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slamgirl [31]
3 years ago
13

PY85

Physics
1 answer:
aliya0001 [1]3 years ago
3 0

Answer:

460 g

Explanation:

Heat lost by the warm water = heat gained by the cold water

-mCΔT = mCΔT

-m (4.184 J/g/K) (37°C − 85°C) = (1000 g) (4.184 J/g/K) (37°C − 15°C)

-m (37°C − 85°C) = (1000 g) (37°C − 15°C)

-m (-48°C) = (1000 g) (22°C)

m = 458 g

Rounded to two significant figures, you need a mass of 460 g of water.

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two point charges of 5*10^-19 C and 20*10^-19C are separated by a distance of 2m. at which point on the line joining them will h
Aneli [31]

Answer:

On that line segment between the two charges, at approximately 0.7\; \rm m away from the smaller charge (the one with a magnitude of 5 \times 10^{-19}\; \rm C,) and approximately 1.3\; \rm m from the larger charge (the one with a magnitude of 20 \times 10^{-19}\; \rm C.)

Explanation:

Each of the two point charges generate an electric field. These two fields overlap at all points in the space around the two point charges. At each point in that region, the actual electric field will be the sum of the field vectors of these two electric fields.

Let k denote the Coulomb constant, and let q denote the size of a point charge. At a distance of r away from the charge, the electric field due to this point charge will be:

\displaystyle E = \frac{k\, q}{r^2}.

At the point (or points) where the electric field is zero, the size of the net electrostatic force on any test charge should also be zero.

Consider a positive test charge placed on the line joining the two point charges in this question. Both of the two point charges here are positive. They will both repel the positive test charge regardless of the position of this test charge.

When the test charge is on the same side of both point charges, both point charges will push the test charge in the same direction. As a result, the two electric forces (due to the two point charges) will not balance each other, and the net electric force on the test charge will be non-zero.  

On the other hand, when the test charge is between the two point charges, the electric forces due to the two point charges will counteract each other. This force should be zero at some point in that region.

Keep in mind that the electric field at a point is zero only if the electric force on any test charge at that position is zero. Therefore, among the three sections, the line segment between the two point charges is the only place where the electric field could be zero.

Let q_1 = 5\times 10^{-19}\; \rm C and q_2 = 20 \times 10^{-19}\; \rm C. Assume that the electric field is zero at r meters to the right of the 5\times 10^{-19}\; \rm C point charge. That would be (2 - r) meters to the left of the 20 \times 10^{-19}\; \rm C point charge. (Since this point should be between the two point charges, 0 < r < 2.)

The electric field due to q_1 = 5\times 10^{-19}\; \rm C would have a magnitude of:

\displaystyle | E_1 | = \frac{k\cdot q_1}{r^2}.

The electric field due to q_2 = 20 \times 10^{-19}\; \rm C would have a magnitude of:

\displaystyle | E_2 | = \frac{k\cdot q_2}{(2 - r)^2}.

Note that at all point in this section, the two electric fields E_1 and E_2 will be acting in opposite directions. At the point where the two electric fields balance each other precisely, | E_1 | = | E_2 |. That's where the actual electric field is zero.

| E_1 | = | E_2 | means that \displaystyle \frac{k\cdot q_1}{r^2} = \frac{k\cdot q_2}{(2 - r)^2}.

Simplify this expression and solve for r:

\displaystyle q_1\, (2 - r)^2 - q_2 \, r^2 = 0.

\displaystyle 5\times (2 - r)^2 - 20\, r^2 = 0.

Either r = -2 or \displaystyle r = \frac{2}{3}\approx 0.67 will satisfy this equation. However, since this point (the point where the actual electric field is zero) should be between the two point charges, 0 < r < 2. Therefore, (-2) isn't a valid value for r in this context.

As a result, the electric field is zero at the point approximately 0.67\; \rm m away the 5\times 10^{-19}\; \rm C charge, and approximately 2 - 0.67 \approx 1.3\; \rm m away from the 20 \times 10^{-19}\; \rm C charge.

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3 years ago
Which creates energy used by the other system to help you exercise?
jasenka [17]

Answer:

There are many different systems involved in when we exercise, the three main ones are the Respiratory system which is involved in breathing the circulatory system which is about circulation of blood around the body and finally the muscular system and finally the Muscular system which is about how we move.

Explanation:

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3 years ago
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Why do conductors transfer heat faster than insulators?
Nesterboy [21]

Answer:

Conduction is usually faster in certain solids and liquids than in gases. Materials that are good conductors of thermal energy are called thermal conductors. Metals are especially good thermal conductors because they have freely moving electrons that can transfer thermal energy quickly and easily.

Heat transfer by convection happens through the air, and there are millions of minuscule air spaces between the fibers. Heat transfer by radiation is also slow since one fiber must radiate its heat to another.

When we give heat then kinetic energy is increase and this heat is transferred from hot metal to cold metal through this free electrons. As in insulator the free electrons are negligible so that the heat is not transferred from hot junction to cold junction due to absence of this free electrons.

Explanation:

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2 years ago
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An object with a mass of 1 kg is attached to one end of a cord 10 m long. The cord will break if its tension exceeds 10 N. The o
faust18 [17]

Answer:

10 m/s

Explanation:

∑F = ma

T = m v² / r

10 N = (1 kg) v² / (10 m)

v = 10 m/s

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3 years ago
What distinguishes a tornado watch from a tornado warning?
tankabanditka [31]
"Watch" . . . atmospheric conditions exist that could
produce tornados, but none have been seen yet.

"Warning" . . . tornados have formed and have been seen.
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