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jeka94
3 years ago
5

A walkman uses four standard 1.5 V batteries. How much resistance is in the circuit if it uses a current of 0.02A? *

Physics
1 answer:
hodyreva [135]3 years ago
7 0

Answer:

75ohms

Explanation:

V= IR

V = 1.5volts

I = 0.02A

1.5 = 0.02×R

Making R the subject

R = 1.5/0.02

R = 75ohms

The resistance in the circuit will be 75ohms

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3 0
3 years ago
When bouncing a ball, the bouncing motion results in the ball ____________.
Alekssandra [29.7K]

Answer: "B" Changing Position

Great Question!

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8 0
3 years ago
An object of mass 300 g, moving with an initial velocity of 5.00i-3.20j m/s, collides with an sticks to an object of mass 400 g,
Alexus [3.1K]

Answer:

Velocity is 2.17 m/s at an angle of 9.03° above X-axis.

Explanation:

Mass of object 1 , m₁ = 300 g = 0.3 kg

Mass of object 2 , m₂ = 400 g = 0.4 kg

Initial velocity of object 1 , v₁ = 5.00i-3.20j m/s

Initial velocity of object 2 , v₂ = 3.00j m/s

Mass of composite = 0.7 kg

We need to find final velocity of composite.

Here momentum is conserved.

Initial momentum = Final momentum

Initial momentum = 0.3 x (5.00i-3.20j) + 0.4 x 3.00j = 1.5 i + 0.24 j kgm/s

Final momentum = 0.7 x v = 0.7v kgm/s

Comparing

1.5 i + 0.24 j = 0.7v

v = 2.14 i + 0.34 j

Magnitude of velocity      

       v=\sqrt{2.14^2+0.34^2}=2.17m/s

Direction,  

       \theta =tan^{-1}\left ( \frac{0.34}{2.14}\right )=9.03^0

Velocity is 2.17 m/s at an angle of 9.03° above X-axis.

7 0
3 years ago
If an electric wire is allowed to produce a magnetic field no larger than that of the Earth (0.503 X 104 T) at a distance of 15
dem82 [27]

Answer:

37.725 A

Explanation:

B = magnitude of the magnetic field produced by the electric wire = 0.503 x 10⁻⁴ T

r = distance from the wire where the magnetic field is noted = 15 cm = 0.15 m

i = magnitude of current flowing through the wire = ?

Magnetic field by a long wire is given as

B = \frac{\mu _{o}}{4\pi }\frac{2i}{r}

Inserting the values

0.503\times 10^{-4} = (10^{-7})\frac{2i}{0.15}

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6 0
3 years ago
A 23.9 g sample of iridium is heated to 89.7 oC, and then dropped into 20.0 g of water in a coffee-cup calorimeter. The temperat
Nikolay [14]

Answer:

The specific heat capacity of iridium = 0.130 J/g°C

Explanation:

Assuming no heat losses to the environment and to the calorimeter,

Heat lost by the iridium sample = Heat gained by water

Heat lost by the iridium sample = mC ΔT

m = mass of iridium = 23.9 g

C = specific heat capacity of the iridium = ?

ΔT = change in temperature of the iridium = 89.7 - 22.6 = 67.1°C

Heat lost by the iridium sample = (23.9)(C)(67.1) = (1603.69 C) J

Heat gained by water = mC ΔT

m = mass of water = 20.0 g

C = 4.18 J/g°C

ΔT = 22.6 - 20.1 = 2.5°C

Heat gained by water = 20 × 4.18 × 2.5 = 209 J

Heat lost by the iridium sample = Heat gained by water

1603.69C = 209

C = (209/1603.69) = 0.130 J/g°C

7 0
3 years ago
Read 2 more answers
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