the south pole of a permanent magnet through it which according to Flemings Right Hand Rule is clockwise

Therefore;

The direction of the induced current in the loop (as viewed from above, looking down the magnet) is clockwise

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You are trying to overhear a juicy conversation, but from your distance of 20.0 m , it sounds like only an average whisper of 30
12345 [234]

Answer:

r₂ = 0.2 m

Explanation:

given,

distance = 20 m

sound of average whisper = 30 dB

distance moved closer = ?

new frequency = 80 dB

using formula

\beta = 10 log(\dfrac{I_1}{I_0})

   I₀ = 10⁻¹² W/m²

now,

30 = 10 log(\dfrac{I_1}{10^{-12}})

\dfrac{I_1}{10^{-12}}= 10^3

I_1= 10^{-8}\ W/m^2

to hear the whisper sound = 80 dB

80 = 10 log(\dfrac{I_2}{10^{-12}})

\dfrac{I_2}{10^{-12}}= 10^8

I_2= 10^{-4}\ W/m^2

we know intensity of sound is inversely proportional to square of distances

\dfrac{I_1}{I_2}=\dfrac{r_2^2}{r_1^2}

\dfrac{10^{-8}}{10^{-4}}=\dfrac{r_2^2}{20^2}

10^{-4}=\dfrac{r_2^2}{20^2}

  r₂ = 0.2 m

6 0
3 years ago
A 1400 kg wrecking ball hangs from a 20-m-long cable. the ball is pulled back until the cable makes an angle of 30.0 ∘ with the
anastassius [24]

From the geometry of the problem, the 20 m-long cable creates the hypotenuse of a right triangle, with the extended of the other two sides of size 20 m * cos(30 deg), which is around 17.3 m. Therefore, the ball has increased by 20 m - 17.3 m = 2.7 m. 

The potential energy will have altered by m*g*h, which is 1400 kg * 9.8 m/s^2 * 1.6 m , or about 37044 joules.

5 0
4 years ago
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klemol [59]
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6 0
3 years ago
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Over an interval of 4.0 s the average rate of change of the concentration of B was measured to be -0.0760 M/s. What is the final
Serhud [2]

Complete question:

Consider the hypothetical reaction 4A + 2B → C + 3D

Over an interval of 4.0 s the average rate of change of the concentration of B was measured to be -0.0760 M/s. What is the final concentration of A at the end of this same interval if its concentration was initially 1.600 M?

Answer:

the final concentration of A is 0.992 M.

Explanation:

Given;

time of reaction, t = 4.0 s

rate of change of the concentration of B =  -0.0760 M/s

initial concentration of A = 1.600 M

⇒Determine the rate of change of the concentration of A.

From the given reaction: 4A + 2B → C + 3D

2 moles of B ---------------> 4 moles of A

-0.0760 M/s of B -----------> x

x = \frac{4(-0.076)}{2} \\\\x = -0.152 \ M/s

⇒Determine the change in concentration of A after 4s;

ΔA = -0.152 M/s  x 4s

ΔA = -0.608 M

⇒ Determine the final concentration of A  after 4s

A = A₀ + ΔA

A = 1.6 M + (-0.608 M)

A = 1.6 M - 0.608 M

A = 0.992 M

Therefore, the final concentration of A is 0.992 M.

5 0
3 years ago
A stone is thrown vertically upward with a speed of 18.0 m/s. How fast is it moving when it reaches a height of 11.0m? How long
steposvetlana [31]
Given: v0= 18.0 m/s, y0=0m, yf=11m, g=-9.81 m/s^2 

v0= initial velocity, vf= final velocity, y0= initial height, yf= final height, g= gravity, sqrt()= square root, ^2=squared 

vf^2=v0^2 + (2)(g)(yf-y0) 
vf^2=(18.0 m/s)^2+(2)(-9.81 m/s^2)(11 m-0m) 
vf^2=18.0 m/s)^2 + (-19.62 m/s^2)(11 m)
vf^2=(324 m^2/s^2) - (215.82 m^2/s^2) 
vf^2=108.18 m^2/s^2 
vf=sqrt(108.18 m^2/s^2) 
vf=10.4 m/s
8 0
3 years ago
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