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2 years ago

12

A capacitor is charged to a potential difference of 3 volt it delivers 30% store energy to lamp what is the final potential diff

erence across the capacitor

1 answer:

nexus9112 [7]2 years ago

4 0

Answer:

3.98V

Explanation:

Given

Pontential difference V as 3v

Energy delivered is 30%,

Recall that Enery E=1/2cv^2 from this E=V^2(since Current C is not provided we can assume a value 2)

So E=V^2

E=3^2=9

At full charge E=9,30%of 9,0.3*9=2.7 energy in capacitor is 9-2.7=6.3

But E=V^2

✓E=V

✓6.3=3.98V

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What is the velocity of a ball dropped from a height of 150 m when it hits the ground? Take the upward direction as positive.

djyliett [7]

Answer:

The velocity of the ball when its hit the ground will be 54.22 m/sec

Explanation:

We have given height from which ball is dropped h = 150 m

Acceleration due to gravity $g=9.8m/sec^2$

As the ball is dropped so initial velocity will be zero so u = 0 m/sec

According to third equation of motion we know that $v^2=u^2+2gh$

$v^2=0^2+2\times 9.8\times 150$

$v=54.22m/sec$

So the velocity of the ball when its hit the ground will be 54.22 m/sec

7 0

3 years ago

A car is traveling at 15m/s on a horizontal road. the brakes are applied and the car skids to a stop in 4.0s . the coefficient o

iren2701 [21]

Answer:

the coefficient of Kinetic friction between the tires and road is 0.38

Option A) .38 is the correct answer

Explanation:

Given that;

final velocity v = 0

initial velocity u = 15m/s

time taken t = 4 s

acceleration a = ?

from the equation of motion

v = u + at

we substitute

0 = 15 + a × 4

acceleration a = -15/4 = - 3.75 m/s²

the negative sign tells us that its a deacceleration so the sign can be ignored.

Deacceleration due to friction a = μ × g

we substitute

3.75 = μ × 9.8

μ = 3.75 / 9.8 = 0.3826 ≈ 0.38

Therefore the coefficient of Kinetic friction between the tires and road is 0.38

Option A) .38 is the correct answer

8 0

2 years ago

The route followed by a hiker consists of three displacement vectors, X, Y and Z. Vector X is along a measured trail and is 1430

poizon [28]

Answer:

magnitude : 1635.43 m
Angle: 130°28'20'' north of east

Explanation:

First, we will find the Cartesian Representation of the $\vec{X}$ and $\vec{Y}$ vectors. We can do this, using the formula

$\vec{A}= | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )$

where $| \vec{A} |$ its the magnitude of the vector and θ the angle. For $\vec{X}$ we have:

$\vec{X}= 1430 m \ ( \ cos( 42 \°) \ , \ sin (42 \°) \ )$

$\vec{X}= ( \ 1062.70 m \ , \ 956.86 m \ )$

where the unit vector $\hat{i}$ points east, and $\hat{j}$ points north. Now, the $\vec{Y}$ will be:

$\vec{Y}= - 2200 m \hat{j} = ( \ 0 \ , \ - 2200 m \ )$

Now, taking the sum:

$\vec{X} + \vec{Y} + \vec{Z} = 0$

This is

$\vec{Z} = - \vec{X} - \vec{Y}$

$(Z_x , Z_y) = - ( \ 1062.70 m \ , \ 956.86 m \ ) - ( \ 0 \ , \ - 2200 m \ )$

$(Z_x , Z_y) = ( \ - 1062.70 m \ , \ 2200 m \ - \ 956.86 m \ )$

$(Z_x , Z_y) = ( \ - 1062.70 m \ , \ 1243.14 m\ )$

Now, for the magnitude, we just have to take its length:

$|\vec{Z}| = \sqrt{Z_x^2 + Z_y^2}$

$|\vec{Z}| = \sqrt{(- 1062.70 m)^2 + (1243.14 m)^2}$

$|\vec{Z}| = 1635.43 m$

For its angle, as the vector lays in the second quadrant, we can use:

$\theta = 180\° - arctan(\frac{1243.14 m}{ - 1062.70 m})$

$\theta = 180\° - arctan( -1.1720)$

$\theta = 180\° - 45\°31'40''$

$\theta = 130\°28'20''$

5 0

3 years ago

Please help on this one?

ruslelena [56]

Answer:

A, 30V

Explanation:

First combine all resistors to an equivalent resistor. Since they are in series, the equivalent resistance is the sum of all resistor

Req = 20 + 40 + 60 = 120Ω

Using Ohm's law, find the current in the circuit

V = I * R

I = V / R

I = 60V / 120Ω

I = 0.5 A

Now the potential drop across the resistor R3 is the current times R3 resistance, therefore:

Vdrop = 0.5A * 60Ω = 30V

So the potential drop across resistor R3 is 30 V

3 0

3 years ago

Read 2 more answers

A pulse can be described as a single wave disturbance that moves through a medium. Consider a pulse that is defined at time t =

Anastaziya [24]

Answer:

a)A= 3 m

b) $y(x)=\dfrac{6}{(x-3t)^2+2}\ m$

c)D= 15 m

Explanation:

Given that

$y(x)=\dfrac{6}{x^2+2}\ m$

v= 3 m/s

a)

The amplitude(A) of the pulse :

When x= 0 ,Then y = A

Put x= 0

$y(x)=\dfrac{6}{x^2+2}\ m$

$y(0)=\dfrac{6}{0^2+2}\ m$

y= A= 3 m

A= 3 m

b)

Distance travel in time t

x= vt

x= 3 t

$y(x)=\dfrac{6}{(x-3t)^2+2}\ m$

c)

The distance covered by pulse in the time 5 s

D = v t

D= 3 x 5

D= 15 m

7 0

3 years ago