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uranmaximum [27]

3 years ago

12

When metals and nonmetals combined, first there is a complete transfer of charge to form cations and anions to complete their va

lence shell and increase stability. Then _________ bonds are formed by electrostatic attraction.

1 answer:

Llana [10]3 years ago

3 0

<span> Then IONIC bonds are formed by electrostatic attraction.</span>

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3. Do Newton's Laws of Motion apply to a Water Spout? If so, how?

PSYCHO15rus [73]

Yes it’s spills out becasue bucket

6 0

3 years ago

What is the potential difference across a parallel-plate capacitor whose plates are separated by a distance of 4.0 mm where each

suter [353]

The potential difference across the parallel plate capacitor is 2.26 millivolts

<h3>Capacitance of a parallel plate capacitor</h3>

The capacitance of the parallel plate capacitor is given by C = ε₀A/d where

ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m,
A = area of plates and
d = distance between plates = 4.0 mm = 4.0 × 10⁻³ m.

<h3>Charge on plates</h3>

Also, the surface charge on the capacitor Q = σA where

σ = charge density = 5.0 pC/m² = 5.0 × 10⁻¹² C/m² and
a = area of plates.

<h3>The potential difference across the parallel plate capacitor</h3>

The potential difference across the parallel plate capacitor is V = Q/C

= σA ÷ ε₀A/d

= σd/ε₀

Substituting the values of the variables into the equation, we have

V = σd/ε₀

V = 5.0 × 10⁻¹² C/m² × 4.0 × 10⁻³ m/8.854 × 10⁻¹² F/m

V = 20.0 C/m × 10⁻³/8.854 F/m

V = 2.26 × 10⁻³ Volts

V = 2.26 millivolts

So, the potential difference across the parallel plate capacitor is 2.26 millivolts

Learn more about potential difference across parallel plate capacitor here:

brainly.com/question/12993474

7 0

2 years ago

const2013 [10]

Answer:

I think the answer 1

Explanation:

im probably wrong too i dont know

5 0

3 years ago

PLEASE HELP NOT A LOT OF TIME LEFT!!!!!

GenaCL600 [577]

D. 65.1 is the answer

8 0

3 years ago

1.)Two objects, one of m=20,000 kg, and another of 12,500 kg, are placed at a distance of 5 meters apart. What is the force of g

Delvig [45]

1) $6.67\cdot 10^{-4} N$

The force of gravitation between the two objects is given by:

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} kg^{-1} m^{3} s^{-2}$ is the gravitational constant

m1 = 20,000 kg is the mass of the first object

m2 = 12,500 kg is the mass of the second object

r = 5 m is the distance between the two objects

Substituting the numbers inside the equation, we find

$F=(6.67\cdot 10^{-11})\frac{(20,000 kg)(12,500 kg)}{(5 m)^2}=6.67\cdot 10^{-4} N$

2) $2.7\cdot 10^{-3} N$

From the formula in exercise 1), we see that the force is inversely proportional to the square of the distance:

$F \sim \frac{1}{r^2}$

this means that if we cut in a half the distance without changing the masses, the magnitude of the forces changes by a factor

$F'\sim \frac{1}{(r/2)^2}=4 \frac{1}{r^2}=4F$

So, the gravitational force increases by a factor 4. Therefore, the new force will be

$F' = 4 F=4(6.67\cdot 10^{-4} N)=2.7\cdot 10^{-3} N$

3) 12.5 Nm

The torque is equal to the product between the magnitude of the perpendicular force and the distance between the point of application of the force and the centre of rotation:

$\tau=Fd$

Where, in this case:

F = 25 N is the perpendicular force

d = 0.5 m is the distance between the force and the center

By using the equation, we find

$\tau=(25 N)(0.5 m)=12.5 Nm$

4) 0.049 kg m^2/s

The relationship between angular momentum (L), moment of inertia (I) and angular velocity ( $\omega$ ) is:

$L=I\omega$

In this problem, we have

$I=0.007875 kgm^2$

$\omega=6.28 rad/s$

So, the angular momentum is

$L=I\omega=(0.007875 kgm^2)(6.28 rad/s)=0.049 kg m^2/s$

6 0

3 years ago