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Pachacha [2.7K]
2 years ago
6

The current in the wires of a circuit is 120.0 milliAmps. If the voltage impressed across the ends of the circuit were doubled (

with no change in its resistance), then its new current would be _____ milliAmps.
Physics
1 answer:
skad [1K]2 years ago
7 0

Answer:

Its new current will be 240.0milliAmps

Explanation:

According to ohms law,

V = IR

V is the supply voltage in a circuit

I is the current

R is the resistance

If the current in the wires of a circuit is 120.0 milliAmps, the voltage will be:

V1 = 120R1

R1 = V1/120...(1)

If the voltage impressed across the ends of the circuit were doubled (with no change in its resistance),

V2 = 2V1

V2 = IR2

2V1 = IR2

R2 = 2V1/I ... (2)

Since there are no changes in resistance

R1 = R2

Equating 1 and 2, we have:

V1/120 = 2V1/I

1/120 = 2/I

Cross multiplying

I = 2×120

I = 240milliAmps

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Answer:

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Explanation:

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3 years ago
Uest<br>1. State Newton's law of cooling.​
garik1379 [7]

Answer:

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When stated in terms of temperature differences, Newton's law (with several further simplifying assumptions, such as a low Biot number and a temperature-independent heat capacity) results in a simple differential equation expressing temperature-difference as a function of time. The solution to that equation describes an exponential decrease of temperature-difference over time. This characteristic decay of the temperature-difference is also associated with Newton's law of cooling

6 0
3 years ago
The angle θ is slowly increased. Write an expression for the angle at which the block begins to move in terms of μs.
Reika [66]

Answer:

tan \theta = \mu_s

Explanation:

An object is at rest along a slope if the net force acting on it is zero. The equation of the forces along the direction parallel to the slope is:

mg sin \theta - \mu_s R =0 (1)

where

mg sin \theta is the component of the weight parallel to the slope, with m being the mass of the object, g the acceleration of gravity, \theta the angle of the slope

\mu_s R is the frictional force, with \mu_s being the coefficient of friction and R the normal reaction of the incline

The equation of the forces along the direction perpendicular to the slope is

R-mg cos \theta = 0

where

R is the normal reaction

mg cos \theta is the component of the weight perpendicular to the slope

Solving for R,

R=mg cos \theta

And substituting into (1)

mg sin \theta - \mu_s mg cos \theta = 0

Re-arranging the equation,

sin \theta = \mu_s cos \theta\\\rightarrow tan \theta = \mu_s

This the condition at which the equilibrium holds: when the tangent of the angle becomes larger than the value of \mu_s, the force of friction is no longer able to balance the component of the weight parallel to the slope, and so the object starts sliding down.

4 0
3 years ago
Two identical charges, 2 m apart, exert forces of magnitude 4 N on each other. The value of each charge is: 1. 9 × 105 C 2. 4.2
lesya692 [45]

Answer:

The value of each charge is 4.22 x 10⁻⁵ C

Explanation:

Given;

distance between the two identical charges, d = 2 m

the force of repulsion between these two charges, F = 4N

Apply Coulomb's law;

F = \frac{kq_1q_2}{r^2} \\\\but \ q_1 =q_2,then \ let \ q_1 =q_2 = q\\\\F = \frac{kq^2}{r^2}\\\\q^2 = \frac{Fr^2}{k}\\\\q^2 = \frac{4*2^2}{9*10^9} \\\\q ^2 = 1.7778*10^{-9}\\\\q = \sqrt{1.7778*10^{-9}}\\\\q =4.22 *10^{-5} C\\\\q= q_1=q_2= 4.22 *10^{-5} C

Therefore, the value of each charge is 4.22 x 10⁻⁵ C

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Alenkinab [10]

Answer:

6.75J

Explanation:

U=1/2KΔx²

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