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3 years ago

What is the purpose of a rearview mirror in a car?

3 answers:

8 0

It is called rearVIEW mirror so

ans is The mirror allows the light to be reflected so the objects behind the car can be seen.

Shalnov [3]3 years ago

5 0

The third choice.

The driver wants to see the object that is behind him. The light reflects off the mirror into the eyes of the driver portraying the object behind him

ratt3 years ago

0 0

C: The mirror allows the light to be reflected so the objects behind the car can be seen.

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The courtroom role that possesses the most discretion is?

KonstantinChe [14]

The role of the Prosecutor's

5 0

3 years ago

Earth orbits 1 AU from the Sun, and the Oort cloud extends from about 10,000 to 100,000 AU from the Sun. If you represent Earth’

sertanlavr [38]

Explanation:

Earth’s orbit around the Sun represented as :

1 AU = 4 inches

Therefore,

Inner edge of the Oort Cloud represented as :

10,000 AU = 40,000 / 63360 = 0.631 miles

Outer edge of the Oort Cloud represented as:

100,000 AU = 400,000 / 63360 = 6.31 miles

4 0

3 years ago

dangina [55]

Answer:

C 2000v its obviously ans because if o is 1000 2 vo is 2000v

7 0

3 years ago

Unless indicated otherwise, assume the speed of sound in air to be v = 344 m/s. A pipe closed at both ends can have standing wav

uranmaximum [27]

Answer:

68.8 Hz

137.6 Hz, 206.4 Hz

Explanation:

L = Length of tube = 2.5 m

v = Velocity of sound in air = 344 m/s

Distance between nodes is given by

$L=\dfrac{\lambda}{2}+m\dfrac{\lambda}{2}\\\Rightarrow \dfrac{\lambda(n+1)}{2}=L\\\Rightarrow \lambda=\dfrac{2L}{n+1}$

Where n = 0, 1, 2, 3, ...

Making n+1 = n

$\lambda=\dfrac{2L}{n}$

where n = 1, 2, 3 .....

For fundamental frequency n = 1

$\lambda=\dfrac{2\times 2.5}{1}\\\Rightarrow \lambda=5\ m$

Frequency is given by

$f=\dfrac{v}{\lambda}\\\Rightarrow f=\dfrac{344}{5}\\\Rightarrow f=68.8\ Hz$

The fundamental frequency is 68.8 Hz

First overtone

$2f=2\times 68.8=137.6\ Hz$

Second overtone

$3f=3\times 68.8=206.4\ Hz$

The overtones are 137.6 Hz, 206.4 Hz

4 0

3 years ago

A simple series circuit consists of a 130 ? resistor, a 30.0V battery, a switch, and a 2.10 pF parallel-plate capacitor (initial

V125BC [204]

Answer with Explanation:

We are given that

Resistance,R=130 ohm

Potential difference, V=30 V

Capacitor,C=2.1pF= $2.1\times 10^{-12} F$

$1pF=10^{-12} F$

$d=5.0 mm=5\times 10^{-3} m$

$1mm=10^{-3} m$

a.Maximum flux

$\phi=EA=\frac{V}{d}\times \frac{Cd}\epsilon_0}=\frac{CV}{\epsilon_0}$

$\epsilon_0=8.85\times 10^{-12}$

$\phi=\frac{2.1\times 10^{-12}\times 30}{8.85\times 10^{-12}}=7.12Vm$

b.Maximum displacement current, $I=\frac{V}{R}=\frac{30}{130}=0.23 A$

c.We have to find electric flux at t=0.5 ns

$t=0.5ns=0.5\times 10^{-9} s$

$1ns=10^{-9}s$

$q=CV(1-e^{-\frac{t}{RC}})$

$q=30\times 2.1\times 10^{-12}(1-e^{-\frac{0.5\times 10^{-9}}{130\times 2.1\times 10^{-9}})$

$q=52.9\times 10^{-12} C$

$\phi=\frac{q}{\epsilon_0}=\frac{52.9\times 10^{-12}}{8.85\times 10^{-12}}=5.98Vm$

d.Displacement current at t=0.5ns

$I=(\frac{V}{R})e^{-\frac{t}{RC}}=\frac{30}{130}e^{-\frac{0.5\times 10^{-9}}{130\times 2.1\times 10^{-12}}}$

$I=0.037 A$

5 0

3 years ago