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3 years ago

Two precautions in images of a convex lens

Physics

2 answers:

yan [13]3 years ago

5 0

Answer:

1 . rays should passs through correct center or points(i.e. optic center or focus)

2.your line representing rays should be straight and u should tell the nature of image

Explanation:

JulsSmile [24]3 years ago

5 0

Your lines resenting should be straight

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A flat coil having 40 turns, each one of cross-sectional area 12.0 cm^2, is oriented with its plane perpendicular to a uniform m

iragen [17]

what means emf ? what it means

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3 years ago

Describe how one plays Dr.Dogeball

JulsSmile [24]

•To play Dr. Dodgeball you need to have 2 teams to verse each other.
•Next, select one person from each team to be the doctor (depending on the size of the teams you can have varying amounts of doctors)
•Continue to play dodgeball how you normally would
•When a player gets hit and is “out” they have to sit on the ground and wait for the doctor to “revive them” (this usually requires the doctor dragging,touching, or moving the player that is out to a “revival place” which is usually decided on by the advisor or person in charge.
•Finally, try to get all the doctors and players out from the other team. Get the doctors first, for they cannot revive themselves. Which means the other players are out after they get hit with a ball since the doctors are out. (Some games are played where if all doctors are out the game ends)
Hope this helped! Play on! And plz mark brainliest lol this was long to write :D

4 0

3 years ago

A damped mass/spring system takes 14.0 s for its amplitude of the oscillator to decrease by a factor of 9. By what factor does t

fiasKO [112]

Answer:

The correct answer is "0.246".

Explanation:

Given that the amplitude is decreased by a factor of 9, then

$A \rightarrow (A-\frac{A}{9} )$

$A \rightarrow \frac{8A}{9}$

As we know,

Energy will be:

⇒ $E_{1}=\frac{1}{2}KA^2$

and,

⇒ $E_{2}=\frac{1}{2}K(\frac{8A}{9} )^2$

$=\frac{64KA^2}{162}$

⇒ $\Delta E=E_1-E_2$

On putting the estimated values, we get

$=\frac{1}{2}KA^2-\frac{64KA^2}{162}$

⇒ $\frac{\Delta E}{E}=\frac{\frac{20}{162}KA^2}{\frac{1}{2}KA^2}$

$=\frac{40}{162}$

$=0.246$

3 0

3 years ago

Car A has a mass of 1,200 kg and is traveling at a rate of 22 km/hr. It collides with car B. Car B has a mass of 1,900 kg and is

anastassius [24]

The car A has a mass of 1200 kg.

The car B has the mass of 1900 kg.

It is given that velocity of car A is given as 22 Km/hr

The car B has the velocity of 25 Km/hr.

Let the mass of two bodies are denoted as $m_{1} \ and\ m_{2}$

Let the velocity of cars A and B are denoted as $v_{1} \ and\ v_{2}$

The momentum before collision is-

$p_{i} =m_{1} v_{1} +m_{2} v_{2}$

[Here p stand for momentum.]

We are asked to calculate the final momentum of the system after collision.

The answer of the question is based law of conservation of linear momentum.

As per law of conservation of linear momentum the sum total linear momentum for an isolated system is always constant.Hence irrespective of the type of collision[elastic and inelastic],the momentum of the system is always constant which is a universal truth.

Let after the collision the velocity of A and B are $v'_{1} \ and\ v'_{2}$