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3 years ago

What is the conversion factor between kg/m^3 and g/m^3. Express your experimental values for the densities in kg/m^3.

Physics

1 answer:

romanna [79]3 years ago

6 0

1000 g = 1 kilog (10³⇒kilo)

so, $\frac{1000 g}{m^{3} }=\frac{1 kg}{m^{3} }$

Since I don't have your values from your experiment, I'm assuming you have your mass values in g. Your teacher wants them in kg, so divide each value (g) by 1000 .

Then divide this mass (now in kg) by your volume in m³. This will make your density units kg/m³

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Hire rubs an inflated balloon on his hair and his hair sticks to the balloon as he moves the balloon away from his head how does

sergey [27]

The options are missing and they are;

A) the electric force increases because the balloon loses its charge.

B) the electric force increases because the distance increases.

C) the electric force decreases because the distance increases.

D) the electric force decreases because his hair loses its charge.

Answer:

Correct answer is option C - the electric force decreases because the distance increases.

Explanation:

The formula for electric force is;

F = k•q1•q2/r²

Where;

K is coulombs constant

q1 and q2 are particle charges

r is distance

So,looking at the formula given earlier, if we increase the distance, the denominator will increase and thus the Force will decrease.

So the correct option is option C

8 0

3 years ago

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A law of motion that states that an object at rest stays at rest and one in motion stays

adoni [48]

Answer:

that is the first law of motion or newtons first law of motion

6 0

3 years ago

A runner starts at position A. He runs 40 m North, 10 m East and 40 m

Alecsey [184]

The answer is C.) 10 m East

8 0

3 years ago

A 2.43ug particle moves at 1.97 x 108 m/s. What is its momentum?

Ira Lisetskai [31]

Answer:

0.48 kgm/s

Explanation:

$m$ = mass of the particle = 2.43 μg = 2.43 x 10⁻⁶ x 10⁻³ kg = 2.43 x 10⁻⁹ kg

$v$ = velocity of the particle = 1.97 x 10⁸ m/s

$p$ = momentum of the particle

momentum of the particle is given as

$p = m v$

inserting the values

$p = (2.43\times 10^{-9})(1.97\times 10^{8})$

$p = 0.48$ kgm/s

4 0

4 years ago

a parallel plate capacitor has square plates that have edge length equal to 100cm and are separated by 1 mm. It is connected to

Flura [38]

Answer:

the energy is stored in the capacitor is 0.32 μJ

Explanation:

Given;

distance of separation, d = 1 mm = 0.001 m

edge length of the square = 100 cm

potential difference across the plates, V = 12 v

let the side of the square = L

This edge length is also the diagonal of the square which makes a right angle with the side of the square.

Applying Pythagoras theorem;

L² + L² = 100²

2L² = 100²

L² = 100²/2

Note area of a square is L²

A = L² = 100²/2 = 5000 cm²

A (m²) = 5000 cm² x 1m²/(100 cm)²

A = 5000 cm² x 1m²/10000 cm²

A = 0.5 m²

Energy stored in a parallel plate capacitor, E= ¹/₂CV²

C = ε₀A/d

where;

ε₀ is permittivity of free space = 8.85 x 10⁻¹² F/m

d is the distance of separation = 0.001 m

A is the area of the plate

C = ε₀A/d = (8.85 x 10⁻¹²)x0.5 / 0.001

C = 4425 x 10⁻¹² F

E = ¹/₂CV² = ¹/₂ x 4425 x 10⁻¹² x ( 12)²

E = 318600 x 10⁻¹² = 0.32 μJ

Therefore, the energy is stored in the capacitor is 0.32 μJ

8 0

3 years ago

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