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3 years ago

An object 5.Ocm in the length is placed at a distance of 20cm in front of convex mirror

Physics

1 answer:

andrew-mc [135]3 years ago

4 0

Answer:

Position = $\frac{60}{7}\ cm$ behind the mirror

Nature = Virtual and Erect

Size = $\frac{15}{7}\ cm$ : Diminished

Explanation:

Sign convention-Distance measured to the left of pole is negative and to the right of pole is positive.

Object distance = u = -20 cm

Focal length = f = Radius of curvature/2 = 30/2 = 15 cm

We have to use mirror formula to find image distance.

$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}\\ \frac{1}{-20}+\frac{1}{v}=\frac{1}{15}\\ \frac{1}{v}=\frac{7}{60}\\v=\frac{60}{7}\ cm$

Since the image distance is positive, it is formed behind the mirror or a virtual image is formed.

Magnification = $=\frac{h_{image}}{h_{object}}=$ $-\frac{v}{u}=\frac{60}{7\times20}=\frac{3}{7}$

Height of the object = 5 cm

Height of the image = $5\times\frac{3}{7}=\frac{15}{7}\ cm$

Since the height of the image is positive and less than the size of object,it is erect and diminished.

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Two planets have masses 2 x 10^23 kg and 5 x 10^22 kg, and the distance between them is 3 x 10^16 m. Calculate the gravitational

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Explanation:

given,

mass of one planet (m1)=2*10^23 kg

mass of another planet (m2)=5*10^22kg

distance between them(d)=3*10^16m

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gravitational force between them(F)=?

we know,

F=Gm1m2/d^2

or, F=6.67*10^-11*2*10^23*5*10^22/(3*10^16)^2

or, F=6.67*2*5*10^-11+23+22/3*3*10^32

or, F=66.7*10^34/9*10^32

or, F=7.41*10^34-32

•°• F=7.41*10^2

thus, the gravitational force between them is 7.14*10^2

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2 years ago

Two particles each of mass m and charge q are suspended by strings of length / from a common point. Find the angle e that each s

ozzi

Answer:

$\theta =\left (\frac{kq^{2}}{4L^{2}\times mg} \right )^{\frac{1}{3}}$

Explanation:

Let the length of the string is L.

Let T be the tension in the string.

Resolve the components of T.

As the charge q is in equilibrium.

T Sinθ = Fe ..... (1)

T Cosθ = mg .......(2)

Divide equation (1) by equation (2), we get

tan θ = Fe / mg

$tan\theta =\frac{\frac{kq^{2}}{AB^{2}}}{mg}$

$tan\theta =\frac{\frac{kq^{2}}{4L^{2}Sin^{\theta }}}}{mg}$

$tan\theta =\frac{kq^{2}}{4L^{2}Sin^{2}\theta \times mg}$

$tan\theta\times Sin^{2}\theta =\frac{kq^{2}}{4L^{2}\times mg}$

As θ is very small, so tanθ and Sinθ is equal to θ.

$\theta ^{3} =\frac{kq^{2}}{4L^{2}\times mg}$

$\theta =\left (\frac{kq^{2}}{4L^{2}\times mg} \right )^{\frac{1}{3}}$

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2 years ago

You are planning to make an open rectangular box from an 8-inch by 15-inch piece of cardboard by cutting congruent squares from

Phoenix [80]

Let us say that x is the cut that we will make on the sides to make a box, therefore the new dimensions are:

l = 15 – 2x

w = 8 – 2x

It is 2x since we cut on two sides.

We know that volume is:

V = l w x

V = (15 – 2x) (8 – 2x) x

V = 120x – 30x^2 – 16x^2 + 4x^3

V = 120x – 46x^2 + 4x^3

Taking the 1st derivative:

dV/dx = 120 – 92x + 12x^2

Set dV/dx = 0 to get maxima:

120 – 92x + 12x^2 = 0

Divide by 12:

x^2 – (92/12)x + 10 = 0

(x – (92/24))^2 = -10 + (92/24)^2

x - 92/24 = ±2.17

x = 1.66, 6

We cannot have x = 6 because that will make our w negative, so:

x = 1.66 inches

So the largest volume is:

V = 120x – 46x^2 + 4x^3

V = 120(1.66) – 46(1.66)^2 + 4(1.66)^3

V = 90.74 cubic inches

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3 years ago

Two 100 kg bumper cars are moving toward each other in opposite directions. Car A is moving at 8 m/s and Car Bat -10 m/s when th

tatiyna

Answer:

b) -10 m/s

Explanation:

In perfectly elastic head on collisions of identical masses, the velocities are exchanged with one another.

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2 years ago

All of the following equations are statements of the ideal gas law except

elena55 [62]

Answer:

a. P = nRTV

Explanation:

The question is incomplete. Here is the complete question.

"All of the following equations are statements of the ideal gas law except a. P = nRTV b. PV/T = nR c. P/n = RT/v d. R = PV/nT"

Ideal gas equation is an equation that describes the nature of an ideal gas. The molecule of an ideal gas moves at a particular velocity depending on the temperature. This gases collides with one another elastically. The collision that an ideal gas experience is a perfectly elastic collision.

The ideal gas equation is expressed as shown:

PV = nRT where:

P is the pressure of the gas

V is the volume

n is the number of moles

R is the ideal gas constant

T is the temperature.

Based on the formula given for an ideal gas, it can be inferred that the equation. P = nRTV is not a statement of an ideal gas equation.

The remaining option will results to an ideal gas equation if they are cross multipled.

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3 years ago