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vampirchik [111]
4 years ago
10

Two equal resistors are connected in series with a 1.50V battery. In order to keep the current at 0.030 A, the resistors much ea

ch have a value of of what( two digits)
Physics
1 answer:
AlexFokin [52]4 years ago
4 0

Answer: 25 Ohms

Explanation:

From this question, the following parameters are given:

Voltage V = 1.5 v

Current I = 0.03A

From Ohm's law;

V = IR

Where R = resultant resistance of the two resistors.

Substitute V and I into the formula and make resultant R the subject of formula.

1.5 = 0.03 × R

R = 1.5/0.03

R = 50 Ohms

From the question, it is given that Thr two equal resistors are connected in series.

R = R1 + R2

But R1 = R2

50 = 2R1

R1 = 50/2

R1 = 25

R1 = R2 = 25 Ohms

Therefore, the resistors must each have a value of 25 Ohms

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Directions: Fill in the blank with the correct word. Choose from the list of possible answers in the word bank below:
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3 years ago
After you pick up a spare, your bowling ball rolls without slipping back toward the ball rack with a linear speed of 2.85 m/s. T
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Answer:

V' = 0.84 m/s

Explanation:

given,

Linear speed of the ball, v = 2.85 m/s

rise of the ball, h = 0.53 m

Linear speed of the ball, v' = ?

rotation kinetic energy of the ball

KE_r = \dfrac{1}{2}I\omega^2

I of the moment of inertia of the sphere

I = \dfrac{2}{5}MR^2

 v = R ω

using conservation of energy

KE_r = \dfrac{1}{2}( \dfrac{2}{5}MR^2)(\dfrac{V}{R})2

KE_r = \dfrac{1}{5}MV^2

Applying conservation of energy

Initial Linear KE + Initial roational KE = Final Linear KE + Final roational KE + Potential energy

\dfrac{1}{2}MV^2 + \dfrac{1}{5}MV^2 = \dfrac{1}{2}MV'^2 + \dfrac{1}{5}MV'^2 + M g h

0.7 V^2 = 0.7 V'^2 + gh

0.7\times 2.85^2 = 0.7\times V'^2 +9.8\times 0.53

V'² = 0.7025

V' = 0.84 m/s

the linear speed of the ball at the top of ramp is equal to 0.84 m/s

6 0
3 years ago
50g of ice at 0°C is mixed with 50g of water at 80°C, what will be the final temperature of a mixture in
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Answer:

0° C

Explanation:

Given that

Mass of ice, m = 50g

Mass of water, m(w) = 50g

Temperature of ice, T(i) = 0° C

Temperature of water, T(w) = 80° C

Also, it is known that

Specific heat of water, c = 1 cal/g/°C

Latent heat of ice, L(w) = 89 cal/g

Let us assume T to be the final temperature of mixture.

This makes the energy balance equation:

Heat gained by ice to change itself into water + heat gained by melted ice(water) to raise its temperature at T° C = heat lost by water to reach at T° C

m(i).L(i) + m(i).c(w)[T - 0] = m(w).c(w)[80 - T], on substituting, we have

50 * 80 + 50 * 1(T - 0) = 50 * 1(80 - T)

4000 + 50T = 4000 - 50T

0 = 100 T

T = 0° C

Thus, the final temperature is 0° C

3 0
3 years ago
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