A truck is traveling at a speed of 50 km/h. What happens when the truck slows down?

A 5.00-V battery charges the parallel plates in a capacitor, with a plate area of 865 mm2 and an air-filled separation of 3.00 m

Westkost [7]

Answer:

W = 3.21x10⁻¹¹ J

Explanation:

The work required to separate the plates can be calculated using the following equation:

$W = U_{2} - U_{1} = \frac{1}{2}(C_{2}V_{2}^{2} - C_{1}V_{1}^{2})$

Where:

U₂: is the final stored energy

U₁: is the initial stored energy

C₂: is the final capacitance

C₁: is the initial capacitance

V₁: is the initial potential difference = 5.00 V

V₂: is the final potential difference

The initial and final capacitance is:

$C_{1} = \epsilon_{0}*\frac{A}{d_{1}}$

Where:

ε₀: is the vacuum permittivity = 8.85x10⁻¹² C²/(N*m²)

d: is the initial distance = 3.00 mm = 3.00x10⁻³ m

A: is the plate area = 865 mm² = 8.65x10⁻⁴ m²

$C_{1} = \epsilon_{0}*\frac{A}{d_{1}} = 8.85 \cdot 10^{-12} C^{2}/(N*m^{2})*\frac{8.65 \cdot 10^{-4} m^{2}}{3.00 \cdot 10^{-3} m} = 2.55 \cdot 10^{-12} F$

Similarly, C₂ is:

$C_{2} = \epsilon_{0}*\frac{A}{d_{2}} = 8.85 \cdot 10^{-12} C^{2}/(N*m^{2})*\frac{8.65 \cdot 10^{-4} m^{2}}{3.00 + 3.00 \cdot 10^{-3} m} = 1.28 \cdot 10^{-12} F$

Now, V₂ can be calculated by finding the initial charge (q₁):

$q_{1} = C_{1}V_{1} = 2.55 \cdot 10^{-12} F*5.00 V = 1.28 \cdot 10^{-11} C$

Since, q₁ is equal to q₂, V₂ is:

$V_{2} = \frac{q_{2}}{C_{2}} = \frac{1.28 \cdot 10^{-11} C}{1.28 \cdot 10^{-12} F} = 10 V$

Finally, we can find the work:

$W = \frac{1}{2}(C_{2}V_{2}^{2} - C_{1}V_{1}^{2}) = \frac{1}{2}(1.28 \cdot 10^{-12} F*(10 V)^{2} - 2.55 \cdot 10^{-12} F(5.00 V)^{2}) = 3.21 \cdot 10^{-11} J$

Therefore, the work required to separate the plates is 3.21x10⁻¹¹ J.

I hope it helps you!

4 0

3 years ago

HELP!!! What role might electrostatic force play in spider dispersal, according to a recent study?

Marat540 [252]

The positively charged atmosphere attracts negatively charged spider silk, might electrostatic force play in spider dispersal, according to a recent study.

Answer: Option C

Explanation:

The positive charge present in upper of the atmosphere and the negative charge on planet’s surface. During cloudless skies days, the air possesses a voltage of nearly around 100 volts for each and every meter from above the ground.

Ballooning spiders process within this planetary electric field. When their silk relieve their bodies then it picks up a negative charge. This oppose the similar negative charges on the surfaces on which the spiders settles and create sufficient force to lift them into the air. And spiders can hike those forces by climbing onto blades of grass,twigs, or leaves.

6 0

3 years ago

How long will a trip take in hours of you travel 450kmat an average speed of 80 km/hr

777dan777 [17]

5.625 hours and it is 450 divided by 80
Have A Good Day

4 0

3 years ago

Find the minimum radius of a helium balloon that will lift her off the ground. the density of helium gas is 0.178 kg/m3, and the

marysya [2.9K]

$Volume of balloon = \frac{4}{3} \pi r^{3}$

Where r is the radius of balloon.

Here mass of woman = 68 kg

Mass of air displaced by a balloon with volume V = 1.29*V

Mass of helium inside balloon = 0.178*V

Total mass to be lifted by balloon = 68 +0.178*V

Buoyant force = 1.29V-0.178V=1.112V

So we have 1.112 V = 68+ 0.178*V

0.934 V = 68

V = 72.81 $m^3$

\frac{4}{3} \pi r^{3}[/tex]= 72.81

r = 2.59 m

So radius of helium balloon = 2.59 m

4 0

3 years ago

A flat sheet of paper of area 0.365 m2 is oriented so that the normal to the sheet is at an angle of 60 ∘ to a uniform electric

andriy [413]

Answer:

A.) 3.65 N*m²/C B) No C) 0º D) 90º

Explanation:

A) The electric flux, when the electric field is uniform across a gausssian surface, can be calculated as the dot product of the electric field vector, and the vector representing the area of the surface (normal to the surface and directed outward it by convention), as follows:

Flux = E*A*cos φ

where E = 20 N/C, A = 0.365 m², φ = 60º.

Replacing by the values, we can get the value of the electric flux, as follows:

Flux = 20 N/C* 0.365 m²*0.5 = 3.65 N*m²/C

B) While the area remains constant, and doesn't change orientation, the value of the flux will be the same, regardless the shape of the sheet.

C) When the normal to the sheet and the electric field are parallel each other, the surface will intercept the maximum number of field lines, i.e. the flux will be directly E*A*cos 0º = E*A (maximum value possible).

D) When the electric field is tangent to the surface, this means that no field lines will be intercepted by the sheet, so the flux is zero.

In this case, φ = 90º, cos φ = 0

⇒ E*A*cos 90º = E*A*0 = 0

5 0

3 years ago