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Kryger [21]
1 year ago
15

What is the resistance of a carbon rod at 25.8 ∘C if its resistance is 0.0200 Ω at 0.0 ∘C ?

Physics
1 answer:
torisob [31]1 year ago
4 0

0.02020 ohm is the resistance of a carbon rod at 25.8 ∘C if its resistance is 0.0200 Ω at 0.0 ∘C.

<h3 /><h3>What is a resistor?</h3>

A resistor is an electrical component that controls or restricts how much electrical current can pass across a circuit in an electronic device. A specified voltage can be supplied via resistors to an active device like a transistor.

The temperature of the resistor varies based on the variation in the temperature. The equation that describes the relationship between the two of them is:

R = R0[1+ alpha(T-T0)]  where:

R is the new resistance we are looking for

alpha is the temperature coefficient of resistance. For carbon rod, alpha = ₋ 4.8 x 10^{-4}(1/°c)

T0 is the standard temperature =25.8°C

R0 is the resistance at T0 = 0.0200 ohms

T is the temperature at which we want to get R = 0

Substitute in the equation to get R as follows:

R = 0.0200 [1+( ₋ 4.8 x 10^{-4}) (0-25.8)] = 0.02020 ohm

To know more about resistance refer to: brainly.com/question/11431009

#SPJ1

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1. You place an object 63 cm in front of a converging lens, with a 40 cm focal length.
NikAS [45]

Answer:

1.

109.6 cm ,  - 1.74 , real

2.

1.5

Explanation:

1.

d₀ = object distance = 63 cm

f = focal length of the lens = 40 cm

d = image distance = ?

using the lens equation

\frac{1}{f} = \frac{1}{d_{o}} + \frac{1}{d}

\frac{1}{40} = \frac{1}{63} + \frac{1}{d}

d = 109.6 cm

magnification is given as

m = \frac{-d}{d_{o}}

m = \frac{-109.6}{63}

m = - 1.74

The image is real

2

d₀ = object distance = a

d = image distance = - (a + 5)

f = focal length of lens = 30 cm

using the lens equation

\frac{1}{f} = \frac{1}{d_{o}} + \frac{1}{d}

\frac{1}{30} = \frac{1}{a} + \frac{1}{- (a + 5)}

a = 10

magnification is given as

m = \frac{-d}{d_{o}}

m = \frac{- (- (a +5))}{a}

m = \frac{(5 + 10)}{10}

m = 1.5

6 0
2 years ago
Compute your average velocity in the following two cases: (a) You walk 50.2 m at a speed of 2.21 m/s and then run 50.2 m at a sp
Readme [11.4K]

Answer:

a) 2.87 m/s

b) 3.23 m/s

Explanation:

The avergare velocity can be found dividing the length traveled d by the total time t.

a)

For the first part we easily know the total traveled length which is:

d = 50.2 m + 50.2 m = 100.4 m

The time can be found dividing the distance by the velocity:

t1 = 50.2 m / 2.21 m/s = 22.7149 s

t2 = 50.2 m / 4.11 m/s = 12.2141 s

t = t1 +t2 = 34.9290 s

Therefore, the average velocity is:

v = d/t =2.87 m/s

b)

Here we can easily know the total time:

t = 1 min + 1.16 min = 129.6 s

Now the distance wil be found multiplying each velocity by the time it has travelled:

d1 = 2.21 m/s * 60 s = 132.6 m

d2 = 4.11 m/s *(1.16 * 60 s) = 286.056 m

d = 418.656 m

Therefore, the average velocity is:

v = d/t =3.23 m/s

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2 years ago
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