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2 years ago

The δe of a system that releases 12.4 j of heat and does 4.2 j of work on the surroundings is __________ j.

1 answer:

5 0

The answer for this problem is clarified through this, the system is absorbing (+). And now see that it uses that the SURROUNDINGS are doing 84 KJ of work. Any time a system is overshadowing work done on it by the surroundings the sign will be +. So it's just 12.4 KJ + 4.2 = 16.6 KJ.

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When Earth’s Northern Hemisphere is tilted toward the Sun during June, some would argue that the cause of our seasons is that th

Ludmilka [50]

Answer:

Distance of Earth from the Sun has nothing to do with the seasons only the tilt is responsible for the change in seasons.

Explanation:

The Earth's tilt does cause the seasons but the distance from the sun and has nothing to do with the change in seasons. In June, when the Northern Hemisphere is tilted in the direction of the Sun during the Northern Hemisphere summer the Earth is actually farthest from the Sun. In January, when the Southern Hemisphere is tilted in the direction of the Sun during the Northern Hemisphere winter the Earth is actually closest to the Sun. This is caused due to the elliptical orbit of the Earth. So, distance of Earth from the Sun has nothing to do with the seasons.

4 0

2 years ago

An LR circuit contains an ideal 60-V battery, a 42-H inductor having no resistance, a 24-ΩΩ resistor, and a switch S, all in ser

s2008m [1.1K]

Answer:

1.6 s

Explanation:

To find the time in which the potential difference of the inductor reaches 24V you use the following formula:

$V_L=V_oe^{-\frac{Rt}{L}}$

V_o: initial voltage = 60V

R: resistance = 24-Ω

L: inductance = 42H

V_L: final voltage = 24 V

You first use properties of the logarithms to get time t, next, replace the values of the parameter:

$\frac{V_L}{V_o}=e^{-\frac{Rt}{L}}\\\\ln(\frac{V_L}{V_o})=-\frac{Rt}{L}\\\\t=-\frac{L}{R}ln(\frac{V_L}{V_o})\\\\t=-\frac{42H}{24\Omega}ln(\frac{24V}{60V})=1.6s$

hence, after 1.6s the inductor will have a potential difference of 24V

3 0

2 years ago

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu

MArishka [77]

Complete Question:

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bury itself just below the surface. What would have to be the mass of this asteroid, in terms of the earth’s mass M, for the day to become 25.0% longer than it presently is as a result of the collision? Assume that the asteroid is very small compared to the earth and that the earth is uniform throughout.

Answer:

m = 0.001 M

For the whole process check the following page: https://www.slader.com/discussion/question/suppose-that-an-asteroid-traveling-straight-toward-the-center-of-the-earth-were-to-collide-with-our/

6 0

2 years ago

Car B is following Car A and has a greater speed than Car A. the two cars are moving in a straight line and in the same directio

ElenaW [278]

Answer:

Collision force will be same in both the cases.

Explanation:

A perfectly inelastic collision is said to take place when a system loses the amount of its Kinetic Energy at its maximum. In a perfectly inelastic collision, the colliding particles stick to each other. In such a collision, kinetic energy is lost by combining the two bodies with each other.

In situation 1:

Speed of Car A, $v_{A} = 10 mph$

Speed of Car B, $v_{B} = 20 mph$

Relative speed of car A and car B, $v = v_{b} - v_{a} = 10 m/s$

Now, in the situation 2:

Speed of car A, $v_{A} = 30 mph$

Speed of car B, $v_{B} = 40 mph$

Relative speed of car A and car B, $v = v_{b} - v_{a} = 10 m/s$

Therefore, Car A and Car B both have the same relative speed, v = 10 m/s

7 0

1 year ago

At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to

mote1985 [20]

Answer:

Ok, the question is incomplete buy ill try to answer this in a general way.

Suppose that you have no-polarized light.

When that light hits one polaroid, the light becomes polarized along some line, and has an intensity I0.

Now, when polarized light hits a polaroid which axis is at an angle θ with respect to the polarization of the light, the intensity of the resulting beam is given by the Malus's law:

I(θ) = I0*cos^2(θ)

For example, if the axis of the polaroid is exactly the same as the one of the polarized light, then we have θ = 0°

and:

I(0°) = I0*cos^2(0°) = I0

So the intensity does not change.

Now, knowing the initial intensity, you can find the angle needed to get a given intensity.

For example, if the question was:

"At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to A"

We should solve:

I(θ) = A = I0*cos^2(θ)

(A/i0) = cos^2(θ)

√(A/I0) = cos(θ)

Acos(√(A/I0)) = θ

6 0

1 year ago