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Pani-rosa [81]
3 years ago
11

What is the name of asteroids that stray outside the main belt and cross outside Mars taking them closer to the Sun?

Physics
1 answer:
Svetradugi [14.3K]3 years ago
6 0

Answer:

b im pretty sure (let me know if this is correct!) :)

Explanation:

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In a standing wave, areas of destructive are the?
Alborosie

antinodes

Hope this helps :)

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3 years ago
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ou have just moved into a new apartment and are trying to arrange your bedroom. You would like to move your dresser of weight 3,
Dimas [21]

So far, since you moved into the apartment until the end of this much of the story, you haven't done ANY work on the dresser yet.

I'll admit that you pushed, groaned and grunted, sweated and strained plenty.  You're physically and mentally exhausted, you're not interested in the dresser at the moment, and right now you just want to snappa cappa brew, crash on the couch, and watch cartoons on TV.  But if you've done your Physics homework, you know you haven't technically done any <u><em>work</em></u> yet.

In Physics, "Work" is the product of Force times Distance.

Since the dresser hasn't budged yet, the Distahce is zero.  So no matter how great the Force may be, it's multiplied by zero, so the <em>Work is zero</em>.

5 0
3 years ago
A 10 gauge copper wire carries a current of 23 A. Assuming one free electron per copper atom, calculate the magnitude of the dri
Reptile [31]

Question:

A 10 gauge copper wire carries a current of 15 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm².)

Answer:

3.22 x 10⁻⁴ m/s

Explanation:

The drift velocity (v) of the electrons in a wire (copper wire in this case) carrying current (I) is given by;

v = \frac{I}{nqA}

Where;

n = number of free electrons per cubic meter

q =  electron charge

A =  cross-sectional area of the wire

<em>First let's calculate the number of free electrons per cubic meter (n)</em>

Known constants:

density of copper, ρ = 8.95 x 10³kg/m³

molar mass of copper, M = 63.5 x 10⁻³kg/mol

Avogadro's number, Nₐ = 6.02 x 10²³ particles/mol

But;

The number of copper atoms, N, per cubic meter is given by;

N = (Nₐ x ρ / M)          -------------(ii)

<em>Substitute the values of Nₐ, ρ and M into equation (ii) as follows;</em>

N = (6.02 x 10²³ x 8.95 x 10³) / 63.5 x 10⁻³

N = 8.49 x 10²⁸ atom/m³

Since there is one free electron per copper atom, the number of free electrons per cubic meter is simply;

n = 8.49 x 10²⁸ electrons/m³

<em>Now let's calculate the drift electron</em>

Known values from question:

A = 5.261 mm² = 5.261 x 10⁻⁶m²

I = 23A

q = 1.6 x 10⁻¹⁹C

<em>Substitute these values into equation (i) as follows;</em>

v = \frac{I}{nqA}

v = \frac{23}{8.49*10^{28} * 1.6 *10^{-19} * 5.261*10^{-6}}

v = 3.22 x 10⁻⁴ m/s

Therefore, the drift electron is 3.22 x 10⁻⁴ m/s

6 0
4 years ago
A 50 kg mass is placed 2 meters from the fulcrum. To balance the lever, a second object is placed 4 meters
julia-pushkina [17]

Answer:

mass =25 kg

using clockwise moment = anticlockwise moment

8 0
3 years ago
Through which medium does a mechanical waves travel most quickly
PIT_PIT [208]
The answer would be solids
6 0
3 years ago
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