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3 years ago

Which statement below best describes the relationship between the voltage, resistance, and current?

Physics

1 answer:

aliina [53]3 years ago

4 0

Answer:

As voltage increases, current increases and resistance stays the same .

Explanation:

Ohm's law gives the relationship between the voltage, resistance, and current. The mathematical form of Ohm's law is given by :

$V\propto I\\\\V=IR$

R is resistance

I is current

V is voltage

So, as voltage increases, current increases and resistance stays the same. The correct option is (A).

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Final velocity will be greater than initial velocity of an object is

Natali [406]

Answer:

accelerating

Explanation:

If we consider(v > u) Acceleration:

final velocity(v)= 14m/s

initial velocity(u)=10m/s

time taken(t)= 2 seconds

a= $\frac{(v-u)}{t} =\frac{(14-10)}{2}$ =2m/s²

If we consider (v<u) Deceleration:

final velocity(v)= 3m/s

initial velocity(u)=9m/s

time taken(t)=2 seconds

a= $\frac{(v-u)}{t}=\frac{(3-9)}{2}$ = -3m/s²

4 0

3 years ago

What does the magnitude of centripetal acceleration depend on ?

MariettaO [177]

Explanation:

Centripetal acceleration ac is the acceleration experienced while in uniform circular motion. It always points toward the center of rotation. It is perpendicular to the linear velocity v and has the magnitude ac=v2r;ac=rω2 a c = v 2 r ; a c = r ω 2 .

3 0

3 years ago

The intensity of the Sun's light in the vicinity of the Earth is about 1000W/m^2. Imagine a spacecraft with a mirrored square sa

NeX [460]

Explanation:

Formula to represent thrust is as follows.

F = $\frac{dP}{dt}$

= $\frac{2p}{dt}$

or, p = $\frac{E}{c}$

$\frac{p}{dt} = \frac{W}{c}$

F = $\frac{2IA}{c}$

= $\frac{2 (1000 W/m^{2})(5.5 \times 10^{3} m)^{2}}{3 \times 10^{8} m/s}$

= 201.67 N

Thus, we can conclude that the thrust is 201.67 N.

8 0

4 years ago

Two polarizers A and B are aligned so that their transmission axes are vertical and horizontal, respectively. A third polarizer

Reptile [31]

Answer:

Explanation:

Let the angle between the first polariser and the second polariser axis is θ.

By using of law of Malus

(a)

Let the intensity of light coming out from the first polariser is I'

$I' = I_{0}Cos^{2}\theta$ .... (1)

Now the angle between the transmission axis of the second and the third polariser is 90 - θ. Let the intensity of light coming out from the third polariser is I''.

By the law of Malus

$I'' = I'Cos^{2}\left ( 90-\theta \right )$

So,

$I'' = I_{0}Cos^{2}\theta Cos^{2}\left ( 90-\theta \right )$

$I'' = I_{0}Cos^{2}\theta Sin^{2}\theta$

$I'' = \frac{I_{0}}{4}Sin^{2}2\theta$

(b)

Now differentiate with respect to θ.

$I'' = \frac{I_{0}}{4}\times 2 \times 2 \times Sin2\theta \times Cos 2\theta$

$I'' = \frac{I_{0}}{2}\times Sin 4\theta$

7 0

3 years ago

A simple series circuit consists of a 150 Ω resistor, a 29 V battery, a switch, and a 2.1 pF parallel-plate capacitor (initially

Rufina [12.5K]

Find the electric flux and the disp at t=0.50ns
Given: 
Resistor R = 160 Ω 
Voltage ε = 22.0 V 
Capacitor C = 3.10 pF = 3.10 * 10^-12 F 
time t = 0.5 ns = 0.5 * 10^-9 s 
ε0 = 8.85 * 10^-12 
Solution: 
ELECTRIC FLUX: 
Φ = Q/ε0 
we have ε0, we need to find Q the charge 
STEP 1: FIND Q 
Q = C ε ( 1 - e^(-t/RC) ) 
Q = { 3.10 * 10^-12 } { 22.0 } { 1 - e^(- 0.5 * 10^-9 / 160 *3.10 * 10^-12 ) } 
Q = { 3.10 * 10^-12 } { 22.0 } { 1 - 0.365 } 
Q = { 3.10 * 10^-12 } { 22.0 } { 0.635 } 
Q = 43.31 * 10^-12 C 
STEP 2: WE HAVE Q AND ε0 > >>> SOLVE FOR ELECTRIC FLUX >>> 
Φ = Q/ε0 
Φ = { 43.31 * 10^-12 C } / { ε0 = 8.85 * 10^-12 } 
Φ = 4.8937 = 4.9 V.m 
DISPLACEMENT CURRENT 
we use the following equation: 
I = { ε / R } { e^(-t/RC) } 
I = { 22 / 160 } { e^(- 0.5 * 10^-9 / 160 *3.10 * 10^-12 ) } 
I = { 0.1375 } { 0.365 } 
I = 0.0502 A = 0.05 A

8 0

3 years ago