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babymother [125]

3 years ago

14

The plates on a vacuum capacitor have a radius of 3.0 mm and are separated by a distance of 1.5 mm. What is the capacitance of t

his capacitor?

1 answer:

Mkey [24]3 years ago

6 0

Link provided in other answer is a SCAM, don’t click on the link !!!!

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An airplane with a mass of 5,000 kg needs to accelerate 5 m/s2 to take off before it reaches the end of the runway. How much for

Anastasy [175]

Answer:

<h2>25000 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 5000 × 5

We have the final answer as

<h3>25000 N</h3>

Hope this helps you

6 0

3 years ago

what do you think is the purpose of a periscope? Are there any other uses for periscopes besides for submarines?

spin [16.1K]

Military personnel also use periscopes in some gun turrets and in armoured vehicles. More complex periscopes using prisms or advanced fibre optics instead of mirrors and providing magnification operate on submarines and in various fields of science

3 0

3 years ago

What is the slope of (0,0)(4,200)

Kruka [31]

Answer:

.02

Explanation:

X-x/ Y-y= 4-0/200-0 =.02

5 0

3 years ago

Read 2 more answers

A 0.15 g honeybee acquires a charge of 22 pC while flying. The electric field near the surface of the earth is typically 100 N/C

Rus_ich [418]

Answer:

$1.50\ *10^{-6} }$

Explanation:

Given

e=100 N/C

M=0.15 g

$q=\ 22\ pC\\=\ 22\ *10^{-2}$

The ratio of the electric force on the bee to the bee's weight can be determined by the following formula

$\frac{fe}{M*9.81}$

$\frac{22*10^{-12\ *\ 100} }{0.15*\ 10^{-3} *\ 9.81}$

$=\ 1.50\ *10^{-6}$

4 0

4 years ago

A cannon with a muzzle speed of 1 000 m/s is used to start an avalanche on a mountain slope. The target is 2 000 m from the cann

Nataliya [291]

Answer:

∅ = 89.44°

Explanation:

In situations like this air resistance are usually been neglected thereby making g= 9.81 m/ $s^{2}$

Bring out the given parameters from the question:

Initial Velocity ( $V_{1}$ ) = 1000 m/s

Target distance (d) = 2000 m

Target height (h) = 800 m

Projection angle ∅ = ?

Horizontal distance = $V_{1x}$ tcos ∅ .......................... Equation 1

where $V_{1x}$ = velocity in the X - direction

t = Time taken

Vertical Distance = y = $V_{1y}$ t - $\frac{1}{2}$ g $t^{2}$ ................... Equation 2

Where $V_{1y}$ = Velocity in the Y- direction

t = Time taken

$V_{1y}$ = $V_{1}$ sin∅

Making time (t) subject of the formula in Equation 1

t = d/( $V_{1x}$ cos ∅)

t = $\frac{2000}{1000coso}$ = $\frac{2}{cos0}$ = $\frac{d}{cos o}$ ...................Equation 3

substituting equation 3 into equation 2

Vertical Distance = d = $V_{1y}$ $\frac{d}{cos o}$ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Vertical Distance = h = sin∅ $\frac{d}{cos o}$ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Vertical Distance = h = dtan∅ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Applying geometry

$\frac{1}{cos o}$ = $tan^{2} o$ + 1

Vertical Distance = h = d tan∅ - 2 g ( $tan^{2} o$ + 1)

substituting the given parameters

800 = 2000 tan ∅ - 2 (9.81)( $tan^{2} o$ + 1)

800 = 2000 tan ∅ - 19.6( $tan^{2} o$ + 1) Equation 4

Replacing tan ∅ = Q .....................Equation 5

In order to get a quadratic equation that can be easily solve.

800 = 2000 Q - 19.6 $Q^{2}$ + 19.6

Rearranging 19.6 $Q^{2}$ - 2000 Q + 780.4 = 0

$Q_{1}$ = 101.6291

$Q_{2}$ = 0.411

Inserting the value of Q Into Equation 5

tan ∅ = 101.63 or tan ∅ = 0.4114

Taking the Tan inverse of each value of Q

∅ = 89.44° ∅ = 22.37°

4 0

3 years ago