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2 years ago

Which direction does heat flow from?

Physics

2 answers:

son4ous [18]2 years ago

8 0

Answer:

cold to hot as we al have must noticed that

lord [1]2 years ago

7 0

cold to hot , hope this helps

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A 15kg boulder is moved from the Earth to the Moon. Explain what happens to the mass and weight of the boulder.

elena-14-01-66 [18.8K]

Answer:

The mass of the boulder remains constant, while its weight decreases with respect to the value of gravitational force on the moon.

Explanation:

The mass of the boulder = 15 kg

On the earth, its mass remains 15 kg. But its weight is;

weight = m x g

= 15 x 9.8

= 147 N

The boulder's weight on the earth is 147 N.

When transferred to the moon, the mass remains constant i.e 15 kg. But its weight decreases due to a change in the value of acceleration due to gravity on the moon. Thus, the boulder becomes lighter in weight.

6 0

3 years ago

Which image illustrates refraction

photoshop1234 [79]

Answer:

it loaded and it is C. buddy sorry about that :)

4 0

4 years ago

Read 2 more answers

A hair dryer draws 11 A when it is connected to 120 V. If electrical energy costs $ 0.09/kW·h, what is the cost of using the hai

Papessa [141]

Answer:

The cost of using the hair dryer for 15 minutes is $\$3.6\bar 6$

Explanation:

The parameters given in the question are;

The electric current drawn by the the air dryer, I = 11 A

The voltage to which the hair dryer is connected, V = 120 V

The duration of usage of the hair dryer = 15 minutes = 60 minutes /4 = 1 hour/4 = 0.25 hour

The electrical energy costs $0.09/kW·h

The power consumed by the hair dryer = I × V = 11 × 120 = 1320 Watts = 1.32 kW

The energy used by the hair dryer in 15 minutes (0.25 hour) = 1.32 × 0.25 0.33 kW·h

The energy used by the hair dryer in 15 minutes (0.25 hour) = 0.33 kW·h

The energy cost = $0.09/(kW·h)

Therefore;

The cost of using the hair dryer for 15 minutes (0.25 hour) = 0.33 kW·h/($0.09/(kW·h)) = $33/9 = $3 2/3 = $3.6 $\bar 6$ .

6 0

3 years ago

The _________ of a line is defined as its rise divided by its run.

Alexeev081 [22]

Answer:

This is the definition of slope.

Step-by-step explanation:

The formula for slope is:

m(slope) = (y2 - y1)/(x2 - x1)

This shows the difference in y, which is rise, divided by the different in x, or run.

3 0

4 years ago

Read 2 more answers

NASA has asked your team of rocket scientists about the feasibility of a new satellite launcher that will save rocket fuel. NASA

kkurt [141]

Answer:

The answer is " $q=0.0945\,C$ ".

Explanation:

Its minimum velocity energy is provided whenever the satellite(charge 4 q) becomes 15 m far below the square center generated by the electrode (charge q).

$U_i=\frac{1}{4\pi\epsilon_0} \times \frac{4\times4q^2}{\sqrt{(15)^2+(5/\sqrt2)^2}}$

It's ultimate energy capacity whenever the satellite is now in the middle of the electric squares:

$U_f=\frac{1}{4\pi\epsilon_0}\ \times \frac{4\times4q^2}{( \frac{5}{\sqrt{2}})}$

Potential energy shifts:

$= U_f -U_i \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{\sqrt{(15)^2+( \frac{5}{\sqrt{2})^2)}}\right ) \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ 15 +( \frac{5}{2})}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{30+5}{2})}}\right )\\\\$

$=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{35}{2})}}\right )\\\\=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{17.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 24.74- 5 }{87.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 19.74- 5 }{87.5}}\right )\\\\ =\frac{4q^2}{\pi\epsilon_0}\left ( 0.2256 }\right )\\\\= \frac{0.28 \times q^2}{ \epsilon_0}\\\\=q^2\times31.35 \times10^9\,J$

Now that's the energy necessary to lift a satellite of 100 kg to 300 km across the surface of the earth.

$=\frac{GMm}{R}-\frac{GMm}{R+h} \\\\=(6.67\times10^{-11}\times6.0\times10^{24}\times100)\left(\frac{1}{6400\times1000}-\frac{1}{6700\times1000} \right ) \\\\ =(6.67\times10^{-11}\times6.0\times10^{26})\left(\frac{1}{64\times10^{5}}-\frac{1}{67\times10^{5}} \right ) \\\\=(6.67\times6.0\times10^{15})\left(\frac{67 \times 10^{5} - 64 \times 10^{5} }{ 4,228 \times10^{5}} \right ) \\\\$

$=( 40.02\times10^{15})\left(\frac{3 \times 10^{5}}{ 4,228 \times10^{5}} \right ) \\\\ =40.02 \times10^{15} \times 0.0007 \\\\$

$\\\\ =0.02799\times10^{10}\,J \\\\= q^2\times31.35\times10^{9} \\\\ =0.02799\times10^{10} \\\\q=0.0945\,C$

This satellite is transmitted by it system at a height of 300 km and not in orbit, any other mechanism is required to bring the satellite into space.

6 0

3 years ago