Please need help on this I know it’s not D an atom.

Design a voltage divider to provide the following approximate voltages with respect to ground using a 30 V source: 8.18 V, 14.7

yarga [219]

Answer:

R₁ = 14.7 10³ Ω , R₂ = 8.18 10³ Ω , R₃ = 1.72 10³ Ω , R₄ = 5.4 10³ Ω 1/8 W resistor

Explanation:

For this exercise we must use a series circuit since the sum of the voltage on each resin is equal to the source voltage (V = 30 V)

Therefore we build a circuit with 4 resistors in series, in such a way that

V = i R

let the voltage

1st resistance

V = i R

R₁ = V / i

R₁ = 14.7 / 1 10⁻³

R₁ = 14.7 10³ Ω

power is

P = V i

P = 14.7 1 10⁻³

P = 14.7 10⁻³ W = 0.0147 W

a resistance of ⅛ W is indicated

2nd resistance

R₂ = 8.18 / 1 10⁻³

R₂ = 8.18 10³ Ω

Power

P = 8.18 1 10⁻³

P = 0.00818W

a 1/8 W resistor

3rd resistance

this resistance is calculated in such a way that

V₁ + V₂ + V₃ = 24.6

V₃ = 24.6 - V₁ -V₂

V₃ = 24.6 - 14.7 - 8.18

V₃ = 1.72 V

R₃ = 1.72 / 1 10⁻³

R₃ = 1.72 10³ Ω

power

P = Vi

P = 1.72 10⁻³

P = 0.00172 W

a resistance of ⅛ W

To obtain the voltage of 24.6 we use this three resistors together

4th resistance

The value of this resistance is calculated so that the sum of all the voltages reaches the source voltage

30 = V₁ + V₂ + V₃ + V₄

V₄ = 30 - V₁ -V₂ -V₃

V₄ = 30 -14.7 - 8.18 - 1.72

V₄ = 5.4 V

R₄ = 5.4 / 1 10⁻³

R₄ = 5.4 10³ Ω

Power

P = V i

P = 5.4 10⁻³

P = 0.0054 W

⅛ W resistance

The values of these resistance are commercially

Let's check the consumption of the circuit

R_total = R₁ + R₂ + R₃ + R₄

R_total = (14.7 + 8.18 + 1.72 + 5.4) 10³

R_total = 30 10³

the current circulating in the circuit is

i = V / R_total

i = 30/30 10³

i = 1 10⁻³ A

therefore it is within the order requirement.

for connections see attached diagram

8 0

3 years ago

A buoy is a floating device that can have many purposes, but often as a locator for ships. Collin constructs a hollow metal buoy

Yuri [45]

Answer:

A) since the density of the buoy is half the density of sea water when the buoy is at rest on an ocean half of the buoy will be submerged in water

B )angular frequency ( w ) = $\sqrt{\frac{3g}{h} }$

c ) h = 1.34 m ( 4 $\pi ^{2}$ / 3g )

Explanation:

A) since the density of the buoy is half the density of sea water when the buoy is at rest on an ocean half of the buoy will be submerged in water this is because the substances with lesser density floats when placed in a substance with a higher density

B ) when the Buoy is at rest ( t = 0 ) and is then pushed down calculate angular frequency ( w ) of small oscillations in terms of the given variables

volume of cone = hA /3.

h = height of cone, A = Base Area.

therefore the total volume of the Buoy above water level ( at rest )

= ( $\pi r^{2} * \frac{h}{3}$ ) * 2 = $\frac{\pi r^{2} h }{6}$

The part of the buoy immersed in water = x

then the net upward force will be

fb ( force of Buoy ) = fg ( force of gravity )

note force = Mass * Acceleration

force of buoy = $\frac{\alpha }{2} *$ $\frac{\pi r^{2} }{h}$ * a = ( force of gravity ) $\alpha * T\frac{r^{2} }{4} * x * g$

therefore a = $\frac{3g}{h} * x$

angular frequency ( w ) = $\sqrt{\frac{3g}{h} }$

C ) height of the each cone

$\frac{2\pi }{w} = 1$ therefore w = 2 $\pi$

back to the angular frequency : $\frac{3g}{h} = 4\pi ^{2}$

therefore h = 1.34 m ( 4 $\pi ^{2}$ / 3g )

4 0

3 years ago

A double-slit experiment is set up using red light (λ = 706 nm). A first order bright fringe is seen at a given location on a sc

Elanso [62]

Answer:

λ = 470.66 nm

Explanation:

for bright fringe $y_m = \frac{m\lambda D}{d}$

D= distance between slit and screen

d= distance between the slits

for first order bright fringe m = 1,

$y_1 = \frac{1\lambda D}{d}$

$y_1 = {706*D}{d}$

for dark fringe,we have

$y_m = {(m + 1/2)\lambda D}{d}$

Now to get the dark fringes at the same location we should have;

(706)D/d = (m + 1/2)λD/d

put m = 1

(1 + 1/2)λ = (706)

λ = 470.66 nm

6 0

3 years ago

If a car is moving to the left with constantvelocity, one can conclude thatthere mustbe no forces applied to the car.the netforc

Allisa [31]

Answer:

the net force applied to the car is zero.

Explanation:

According to Newton's second law, the acceleration of an object (a) is directly proportional to the net force applied (F):

$a=\frac{F}{m}$

where m is the object's mass.

In this problem, the car is moving with constant velocity: this means that the acceleration is zero, a = 0. Therefore, according to the previous equation, the net force must also be zero: F = 0. So, the correct answer is

the net force applied to the car is zero.

3 0

3 years ago

A 65.0 kg diver is 4.90 m above the water, falling at speed of 6.40 m/s. Calculate her kinetic energy as she hits the water. (Ne

mojhsa [17]

Answer:

4452.5 J.

Explanation:

The diver have both kinetic and potential energy.

Ek = 1/2mv² ................. Equation 1

Where Ek = Kinetic Energy of the diver, m = mass of the diver, v = velocity of the diver.

Given: m = 65 kg, v = 6.4 m/s.

Substitute into equation 1

Ek = 1/2(65)(6.4²)

Ek = 1331.2 J.

Also,

Ep = mgh ............................ Equation 2

Where Ep = Potential energy of the diver when its above the water, h = height of the diver above the water, g = acceleration due to gravity.

Given: m = 65 kg, h = 4.9 m, g = 9.8 m/s²

Substitute into equation 2.

Ep = 65(4.9)(9.8)

Ep = 3121.3 J.

Note: When she hits the water, the potential energy is converted to kinetic energy.

E = Ek+Ep

Where E = Kinetic energy of the diver when she hits the water.

E = 1331.2+3121.3

E = 4452.5 J.

3 0

3 years ago