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3 years ago

What is a concave lens?

Physics

1 answer:

ozzi3 years ago

5 0

A concave lens is a lens that has at least one of its surfaces or both surfaces curved inwards. Due to this reason, this lens diverges the light that falls on it and hence is also called a diverging lens. The concave lens is thinner in the middle compared to its edges. These are used in flashlights, binoculars, telescopes, etc.
Please see attached image for reference.

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1.Explain frame of reference. How is it different when you are riding inside a car or standing by the highway?

galben [10]

Answer:

See explanation

Explanation:

A frame of reference simply refers to a set of coordinates which is used to determine the position and velocity of objects all the objects found in that particular frame. The person inside the car and standing by the highway are in different frames of reference.

Newtons first law of motion states that an object will continue to be at rest or in a state of uniform velocity until it is acted upon by an external force.

Newtons second law states that rate of change of momentum is directly proportional to the impressed force.

Newtons third law states that action and reaction are equal and opposite.

The action of applying brakes leads to an equal reaction of a sudden forward movement. Seat belts help to prevent the occupants of a car from being injured by the sudden stoppage of the car.

Force is simply defined as a push or pull.

Balanced forces do not lead to acceleration of a body but an unbalanced for leads to acceleration of a body.

strong nuclear force, electromagnetic force, weak nuclear force and gravity

Power is the rate of doing work while week is said to be done when the force applied moves a distance in the direction of the force. Power is defined as work/time.

8 0

3 years ago

What is your REASONING for this CLAIM? EXPLAIN WHY your claim is true.

jok3333 [9.3K]

I DONT know FiGURE it out YOURSELF

7 0

3 years ago

I need hellp with thisss plssss????????!!!!

Leto [7]

Answer:

The graph appears to be in error.

The actual figure appears to be a rhombus with sides of 5 and 15 with a height of 5

The work done (F * S) is the area of the rhombus

1/2 * (5 +15) * 5 = 50 J

8 0

2 years ago

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu

vlada-n [284]

Answer:

$\frac{1}{10}$ M

Explanation:

To apply the concept of angular momentum conservation, there should be no external torque before and after

As the asteroid is travelling directly towards the center of the Earth, after impact ,it does not impose any torque on earth's rotation, So angular momentum of earth is conserved

⇒ $I_{1} \times W_{1} =I_{2} \times W_{2}$

$I_{1}$ is the moment of interia of earth before impact
$W_{1}$ is the angular velocity of earth about an axis passing through the center of earth before impact
$I_{2}$ is moment of interia of earth and asteroid system
$W_{2}$ is the angular velocity of earth and asteroid system about the same axis

let $W_{1}=W$

since $\text{Time period of rotation}∝$ $\frac{1}{\text{Angular velocity}}$

⇒ if time period is to increase by 25%, which is $\frac{5}{4}$ times, the angular velocity decreases 25% which is $\frac{4}{5}$ times

therefore $W_{1}$ $=$ $\frac{4}{5} \times W_{1}$

$I_{1}=\frac{2}{5} \times M\times R^{2}$ (moment of inertia of solid sphere)

where M is mass of earth

R is radius of earth

$I_{2}=\frac{2}{5} \times M\times R^{2}+M_{1}\times R^{2}$

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)

where $M_{1}$ is mass of asteroid

⇒ $\frac{2}{5} \times M\times R^{2} \times W_{1}=}(\frac{2}{5} \times M\times R^{2}$ $+$ $M_{1}\times R^{2})\times(\frac{4}{5} \times W_{1})$

$\frac{1}{2} \times M\times R^{2}$ = $(\frac{2}{5} \times M\times R^{2}$ + $M_{1}\times R^{2})$

$M_{1}\times R^{2}=$ $\frac{1}{10} \times M\times R^{2}$

⇒ $M_{1}=}$ $\frac{1}{10} \times M$

3 0

3 years ago

A 4kg ball moving at 8m\sec

Llana [10]

Conservation of momentum: total momentum before = total momentum after

Momentum = mass x velocity

So before the collision:
4kg x 8m/s = 32
1kg x 0m/s = 0
32+0=32

Therefore after the collision
4kg x 4.8m/s = 19.2
1kg x βm/s = β
19.2 + β = 32

Therefore β = 12.8 m/s

8 0

3 years ago