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2 years ago

Light waves reflect off the objects that you see. Where do those light waves originate?

Physics

1 answer:

xxMikexx [17]2 years ago

3 0

Answer:

The Sun is a natural source for visible light waves and our eyes see the reflection of this sunlight off the objects around us.

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Two thermometers are calibrated, one in de

zlopas [31]

-40 c = -40 f but k would be 233.15

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3 years ago

Coach Hogue is riding his motorcycle in a circle on wet pavement. Suddenly the bike slides out from under him. What failed to pr

Arisa [49]

Answer:

The correct option is;

Force of Friction

Explanation:

As coach Hogue rode his motorcycle round in circle on the wet pavement, the motorcycle and the coach system tends to move in a straight path but due to intervention by the coach they maintain the circular path

The motion equation is

v = ωr and we have the centripetal acceleration given by

α = ω²r and therefore centripetal force is then

m×α = m × ω²r = m × v²/r

The force required to keep the coach and the motorcycle system in their circular path can be obtained by the impressed force of friction acting towards the center of the circular motion.

5 0

3 years ago

A 1.90-m-long barbell has a 25.0 kg weight on its left end and a 37.0 kg weight on its right end. if you ignore the weight of th

gregori [183]

Answer: The center of gravity is 1.1338 m away from the left side of the barbell

Explanation:

Length of the barbell = 1.90 m

The distance center of gravity from left = x

Mass on the left side = 25 kg

The distance center of gravity from right = 1.90 - x

Mass on the right side = 37 kg

At the balance point: $m_1x_1=m_2x_2$

$25 kg\times x=37 kg\times (1.90-x)$

$x=1.1338 m$

The center of gravity is 1.1338 m away from the left side of the barbell

7 0

3 years ago

Read 2 more answers

If Ike notices that there is a new moon tonight, when should he expect there to be a new moon again?

podryga [215]

Whatever phase of the moon Ike sees, he can expect to see
the same phase of moon again, after 29.53 days later.

5 0

3 years ago

Under ideal conditions (no atmospheric interference of any kind), if I hit a golf ball at an angle of 25 degrees at an initial s

g100num [7]

Answer:

The required angle is (90-25)° = 65°

Explanation:

The given motion is an example of projectile motion.

Let 'v' be the initial velocity and '∅' be the angle of projection.

Let 't' be the time taken for complete motion.

Let 'g' be the acceleration due to gravity

Taking components of velocity in horizontal(x) and vertical(y) direction.

$v_{x}$ = v cos(∅)

$v_{y}$ = v sin(∅)

We know that for a projectile motion,

t = $\frac{2vsin(∅)}{g}$

Since there is no force acting on the golf ball in horizonal direction.

Total distance(d) covered in horizontal direction is -

d = $v_{x}$ ×t = vcos(∅)× $\frac{2vsin(∅)}{g}$ = $\frac{v^{2}sin(2∅) }{g}$ .

If the golf ball has to travel the same distance 'd' for same initital velocity v = 23m/s , then the above equation should have 2 solutions of initial angle 'α' and 'β' such that -

α +β = 90° as-

d = $\frac{v^{2}sin(2α) }{g}$ = $\frac{v^{2}sin(2[90-β]) }{g}$ = $\frac{v^{2}sin(180-2β) }{g}$ = $\frac{v^{2}sin(2β) }{g}$ .

∴ For the initial angles 'α' or 'β' , total horizontal distance 'd' travelled remains the same.

∴ If α = 25° , then

β = 90-25 = 65°

∴ The required angle is 65°.

5 0

3 years ago

Read 2 more answers