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3 years ago

Describe the image with a object distance of 65cm and a focal length of 20cm

Physics

1 answer:

sukhopar [10]3 years ago

4 0

Answer:

Imag distance = 28.9 cm

Explanation:

Given that,

Object distance, u = -65 cm

The focal length of the convex les, f = +20 cm

We need to find the image distance. The lens formula gives the relation between focal length, object distance and image distance is given by :

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{20}+\dfrac{1}{-65}\\\\v=28.9\ cm$

So, the image distance is 28.9 cm

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An airplane flying at an altitude of 6 miles passes directly over a radar antenna. When the airplane is 10 miles away (s = 10),

Novosadov [1.4K]

Answer:

Explanation:

Given

altitude of the Plane $h=6\ miles$

When Airplane is $s=10\ miles$ away

Distance is changing at the rate of $\frac{\mathrm{d} s}{\mathrm{d} t}=290\ mph$

From diagram we can write as

$h^2+x^2=s^2$

differentiate above equation w.r.t time

$2h\frac{\mathrm{d} h}{\mathrm{d} t}+2x\frac{\mathrm{d} x}{\mathrm{d} t}=2s\frac{\mathrm{d} s}{\mathrm{d} t}$

as altitude is not changing therefore $\frac{\mathrm{d} h}{\mathrm{d} t}=0$

$0+x\frac{\mathrm{d} x}{\mathrm{d} t}=s\frac{\mathrm{d} s}{\mathrm{d} t}$

at $s=10\ miles\ and\ h=6\ miles$

substitute the value we get $x=\sqrt{10^2-6^2}=8\ miles$

$8\times \frac{\mathrm{d} x}{\mathrm{d} t}=10\times 290$

$\frac{\mathrm{d} x}{\mathrm{d} t}=362.5\ mph$

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4 years ago

The function of a capacitor in an electric circuit is to

jok3333 [9.3K]

Answer: B. store electric charges.

Explanation: I JUST TOOK THE PF EXAM AND I GOT IT CORRECT!!!!

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4 years ago

A 2.5-kg ball and a 5.0-kg ball have an elastic collision. Before the collision, the 2.5-kg ball was at rest and the other ball

lesantik [10]

The kinetic energy of 2.5 kg ball after collision is 27.09 J.

Answer:

Explanation:

In elastic collision, the sum of momentum of the objects before collision will be equal to the sum of momentum of the objects after collision.

We know that momentum is the product of mass and velocity acting on any object.

So, the conservation of energy in elastic collision leads to following equation:

$M_{1} u_{1} +M_{2} u_{2}=M_{1} v_{1}+M_{2} v_{2}$

Since, the momentum is conserved ,the kinetic energy will also be conserved in elastic collision. So

$M_{1} u_{1} ^{2}+M_{2} u_{2} ^{2}=M_{1}v_{1} ^{2}+ M_{2}v_{2} ^{2}$

Since initial velocity for M1 ball is zero, then

$M_{2} u_{2}=M_{1} v_{1}+M_{2} v_{2}$

and

$M_{2} u_{2} ^{2}=M_{1}v_{1} ^{2}+ M_{2}v_{2} ^{2}$

So, on solving all the above equation, we get an equation for velocity and that is

$\frac{2M_{2}u_{2} }{(M_{1}+M_{2} }$ =final velocity of ball with mass 2.5 kg

$v = \frac{2(5*3.5)}{2.5+5}=4.67 m/s$

So kinetic energy will be 1/2 mv2

Kinetic energy of 2.5 kg ball is $\frac{1}{2}*2.5*(4.67)^{2} =27.09 J$

So the kinetic energy of 2.5 kg ball after collision is 27.09 J.

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4 years ago

What is the number of waves that pass a fixed ping in a given amount of time?

Viktor [21]

The number of waves that pass a fixed point in a given amount of time is wave frequency. Wave frequency can be measured by counting the number of crests (high points) of waves that pass the fixed point in 1 second or some other time period. The higher the number is, the greater the frequency of the waves. :)

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3 years ago

If a book has a a mass of 2 kg, how much does the book weigh?

Anestetic [448]

Answer:

A. 20 N

Explanation:

the weight of book is 2×10= 20N

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3 years ago