What is the first step for both loading and unloading a firearm?

An capacitor consists of two large parallel plates of area A separated by a very small distance d. This capacitor is connected t

scoundrel [369]

Answer:

Will be doubled.

Explanation:

For a capacitor of parallel plates of area A, separated by a distance d, such that the charges in the plates are Q and -Q, the capacitance is written as:

$C = \frac{Q}{V} = e_0\frac{A}{d}$

where e₀ is a constant, the electric permittivity.

Now we can isolate V, the potential difference between the plates as:

$V = \frac{Q}{e_0} *\frac{d}{A}$

Now, notice that the separation between the plates is in the numerator.

Thus, if we double the distance we will get a new potential difference V', such that:

$V' = \frac{Q}{e_0} *\frac{2d}{A} = 2*( \frac{Q}{e_0} *\frac{d}{A}) = 2*V\\V' = 2*V$

So, if we double the distance between the plates, the potential difference will also be doubled.

6 0

2 years ago

David is driving a steady 30.0 m/s when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady

SpyIntel [72]

Answer:

Explanation:

Let after time t , Tina catches up David .

Distance travelled by them are equal ,

Distance travelled by Tina

s = ut + 1/2 a t²

= .5 x 2.10 t²

= 1.05 t²

Distance travelled by David

= 30 t ( because of uniform velocity )

1.05 t² = 30t

t = 28.57 s

Distance travelled by Tina

= 1/2 a t²

= .5 x 2.10 x 28.57²

= 857 m approx.

7 0

2 years ago

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What force in Newton is required to accelerate a car starting from rest to 20 m/s in 15 seconds if the mass of the car is 2500 k

Lyrx [107]

We will solve this question using the second law of motion which states that force is directly equal to the product of mass and acceleration.

$\sf \: F=ma$

Where,

F is force
m is mass
a is acceleration

In our case,

F = ?
m = 2500 kg
a = 20m/s

$\tt \: F_{net} = 2500 \times 20 \\ \tt= 50000$

<em>Thus, The force of 50000 Newton is required to accelerate a car of 2500 kg...~</em>

3 0

2 years ago

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In Millikan's experiment, an oil drop of radius 1.362 μm and density 0.888 g/cm3 is suspended in chamber C when a downward-point

Misha Larkins [42]

Answer:

The charge on the oil drop is 3e

Explanation:

F = qE

Where;

F is the applied force in Newton

E is the electric field potential N/C

q is charge in C

Given;

Radius, r = 1.362 μm = 1.362 X 10⁻⁶ m

density, ρ = 0.888 g/cm³ = 0.888 X 10³ kg/m³

Electric field potential = 1.92 ✕ 10⁵ N/C

F =mg

mass of the oil drop = density, ρ X volume of the oil drop

volume of the oil drop (spherical) = (4/3)πr³ = 1.3333π(1.362 X 10⁻⁶)³

⇒ volume of the oil drop = 10.584 X 10⁻¹⁸ m³

mass of the oil drop = 0.888 X 10³ (kg/m³) X 10.584 X 10⁻¹⁸ (m³)

⇒ mass of the oil drop = 9.399 X 10⁻¹⁵ kg

⇒ F =mg = 9.399 X 10⁻¹⁵ kg X 9.8 = 9.21 X10⁻¹⁴ N

F = qE

q = F/E

q = (9.21 X10⁻¹⁴)/(1.92 ✕ 10⁵) = 4.797 X 10⁻¹⁹ C

In terms of e

1e = 1.6 X10⁻¹⁹ C

= (4.797 X 10⁻¹⁹ C)/(1.6 X10⁻¹⁹ C) = 3e

Therefore, the charge on the oil drop is 3e

7 0

3 years ago

Mia wanted to know how Earth’s movements created new landforms. She decided to read about folded mountains and their characteris

kifflom [539]

Answer:

I’m pretty sure it’s D

Explanation:

the crust breaks and shifts when that happens

hope dis helps ^-^

5 0

3 years ago

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