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3 years ago

14

What is the simplest requirement for an electromagnet

1 answer:

nexus9112 [7]3 years ago

4 0

A length of of electrified wire.

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An aurora occurs when ____

ASHA 777 [7]

The short answer to how the aurora happens is that energetic electrically charged particles (mostly electrons) accelerate along the magnetic field lines into the upper atmosphere, where they collide with gas atoms, causing the atoms to give off light.

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3 years ago

What is the peacocks displacement from 2 to 3 seconds

mash [69]

1.5 second difference

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2 years ago

At room temperature, sound travels at a speed of about 344 m/s in air. You see a distant flash of lightning and hear the thunder

beks73 [17]

Answer:

$d=1.67mi$

Explanation:

Assuming the light takes essentially no time to reach you, the distance at which the lightning occurred can be calculated by multiplying the speed of sound by the time it takes to hear the thunder:

$d=v_st\\d=344\frac{m}{s}(7.8s)\\d=2683.2m*\frac{1mi}{1609.34m}=1.67mi$

4 0

3 years ago

These capacitors are then disconnected from their batteries, and the positive plates are now connected to each other and the neg

Kay [80]

Answer:

Following are the solution to the given question:

Explanation:

For charging plates that are connected in a similar manner:

Calculating the total charge:

$\to q =q_1 + q_2 = C_1V_1 +C_2V_2 =1320 + 2714 = 4034 \mu C$

Calculating the common potential:

$\to V = \frac{q}{C}= \frac{q}{(C_1 + C_2)} =\frac{4034}{6.8} = 593 \ V\\\\$

Calculating the charge after redistribution:

$When: \\\\q = q_{1}' + q_{2}' = q_1 + q_2$

$\to q_{1}' = C_1V = 2.2 \times 593 = 1305\ \mu C\\ \\ \to q_{2}' = C_2V = 4.6 \times 593 = 2729 \ \mu C$

6 0

3 years ago

A person can jump a maximum horizontal distance (by using a 45◦ projectile angle) of 5 m on Earth. The acceleration of gravity i

snow_lady [41]

Answer:30 m

Explanation:

Given

Maximum Horizontal distance is 5 m on earth

launching angle $=45^{\circ}$

Acceleration due to gravity on earth is $9.8 m/s^2$

Acceleration due to gravity on moon is $\frac{9.8}{6}=1.63 m/s^2$

Range of projectile is given by

$R=\frac{u^2\sin 2\theta }{g}$

$R_{earth}=\frac{u^2\sin 2\theta }{g}=5$ ----1

$R_{moon}=\frac{u^2\sin 2\theta }{\frac{g}{6}}$ -----2

Divide 1 & 2

$\frac{5}{R_{moon}}=\frac{1}{6}$

$R_{moon}=30 m$

4 0

3 years ago