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3 years ago

Refer to the diagram to answer the question. Which part of the eye enables it to focus? A B C D E

Physics

1 answer:

Ilia_Sergeevich [38]3 years ago

8 0

Answer:

what are the A B C D E

Explanation:

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A ball is dropped from a height of 20m. If its velocity increases uniformly at the rate 10m/s2 with what velocity and after what

belka [17]

Answer: 2 seconds

Explanation:

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3 years ago

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According to the law of refraction, light passing from air into a piece of glass at an angle of 30 degrees will cause the light

pantera1 [17]

Answer:

bend toward the normal line

Explanation:

When light passes from a less dense to a more dense substance, (for example passing from air into water), the light is refracted (or bent) towards the normal. In your question the light is moving from rarer to denser medium

4 0

4 years ago

You are exploring a distant planet. When your spaceship is in a circular orbit at a distance of 630 km above the planet's surfac

Anarel [89]

Answer:

$R = 24.3 m$

Explanation:

As we know that the orbital speed is given as

$v = \sqrt{\frac{GM}{R + h}}$

here we know that

v = 5500 m/s

$R = 4.48 \times 10^6 m$

$h = 630 km$

now we have

$5500 = \sqrt{\frac{(6.6 \times 10^{-11})M}{4.48 \times 10^6 + 6.30\times 10^5}}$

$M = 2.34 \times 10^24 kg$

now acceleration due to gravity of planet is given as

$a = \frac{GM}{R^2}$

$a = \frac{(6.6 \times 10^{-11})(2.34 \times 10^{24})}{(4.48\times 10^6)^2}$

$a = 7.7 m/s^2$

now range of the projectile on the surface of planet is given as

$R = \frac{v^2 sin2\theta}{g}$

$R = \frac{14.6^2 sin(2\times 30.8)}{7.7}$

$R = 24.3 m$

3 0

4 years ago

A proton is going 2,000 m/s into a magnetic field of 300 T. How much force does it feel. The charge of a proton = 1.602 x 10^-19

NemiM [27]

This depends on the direction of the velocity vector to the magnetic field vector. The force is F=q(VxB) ("x" is the cross product.) The max force is when V and B are perpendicular. Then F=qVB = (1.602e-19)(2000)(300) = 9.612e-14 N

7 0

3 years ago

2. Three blocks, A,B and C of mass 2kg. 3kg. 5kg respectively kept side by side with one another are accelerated at 2m/s2 across

gulaghasi [49]

Answer:

Total mass of combination = 2+3+5 = 10kg.

Acceleration produced = 2m/s^2

hence force =( total mass × acceleration)= (2×10)= 20 N.

Net force on 3kg block = acceleration × mass = (2 × 2 )= 4 N

applied force on 2 kg block = 20N

Force between 2 kg and 3 kg block = (20-4) = 16N. ans

Net force on 3 kg block = 3 × 2 =6N.

Applied force on 3 kg block due to 2 kg block = 16N.

hence, force between 3 kg and 5 kg block = (16-6) = 10N .

answers:-

(a) 20 N

(b) 16N

4 0

1 year ago