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alexgriva [62]
3 years ago
8

What is the mass of air in a room

Physics
1 answer:
Karo-lina-s [1.5K]3 years ago
7 0

Answer:

Just do 1000 times 1.8

Explanation:

This is a good answer

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What part of an atom is involved in electricity and magnetism?*
-BARSIC- [3]

Answer:

Electrons

Explanation:

Electrons are very important in the world of electronics. The very small particles can stream through wires and circuits, creating currents of electricity. The electrons move from negatively charged parts to positively charged ones.

I hope this helps!

6 0
3 years ago
Read 2 more answers
What does the atomic number tell us about an atom of a certain element?
Setler79 [48]
It tells us the number of protons that are present in the nucleus, the positively charged region of that atom.
4 0
3 years ago
Particles 1 and 2 of charge q1 = q2 = +3.20 × 10−19 C are on a y axis at distance d = 17.0 cm from the origin. Particle 3 of cha
mel-nik [20]

Answer:

(a) 0.17 m

(b) 5.003 m

(c) 6.38 × 10^{-26} N

(d) 7.37 ×10^{-29} N

Explanation:

(a) The minimum value of x will occur when q3 = 0 m or at origin and q1, q2 are at 0.17 m so the distance between q3 and q1, q2 is 0.17 m, therefore the <em>minimum value of x= 0.17 m</em>.

(b) The maximum value of x will occur when q3 = 5 m because it is said in the question that 5 is the maximum distance travelled by q3. To find the hypotenuse i.e. the distance between q3 and q1,q2, we use Pythagoras theorem.

h^{2} = b^{2} + p^{2}

h^{2} = 5^{2} + 0.17^{2}   \\h = \sqrt{} 25.03\\h= 5.002 m

<em>Hence, the maximum distance is 5.002 m</em>

(c) For minimum magnitude we use the minimum distance calculated in (a)

Minimum Distance = 0.17 m

For electrostatic force=     F=\frac{kq1q2}{x^{2} }

F=\frac{9 x 10^{9} x3.2x10^{-19}x 6.4x10^{-19}  }{0.17^{2} }

F= 6.38×10^{-26} N

(d) For maximum magnitude, we use the maximum distance calculated in (b)

Maximum Distance = 5.002 m

Using the formula for electrostatic force again:

F =  \frac{9x10^{9}x3.2x10^{-19}x6.4x10^{-19}   }{5.002^{2} } }

F= 7.37×10^{-29 N

4 0
3 years ago
A 2.13-kg object on a frictionless horizontal track is attached to the end of a horizontal spring whose force constant is 5.00 N
ANTONII [103]

Answer:

17.54N in -x direction.

Explanation:

Amplitude (A) = 3.54m

Force constant (k) = 5N/m

Mass (m) = 2.13kg

Angular frequency ω = √(k/m)

ω = √(5/2.13)

ω = 1.53 rad/s

The force acting on the object F(t) = ?

F(t) = -mAω²cos(ωt)

F(t) = -2.13 * 3.54 * (1.53)² * cos (1.53 * 3.50)

F(t) = -17.65 * cos (5.355)

F(t) = -17.57N

The force is 17.57 in -x direction

5 0
4 years ago
A train increase its speed steadily from 10m/s to 20m/s in 1minutes A what is the average speed during this time in m/s B how fa
Ipatiy [6.2K]

If it increased its speed steadily at a constant rate, then the average speed for the minute was

(1/2)(10m/s + 20m/s) = 15 m/s .

Rolling at an average speed of 15 m/s for 1 minute (60 seconds), it travels

(15 m/s) (60 sec) = 900 meters

3 0
4 years ago
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