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ladessa [460]
4 years ago
7

If Faraday had used a more powerful battery in his experiments with electromagnetic induction, what effect would this have had o

n his galvanometer's measurements of current when the battery was fully connected? Explain your reasoning for brainliest.
Physics
1 answer:
dmitriy555 [2]4 years ago
6 0

Answer:

The value of current generated would increase.

Explanation:

Electromagnetic induction is the process by which an electromotive force is induced due to a variation of magnetic field.

The induced current is directly proportional to rate at which the coil cuts the magnetic field. Using more powerful battery in the experiment would increase the rate at the the coil cuts the magnetic field, therefore increasing the rate of variation in the magnetic field. This effect would cause a greater deflection on the galvanometer's scale, showing an increase in the current generated.

This experiment proves that an alternating current can be produced from magnetic field.

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A balloon having an initial temperature of 17.8°C is heated so that the volume doubles while the pressure is kept fixed. What i
Rama09 [41]

Answer:

T = 308.6 ^0 C

Explanation:

Here by ideal gas equation we can say

PV = nRT

now we know that pressure is kept constant here

so we will have

V = \frac{nR}{P} T

since we know that number of moles and pressure is constant here

so we have

\frac{V_2}{V_1} = \frac{T_2}{T_1}

now we know that initial temperature is 17.8 degree C

and finally volume is doubled

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so final temperature will be

T_2 = 581.6 k

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9. If you placed the contents of a packet of powdered iced tea mix into a bottle of water and shook it, which of the following w
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The range of a projectile fired at an angle with the horizontal and with an initial velocity of feet per second is where r is me
77julia77 [94]

Answer:

θ = 8.50°

To the nearest angle

θ = 9.0°

the golfer must hit the ball at angle 9° so that it travels 120 feet.

Explanation:

The range of a projectile is the horizontal distance covered by a projectile, which can be written as;

r = (u^2× sin2θ)/g

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g = acceleration due to gravity

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sin2θ = rg/u^2

θ = 1/2 × sin⁻¹(rg/u^2) ....1

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r = 120 ft

u = 115 ft/s

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θ = 1/2 × sin⁻¹(120×32.2/115^2)

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θ = 1/2 × 17.00

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5 0
3 years ago
A 6-in-wide polyamide F-1 flat belt is used to connect a 2-in-diameter pulley to drive a larger pulley with an angular velocity
Likurg_2 [28]

Answer:

a) Fc = 4.15 N, Fi = 435.65 N, (F1)a = 640 N, and F2  = 239.6 N,

b) Ha = 1863.75 N, nfs = 1 , length = 11.8 mm

Explanation:

Given that:

γ= 9.5 kN/m³ = 9500N/m3

b = 6 inches = 0.1524 m

t = 0.0013 mm

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L = 9 ft = 2.7432 m

Ks = 1.25

g = 9.81 m/s²

a)

w=\gamma b t = 9500* 0.1524*0.0013=1.88N/m

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F_c=\frac{wV^2}{g}=1.88*4.65^2/9.81=4.15N

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T=\frac{H_{nom}n_dK_s}{2\pi n}= \frac{1491*1.25*1}{2*\pi*1750/60}=10.17Nm

F_2=(F_1)_a-\frac{2T}{D}= 640-\frac{2*10.17}{0.0508} =239.6N

F_i=\frac{(F_1)_a+F_2}{2} -F_c=435.65N

b)

H_a=1491*1.25=1863.75W

n_f_s=\frac{H_a}{H_{nom}K_S }=1

dip = \frac{L^2w}{8F_i} =\frac{2.7432*1.88}{435.65}=11.8mm

7 0
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