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Anna35 [415]
3 years ago
10

Car 1 drives 20 mph to the south, and car 2 drives 30 mph to the north. From

Physics
1 answer:
Tju [1.3M]3 years ago
7 0

Answer:

10 mph faster than car 1 is going

Explanation:

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A gas is confined to a cylinder under constant atmospheric pressure, as illustrated in the following figure. When the gas underg
GarryVolchara [31]
From the first law of thermodynamics, we use the equation expressed as:

ΔH = Q + W

where Q is the heat absorbed of the system and W is the work done.

We calculate as follows:

ΔH = Q + W
ΔH = 829 J + 690 J = 1519 J

Hope this answers the question. Have a nice day.
7 0
3 years ago
Read 2 more answers
Calculate the orbital period for Jupiter's moon Io, which orbits 4.22×10^5km from the planet's center (M=1.9×10^27kg) .
Verdich [7]

According to the <u>Third Kepler’s Law of Planetary motion</u> “<em>The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.</em>



In other words, this law states a relation between the orbital period T of a body (moon, planet, satellite) orbiting a greater body in space with the size a of its orbit.



This Law is originally expressed as follows:



<h2>T^{2} =\frac{4\pi^{2}}{GM}a^{3}    (1) </h2>

Where;


G is the Gravitational Constant and its value is 6.674(10^{-11})\frac{m^{3}}{kgs^{2}}



M=1.9(10^{27})kg is the mass of Jupiter


a=4.22(10^{5})km=4.22(10^{8})m  is the semimajor axis of the orbit Io describes around Jupiter (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)



If we want to find the period, we have to express equation (1) as written below and substitute all the values:



<h2>T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}    (2) </h2>

T=\sqrt{\frac{4\pi^{2}}{6.674(10^{-11})\frac{m^{3}}{kgs^{2}}1.9(10^{27})kg}(4.22(10^{8})m)^{3}}    



T=\sqrt{\frac{2.966(10^{27})m^{3}}{1.268(10^{17})m^{3}/s^{2}}}    



T=\sqrt{2.339(10^{10})s^{2}}    



Then:


<h2>T=152938.0934s    (3) </h2>

Which is the same as:



<h2>T=42.482h     </h2>

Therefore, the answer is:



The orbital period of Io is 42.482 h



7 0
4 years ago
A CFL bulb has an efficiency of 8.9% and a power of 22 W. How much light energy does the lightbulb produce in 1 second
hjlf
The power that the light is able to utilize out of the supply is only 0.089 of the given.
                           Power utilized = (0.089)(22 W)
                                                  = 1.958 W
                                                  = 1.958 J/s
The energy required in this item is the product of the power utilized and the time. That is,
                           Energy = (1.958 J/s)(1 s) = 1.958 J
Thus, the light energy that the bulb is able to produce is approximately 1.958 J. 
6 0
4 years ago
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in a closed system three objects have the following momentum: 11 kg* m/s, -65 kg*m/s and -100 kg m/s. the objects collide and mo
guajiro [1.7K]

Explanation:

The momentum of the three objects are as follow :

11 kg-m/s, -65 kg-m/s and -100 kg-m/s

Before collision, the momentum of the system is :

P_i=11+(-65)+(-100)\\\\P_i=-154\ kg-m/s

After collison, they move together. It means it is a case of inelastic collision. In this type of collision, the momentum of the system remains conserved.

It would mean that, after collision, momentum of the system is equal to the initial momentum.

Hence, final momentum = -154 kg-m/s.

4 0
4 years ago
Sort the resources into the correct category: renewable or nonrenewable.
alex41 [277]

Answer:

forests

renewable resources

nonrenewable resources

water

iron

oll

wind

silver

Renewable resources- Water

Forests

Wind

Non-renewable resources-

Oil

Iron

Silver

5 0
3 years ago
Read 2 more answers
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