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3 years ago

A plane travelling at 63 m/s[S] down a runway begins accelerating uniformly at 2.8 m/s?[S]. How far does it travel

Physics

1 answer:

AVprozaik [17]3 years ago

6 0

Answer:

270 m

Explanation:

Given:

v₀ = 63 m/s

a = 2.8 m/s²

t = 4.0 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (63 m/s) (4.0 s) + ½ (2.8 m/s²) (4.0 s)²

Δx = 274.4 m

Rounded to two significant figures, the displacement is 270 meters.

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Read 2 more answers

The rectangular plates in a parallel-plate capacitor are 0.063 m x 5.4 m. A distance of 3.5 x 10^-5m separate the plates. The pl

Aleks04 [339]

The capacitance of the capacitor is $1.8\cdot 10^{-9}F$

Explanation:

The capacitance of a parallel-plate capacitor is given by the equation

$C=\frac{k\epsilon_0 A}{d}$

where

k is the dielectric constant of the medium

$\epsilon_0 = 8.85\cdot 10^{-12} F/m$ is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

For the capacitor in this problem, we have:

k = 2.1 is the dielectric constant

$d=3.5\cdot 10^{-5}m$ is the separation between the plates

$A=0.063m \cdot 0.054m = 0.0034 m^2$ (I assumed that 5.4 m is a typo, since it is not a realistic size for the side of the plate)

Therefore, the capacitance of the capacitor is

$C=\frac{(2.1)(8.85\cdot 10^{-12})(0.0034)}{3.5\cdot 10^{-5}}=1.8\cdot 10^{-9}F$

Learn more about capacitors:

brainly.com/question/10427437

brainly.com/question/8892837

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3 years ago

A single point charge is placed at the center of an imaginary cube that has 10 cm long edges. The electric flux out of one of th

12345 [234]

Answer:

Explanation:

<u>Given:</u>

Length of each side of the cube, $L=10\ cm$

The Elecric flux through one of the side of the cube is, $\phi =-1 kNm^2/C.$

The net flux through a closed surface is defined as the total charge that lie inside the closed surface divided by $\epsilon_0$

Since Flux is a scalar quantity. It can added to get total flux through the surface.

$\phi_{total}=\dfrac{Q_{in}}{\epsilon_0}\\6\times {-1}\times 10^3=\dfrac{Q_{in}}{\epsilon_0}\\\\Q_{in}=-6}\times 10^3\epsilon_0\\Q_{in}=-6}\times 10^3\times8.85\times 10^{-12}\\Q_{in}=-53.1\times10^{-9}\ C$

So the the charge at the centre is calculated.

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3 years ago