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3 years ago

If two objects at different temperatures are in contact with each other, what happens to their temperatures?

1 answer:

6 0

In that case, heat energy flows from the warmer object to the cooler one.
As heat flows from one to the other, the temperature of the warmer object
falls, and the temperature of the cooler object rises. When the temperatures
are equal, the flow of heat energy from one to the other stops.

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A 1.0 kg cart moving right at 5.0 m/s on a frictionless track collides with a second cart moving at 2.0m/s. The 1.0 kg cart has

____ [38]

Answer:

m₂ = 3kg

Explanation:

The question wasn't clear about what direction the initial velocity of the second cart was, so I'll assume it was going left at 2.0m/s.

Anyway, this is a conservation of momentum problem. The equation you need to use is the one written in blue. They want you to solve for the mass of the second cart, so do some algebra and rearrange that blue equation in term of m₂.

Now that you have the equation for m₂, plug in all the values given from the question and you'll get 3kg.

3 0

4 years ago

The intensity of sunlight reaching the earth is 1360 w/m2. Assuming all the sunligh is absorbed, what is the radiation pressure

krok68 [10]

(a) $1.15\cdot 10^9 N$

For an electromagnetic wave incident on a surface, the radiation pressure is given by (assuming all the radiation is absorbed)

$p=\frac{I}{c}$

where

I is the intensity

c is the speed of light

In this problem, $I=1360 W/m^2$ ; substituting this value, we find the radiation pressure:

$p=\frac{1360 W/m^2}{3\cdot 10^8 m/s}=4.5\cdot 10^{-6} Pa$

the force exerted on the Earth depends on the surface considered. Assuming that the sunlight hits half of the Earth's surface (the half illuminated by the Sun), we have to consider the area of a hemisphere, which is

$A=2pi R^2$

where

$R=6.37\cdot 10^6 m$

is the Earth's radius. Substituting,

$A=2\pi (6.37\cdot 10^6 m)^2=2.55\cdot 10^{14}m^2$

And so the force exerted by the sunlight is

$F=pA=(4.5\cdot 10^{-6} Pa)(2.55\cdot 10^{14} m^2)=1.15\cdot 10^9 N$

(b) $3.2\cdot 10^{-14}$

The gravitational force exerted by the Sun on the Earth is

$F=G\frac{Mm}{r^2}$

where

G is the gravitational constant

$M=1.99\cdot 10^{30}kg$ is the Sun's mass

$m=5.98\cdot 10^{24}kg$ is the Earth's mass

$r=1.49\cdot 10^{11} m$ is the distance between the Sun and the Earth

Substituting,

$F=(6.67\cdot 10^{-11} )\frac{(1.99\cdot 10^{30}kg)(5.98\cdot 10^{24} kg)}{(1.49\cdot 10^{11} m)^2}=3.58\cdot 10^{22}N$

And so, the radiation pressure force on Earth as a fraction of the sun's gravitational force on Earth is

$\frac{1.15\cdot 10^9 N}{3.58\cdot 10^{22}N}=3.2\cdot 10^{-14}$

6 0

4 years ago

Earth’s atmosphere is constantly bombarded by cosmic ray protons that originate somewhere in space. If the protons all passed th

gulaghasi [49]

Answer:

The current would be 0.12A

Explanation:

Well, the formula for electric current is

$i=\frac{q}{t}$

where i is current, q is charge and t is time. Then we have to calculate the charge. If we know that the charge of a proton is $1.6*10^{-19} C$ , then we calculate the charge of 1500 protons as it follows: