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Paraphin [41]
3 years ago
5

How much current will pass through a 12.5 ohm resistor when it is connected to ta 115 volt source of electrical potential?

Physics
1 answer:
Marrrta [24]3 years ago
8 0

Answer:

9.2 amperes

Explanation:

Ohm's law states that the voltage V across a conductor of resistance R is given by V = R I

Here, voltage V is proportional to the current I.

For voltage, unit is volts (V)

For current, unit is amperes (A)

For resistance, unit is Ohms (Ω)

Put R = 12.5 and V = 115 in V=RI

115=12.5I\\I=\frac{115}{12.5}\\ =9.2\,\,amperes

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You just found a metallic object in a creek bed. you have found its mass to be 96.5 g, and it displaces 5mL of water. what is th
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Two slits are illuminated with green light (λ = 540 nm). The slits are 0.05 mm apart and the distance to the screen is 1.5 m. At
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Answer:

0.21486 mm

Explanation:

The formula for the maximum intensity is given by;

I = I_o•cos²(Φ/2)

Now,we are not given Φ but it can be expressed in terms of what we are given as; Φ = πdy/(λL)

Where;

y is the distance from the central maximum

d is the distance between the slits

λ is the wavelength

L is the distance to the screen

Thus;

I = I_o•πdy/(λL)

We are given;

d = 0.05 mm = 0.5 × 10^(-3) m

λ = 540 nm = 540 × 10^(-9) m

L = 1.25 m

I/I_o = 50% = 0.5

From earlier, we saw that;

I = I_o•πdy/(λL)

We have I/I_o = 0.5

Thus;

I/I_o = πdy/(λL)

Plugging in the relevant values;

0.5 = (π × 0.5 × 10^(-3) × y)/(540 × 10^(-9) × 1.25)

Making y the subject, we have;

y = (0.5 × 540 × 10^(-9) × 1.25)/(π × 0.5 × 10^(-3))

y = 0.00021486 m

Converting to mm, we have;

y = 0.21486 mm

7 0
3 years ago
The structure of which substances tends to exhibit the greatest hardness?
bagirrra123 [75]

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3 0
3 years ago
A dc motor with its rotor and field coils connected in series has an internal resistance of 3.2 Ω. When running at full load on
pochemuha

Answer:

The current drawn by the motor from the line is 4.68 A.

Explanation:

Given that,

Internal resistance of the dc motor, r = 3.2 ohms

Voltage, V = 120 V

Emf in the motor, \epsilon=105\ V

We need to find the current drawn by the motor from the line. A dc motor with its rotor and field coils connected in series, applying loop rule we get :

V=\epsilon+Ir

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I=\dfrac{V-\epsilon}{r}\\\\I=\dfrac{120-105}{3.2}\\\\I=4.68\ A

So, the current drawn by the motor from the line is 4.68 A. Hence, this is the required solution.

7 0
3 years ago
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