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rewona [7]
3 years ago
8

A newton is equivalent to which of the following questions?​

Physics
1 answer:
taurus [48]3 years ago
7 0
D, because kg x (m/s)2 is basically acceleration and mass. If it’s not d then it’s c. But units of m/s2 are mostly in parentheses.
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Given two different resistances, how does the rate of joule heating in them differ if they are connected to a fixed voltage sour
Mkey [24]

In series combination less rate of joule hating should be observed as compared to when connected in parallel combination .

Joule heating, also known as resistive, resistance, or Ohmic heating, is the process by which the passage of an electric current through a conductor produces heat.

Joule's first law states that the power of heating generated by an electrical conductor equals the product of its resistance and the square of the current:

When two or more resistors are connected end to end consecutively, they are said to be connected in series combination. The combined resistance of any number of resistances connected in series is equal to the sum of the individual resistances.

When two or more resistances are connected between the same two points, they are said to be connected in parallel combination. The reciprocal of the combined resistance of a number of resistances connected in parallel is equal to the sum of the reciprocals of all the individual resistances.

Rate of joule heating can be calculated by the formula

Power = V^{2} / R

V = voltage

R = resistance

When both are connected in series than their equivalent resistance will be more as compared to when they are connected in parallel.

Power is inversely related to resistance , if voltage is constant

In series combination less rate of joule hating should be observed as compared to when connected in parallel

To learn more about series combination here

brainly.com/question/16800236

#SPJ4

3 0
2 years ago
Why do two resistors in parallel together contain less resistance than the same two resistors in series?
stira [4]

Answer : When resistors are connected in parallel, the supply of current is equal to the sum of the currents passing through each of the resistors.

To be specific, the currents in the branches of a parallel circuit add up to the supply current. When resistors are connected in parallel, they have the same potential difference across them.

When resistors are connected in series the total resistance is greater than the individual resistances, hence the currents flow is less.

When resistors are connected in parallel, more current flows from the source than would flow for any of them individual resistors, so the total resistance is lower.

4 0
3 years ago
Read 2 more answers
Please help! last test of the week, and i can have a long weekedn! Friday and Sun and Sat.
ludmilkaskok [199]

It must have a medium to travel through.

Happy to help. Marking me the brainliest would really be appreciated.

6 0
3 years ago
A small but measurable current of 1.2 10-10A exists in a copper wire whose diameter is 2.0 mm. Assume the current is uniform. (a
Rom4ik [11]

Q: A small but measurable current of 1.2×10⁻¹⁰A exists in a copper wire whose diameter is 2.0 mm. Assume the current is uniform. (a) Calculate the current density.

Answer:

3.82×10⁻⁵ A/m

Explanation:

Current density: This can be defined as the amount of charge passing through a conductor, per unit area, per unit time. The S.I unit of current density is A/m²

From the question above, the expression for current density is given as,

τ = I/A............... Equation 1

Where τ = current density of the copper wire, I = current flowing  through the copper wire, A = cross sectional area of the copper wire

But,

A = πd²/4................. Equation 2

Substitute equation 2 into equation 1

τ = 4I/(πd²)............ Equation 3

Given: I = 1.2×10⁻¹⁰ A, d = 2 mm = 2×10⁻³ m, π = 3.14

Substitute into equation 3

τ = 4(1.2×10⁻¹⁰)/[3.14×(2×10⁻³)²]

τ = (4.8×10⁻¹⁰)/(1.256×10⁻⁵)

τ = 3.82×10⁻⁵ A/m

5 0
3 years ago
Read 2 more answers
A 1.75 kg box is pushed with a 8.35 N force across ground where k = 0.267. What is the net force on the box?
lubasha [3.4K]

The net force acting on the box of mass 1.75kg when it is pushed across a ground with coefficient of friction k=0.267 with a force of 8.35N, is 3.77N.

To find the answer, we have to know more about the basic forces acting on a body.

<h3>How to find the net force on the box?</h3>
  • Let us draw the free body diagram of the given box with the data's given in the question.
  • From the diagram, we get,

                                 N=mg\\F_t=ma\\F_t=F-f

where, N is the normal reaction, mg is the weight of the box, F_t is the net force, f is the kinetic friction.

  • We have the expression for kinetic friction as,

                      f=kN=kmg=0.267*1.75*9.8= 4.58N

  • Thus, the net force will be,

                        F_t= 8.35-4.58=3.77N

Thus, we can conclude that, the net force acting on the box of mass 1.75kg when it is pushed across a ground with coefficient of friction k=0.267 with a force of 8.35N, is 3.77N.

Learn more about the basic forces on a body here:

brainly.com/question/28061293

#SPJ1

8 0
2 years ago
Read 2 more answers
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