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5 days ago

11

You read that light waves travel through air and water at different speeds. What is one way this can affect what you see around

you?

1 answer:

kicyunya [3.1K]5 days ago

5 0

Answer:

Light waves experience a reduction in speed when traveling through water.

Explanation:

Light can bounce off various surfaces, and its velocity decreases in water.

For instance, in a completely dark room, there’s no reflection from the walls, which is why it appears so dark.

Hope this is helpful!

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In an amusement park rocket ride, cars are suspended from 3.40-m cables attached to rotating arms at a distance of 5.90 m from t

ValentinkaMS [3340]

Answer:

The rotational angular speed is measured at 1.34 rad/s.

Explanation:

Considering the following parameters,

Length = 3.40 m

Distance = 5.90 m

Angle = 45.0°

We are tasked with finding the angular speed of rotation

Using the balance equation

Horizontal component

$T\cos\theta=mg$

$T=\dfrac{mg}{\cos\theta}$

Vertical component

$T\sin\theta=m\omega^2 r$

Substituting the tension value

$mg\tan\theta=m\omega^2(d+L\sin\theta)$

$\omega=\sqrt{\dfrac{g\tan\theta}{(d+L\sin\theta)}}$

Substituting the value into the equation

$\omega=\sqrt{\dfrac{9.8\tan45.0}{5.90+3.40\sin45.0}}$

$\omega=1.34\ rad/s$

Thus, the angular speed of rotation computes to 1.34 rad/s.

7 0

1 month ago

You're riding a unicorn at 25 m/s and come to a uniform stop at a red light 20m away. What's your acceleration?

Ostrovityanka [3056]

The result is -15.625 m/s².

Acceleration signifies the alteration of velocity over a specified duration. It can be calculated with this formula:

$a = \dfrac{vf-vi}{t}$

Where:

vf = final velocity

vi = initial velocity

t = time

Let’s examine the information provided in your query:

Initially, the vehicle was traveling at 25 m/s before coming to a halt. Thus, it was in motion and subsequently ceased moving, indicating that the final velocity is 0 m/s.

However, we notice that the problem does not provide a time value. We need to determine the time taken from when it was in motion to when it reached the traffic light located 20 m away.

The time can be calculated using the kinematics equation:

$d = \dfrac{vi+vf}{2} *t$

We derive the equation by substituting the known values first.

$20m = \dfrac{25m/s+0m/s}{2}(t)$

$20m = 12.5m/s{2}(t)$

$\dfrac{20m}{12.5m/s}=t$
$1.6s=t$

The duration from when it was in motion until it stopped is 1.6s. Now we can utilize this in our acceleration calculation.

$a = \dfrac{0m/s-25m/s}{1.6s}$

$a = \dfrac{-25m/s}{1.6s}$

$a = -15.625m/s^{2}$

It is important to note that the acceleration is negative, indicating the vehicle slowed down.

8 0

25 days ago

Read 2 more answers

The position of an object is given by x = at3 - bt2 + ct,where a = 4.1 m/s3, b = 2.2 m/s2, c = 1.7 m/s, and x and t are in SI un

serg [3462]

Answer:

The response to your inquiry is: 15 m/s²

Explanation:

Equation x = at³ - bt² + ct

a = 4.1 m/s³

b = 2.2 m/s²

c = 1.7 m/s

First we calculate x at t = 4.1 s

x = 4.1(4.1)³ - 2.2(4.1)² + 1.7(4.1)

x = 4.1(68.921) - 2.2(16.81) + 6.97

x = 282.58 - 36.98 + 6.98

x = 252.58 m

Now we calculate speed

v = x/t = 252.58/ 4.1 = 61.6 m/s

Finally

acceleration = v/t = 61.6/4.1 = 15 m/s²

6 0

1 month ago

A supplier wants to make a profit by buying metal by weight at one altitude and selling it at the same price per pound weight at

Ostrovityanka [3056]

The appropriate choice is C.

In physics, the law of gravity helps us understand how gravity varies with height. As altitude increases, so too does the experience of gravity. Changes in altitude also result in variations in weight, though these differences are not particularly significant. Consequently, weighing metals at different heights shows negligible variance as the impact of gravity remains constant across them.

3 0

8 days ago

A student solving a physics problem for the range of a projectile has obtained the expression r= v20sin(2θ)g where v0=37.2meter/

ValentinkaMS [3340]

The formula for range is:

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

Given values are:

$v_0=37.2m/s$

where θ equals 14.1 degrees

$g=9.80m/s^2$

Using the equation above,

$R = \frac{37.2^2 sin2*14.1}{9.80}$

The calculated range is 66.7 meters.

Therefore, the range is approximately 66.1 meters.

5 0

1 month ago

Read 2 more answers