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Reika [66]
3 years ago
15

Volcanic eruptions are caused primarily by the movement of

Physics
1 answer:
Assoli18 [71]3 years ago
4 0
The answer is Tectonic Plates
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False. Most of the bright stars in our galaxy are located in the disk.

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How does temperature influence the appearance of a star? Give an example in your response.
loris [4]

The cooler the temp of the star the star would appear to glisten a red. if the star is warmer the star would be a light blue or blue. An example would be the tiny red dwarf stars. they only have 7% of the suns heat so they appear red.

5 0
3 years ago
hree identical resistors are connected in series. When a certain potential difference is applied across the combination, the tot
pav-90 [236]

Answer:

The power dissipated if the three resistors were connected in parallel across the same potential difference is 405 W

Explanation:

Given;

three identical resistors connected in series

let the first resistor = R₁

let the second resistor = R₂

let the third resistor = R₃

Rt = R₁ + R₂ + R₃

Since the resistors are identical, thus, R₁ = R₂ = R₃ = R

Rt = 3R

Power is given as;

P = IV = V² / R

P = \frac{V^2}{R_t} = \frac{V^2}{3R} \\\\3P = \frac{V^2}{R} ------equation(i)

If the 3 identical resistor connection were changed to parallel, then the equivalent resistance in the circuit will be;

\frac{1}{R_t} = \frac{1}{R_1} +\frac{1}{R_2} + \frac{1}{R_3} \\\\But, R_1 = R_2 = R_3\\\\\\frac{1}{R_t} = \frac{1}{R} +\frac{1}{R} + \frac{1}{R} \\\\\frac{1}{R_t} =\frac{3}{R} \\\\R_t = \frac{R}{3} \\\\P = \frac{V^2}{R_t} = \frac{3V^2}{R} \\\\P_{parallel} = \frac{3V^2}{R} ---------equation (ii)\\\\From \ equation \ (i), 3P_{series} = \frac{V^2}{R}, Substitute \ this \ into \ equation \ (ii)\\\\P = 3(\frac{V^2}{R} )\\\\P = 3(3P)\\\\P_{parallel} = 9P_{series}\\\\P_{parallel} = 9(45)\\\\

P_{parallel} = 405 \ W

Therefore, the power dissipated if the three resistors were connected in parallel across the same potential difference is 405 W

4 0
3 years ago
A freight train has a mass of [02] kg. The wheels of the locomotive push back on the tracks with a constant net force of 7.50 ×
otez555 [7]

Answer:

t = 300.3 seconds

Explanation:

Given that,

The mass of a freight train, m=1.01\times 10^7\ kg

Force applied on the tracks, F=7.5\times 10^5\ N

Initial speed, u = 0

Final speed, v = 80 km/h = 22.3 m/s

We need to find the time taken by it to increase the speed of the train from rest.

The force acting on it is given by :

F = ma

or

F=\dfrac{m(v-u)}{t}\\\\t=\dfrac{m(v-u)}{F}\\\\t=\dfrac{1.01\times 10^7\times (22.3-0)}{7.5\times 10^5}\\\\t=300.3\ s

So, the required time is 300.3 seconds.

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3 years ago
A 25.0 kg object is held 8.50 m above the ground. Calculate its PE
pantera1 [17]

Answer:=mgh

Explanation:

6 0
3 years ago
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