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2 years ago

What is The splitting of white light into its component colours is termed as?

Physics

1 answer:

aliina [53]2 years ago

4 0

Answer:

dispersion

As per your query the splitting of white or visible light into its component colours is known as dispersion. The splitting of white light when it passes through a glass prism into its constituent spectrum of colours is hereby known as dispersion.

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2 years ago

Imagine a large block that sinks in a tub of water. You hold a much smaller block of the exact same material in your hand, above

goldenfox [79]

It will sink because mass does not affect the physical properties of the object.

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3 years ago

A kid is swinging on a swing at the park with an angle of 25°, is this considered simple

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3 years ago

A ball is thrown upward at time t=0 from the ground with an initial velocity of 60 m/s (~ 134 mph). Assume that g = 10 m/s2.At w

Shtirlitz [24]

Answer:

6 second

Explanation:

initial velocity of ball, u = 60 m/s

g = 10 m/s^2

Let the ball takes time t to reach at the maximum height

We know that at maximum height, the velocity of ball is zero.

v = 0 m/s

Use first equation of motion

v = u + gt

0 = 60 - 10 x t

t = 6 second

Thus, the ball takes 6 second to reach to maximum height.

7 0

3 years ago

With full explaniation

zzz [600]

A. The ball's (vertical) velocity $v$ at time $t$ is

$v(t) = 30\dfrac{\rm m}{\rm s} - gt$

so that after 4 seconds, the ball's speed is

$|v(4\,\mathrm s)| = \left|30\dfrac{\rm m}{\rm s} - \left(10\dfrac{\rm m}{\mathrm s^2}\right) (4\,\mathrm s)\right| = \boxed{10\frac{\rm m}{\rm s}}$

(The velocity is -10 m/s, so the ball is falling back down at this point.)

B. At maximum height, the ball has zero velocity, so it takes

$30\dfrac{\rm m}{\rm s} - gt = 0 \implies t = \dfrac{30\frac{\rm m}{\rm s}}g = \boxed{3\,\mathrm s}$

for the ball to reach this height.

C. The height of the ball $y$ at time $t$ is

$y(t) = \left(30\dfrac{\rm m}{\rm s}\right) t - \dfrac g2 t^2$

The maximum height is attained by the ball at 3 seconds after it's thrown, so

$y_{\rm max} = \left(30\dfrac{\rm m}{\rm s}\right) (3\,\mathrm s) - \dfrac{10\frac{\rm m}{\mathrm s^2}}2 (3\,\mathrm s)^2 = \boxed{45\,\mathrm m}$

D. The time it takes for the ball to reach its maximum height is half the time it spends in the air. So the total airtime is $\boxed{6\,\mathrm s}$ .

Put another way: When the ball returns to the height from which it was thrown, its final velocity has the same magnitude as its initial velocity but points in the opposite direction. This is to say, after the total time the ball is in the air, it's final velocity will be -30 m/s. Then the total airtime is

$30\dfrac{\rm m}{\rm s} - gt = -30\dfrac{\rm m}{\rm s} \implies t = \dfrac{60\frac{\rm m}{\rm s}}g = \boxed{6\,\mathrm s}$

Put yet another way: Solve $y(t) = 0$ for $t$ . I don't see a need to elaborate...

5 0

2 years ago