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3 years ago

14

2 Calculate the distance from a srtaight wire where the magnetuc field is 0.66 T. the current in the wire is 4A and K = 2 x 10^-

7. [

1 answer:

nirvana33 [79]3 years ago

7 0

Answer:I don't know!!!!I don't know!!!!I don't know!!!!I don't know!!!!I don't know!!!!

Explanation:

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We often use a spoon to cool coffee by stirring. Some of the heat is conducted from the coffee to the air by this method, but th

Ilia_Sergeevich [38]

Answer:

Explanation:

heat lost by coffee = heat gain by aluminum spoon

heat lost by coffee = mc ( 85° C - T)

175 ml of water = 175 grams

175 grams = (175 / 1000) kg = 0.175 kg

heat loss by coffee = 0.175 × 4182 J/Kgk ( 85 - T) = 731.85 ( 85-T) = 62207.25 - 731.85 T

heat gain by aluminum spoon = 0.012kg × 897 J/kgK ( T - 22) = 10.764 T - 236.808

10.764 T - 236.808 = 62207.25 - 731.85 T

10.764 T + 731.85 T = 62207.25 + 236.808

T = 84.086 ° C approx 84°C

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3 years ago

You have a horizontal cathode ray tube (CRT) for which the controls have been adjusted such that the electron beam should make a

Dvinal [7]

Answer:

Explanation:

Rotate it slowly to establish whether the spot moves. If it is broken, it will stay as it is. By rotating it the electrons' path to the centre of the screen will be tilted to another position by any external disturbing field.

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3 years ago

Mrs. Miller is acting a little strange and is pushing as hard as she can against a wall. Using Newton's Laws, what could be say

deff fn [24]

Newton’s 3rd Law of Motion

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2 years ago

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A red laser beam goes from crown glass with refraction index n=1.3 to air (n=1) with an incident angle of 0.23 radians. What is

Klio2033 [76]

Answer:

θ = 10.28º

Explanation:

To find the angle of refraction use the equation of refraction

n₁ sin θ₁ = n₂ sin θ₂

where index 1 is for incident light and index 2 is for refracted light.

sin θ₂ = n₁ / n₂ sin θ

let's calculate

sin = 1 / 1.3 sin 0.23

sin = 0.175

θ= 0.17528 rad

let's reduce to degrees

θ = 0.17528 rad (180ª / pi rad)

θ = 10.28º

6 0

3 years ago

A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 V. It is used to charge two storage batt

Natali [406]

Complete Question

A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 $\Omega$ . It is used to charge two storage batteries connected in series, each having an emf of 6.00 V and internal resistance of 0.300 $\Omega$ . If the charging current is to be 4.00 A, (a) what additional resistance should be added in series? At what rate does the internal energy increase in (b) the supply, (c) in the batteries, and (d) in the added series resistance? (e) At what rate does the chemical energy increase in the batteries?

Answer:

a

The additional resistance is $R_z = 4.4 \Omega$

b

The rate at which internal energy increase at the supply is $Z_1 = 32 W$

c

The rate at which internal energy increase in the battery is $Z_1 = 32 W$

d

The rate at which internal energy increase in the added series resistance is $Z_3 = 70.4 W$

e

the increase rate of the chemically energy in the battery is $C = 48 W$

Explanation:

From the question we are told that

The open circuit voltage is $V = 40.0V$

The internal resistance is $R = 2 \Omega$

The emf of each battery is $e = 6.00 V$

The internal resistance of the battery is $r = 0.300V$

The charging current is $I = 4.00 \ A$

Let assume the the additional resistance to to added to the circuit is $R_z$

So this implies that

The total resistance in the circuit is

$R_T = R + 2r +R_z$

Substituting values

$R_T = 2.6 +R_z$

And the difference in potential in the circuit is

$E = V -2e$

=> $E = 40 - (2 * 6)$

$E = 28 V$

Now according to ohm's law

$I = \frac{E}{R_T}$

Substituting values

$4 = \frac{28}{R_z + 2.6}$

Making $R_z$ the subject of the formula

So $R_z = \frac{28 - 10.4}{4}$

$R_z = 4.4 \Omega$

The increase rate of internal energy at the supply is mathematically represented as

$Z_1 = I^2 R$

Substituting values

$Z_1 = 4^2 * 2$

$Z_1 = 32 W$

The increase rate of internal energy at the batteries is mathematically represented as

$Z_2 = I^2 r$

Substituting values

$Z_2 = 4^2 * 2 * 0.3$

$Z_2 = 9.6 \ W$

The increase rate of internal energy at the added series resistance is mathematically represented as

$Z_3 = I^2 R_z$

Substituting values

$Z_3 = 4^2 * 4.4$

$Z_3 = 70.4 W$

Generally the increase rate of the chemically energy in the battery is mathematically represented as

$C = 2 * e * I$

Substituting values

$C = 2 * 6 * 4$

$C = 48 W$

6 0

3 years ago