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3 years ago

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When light propagates from a material with a given index of refraction into a material with a smaller index of refraction, the s

peed of the light
A) incease
B) decrease
C) remain constant

1 answer:

SSSSS [86.1K]3 years ago

7 0

Answer: A) increase

Explanation:

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What is a horizontal segment meaning on a position vs. time graph?

sergeinik [125]

The velocity of the object is zero (the object is at rest)

Explanation:

A position vs time graph represents the motion of an object; in particular:

The position of the object x(t) is represented on the y-axis
The time t is represented on the x-axis

For a position-time graph, the slope of the graph is given by

$m=\frac{\Delta x}{\Delta t}$

where

$\Delta x$ is the change in position

$\Delta t$ is the change in time

However, we see that this is equivalent to the definition of velocity:

$v=\frac{\Delta x}{\Delta t}$

Therefore, the slope of a position-time graph is equivalent to the velocity of the object.

And so, a horizontal segment on a position vs time graph means that the object has zero velocity (because the slope is zero).

Learn more about velocity:

brainly.com/question/5248528

#LearnwithBrainly

4 0

4 years ago

A piece of clay sits 0.10 m from the center of a potter’s wheel. If the potter spins the wheel at an angular speed of 15.5 rad/s

finlep [7]

Answer:

$a=24.025\ m/s^2$

Explanation:

Given that

Distance from the center ,r= 0.1 m

The angular speed ,ω = 15.5 rad/s

We know that centripetal acceleration is given as

a=ω² r

a=Acceleration

r=Radius

ω=angular speed

a=ω² r

Now by putting the values in the above equation we get

$a=15.5^2\times 0.1\ m/s^2$

$a=24.025\ m/s^2$

Therefore the acceleration of the clay will be $a=24.025\ m/s^2$ .

4 0

3 years ago

Please help me so I don't fail.

NikAS [45]

Answer:

3

Explanation:

4 0

3 years ago

At the base of a frictionless icy hill that rises at 25.0∘ above the horizontal, a toboggan has a speed of 11.9 m/s toward the h

forsale [732]

Answer:

17.10 m

Explanation:

When the toboggan stops, we have:

$N-w_y=0\\F-w_x=0$

The x-component of weight is the product of the weight and the sine of the angle above the horizontal, so the y-component of W is the product of the weight and the cosine of the angle above the horizontal.

$N-mgcos(25^\circ)=0\\F-mgsin(25^\circ)=0\\F=mgsin(25^\circ)(1)$

The work-energy principle states that the change in the kinetic energy of an object is equal to the net work done on the object.

$\Delta K=W\\K_f-K_i=F\cdot h\\\frac{m(v_f)^2}{2}-\frac{m(v_i)^2}{2}=Fhcos(180^\circ)$

Recall that $v_f=0$ , replacing (1):

$-\frac{m(v_i)^2}{2}=mgsin(25^\circ)h(-1)\\h=\frac{(v_i)^2}{2gsin(25^\circ)}\\h=\frac{(11.9\frac{m}{s})^2}{2(9,8\frac{m}{s^2})sin(25^\circ)}\\h=17.10 m$

7 0

3 years ago

While playing at the swimming pool, a 40-kilogram boy jumps off the side of the pool. When he hits the water, he is traveling 3

otez555 [7]

K.E. = mv² / 2
Here, m = 40 Kg
v = 3 m/s

Substitute their values,
K.E. = 40 * (3)² / 2
K.E. = 40 * 9 / 2
K.E. = 360/2
K.E. = 180 J

In short, Your Answer would be Option A

Hope this helps!

8 0

3 years ago

Read 2 more answers