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2 years ago

2. Explain brightness of light using the wave model of light.

2 answers:

7 0

Answer: A wave model of light is useful for explaining brightness, color, and the frequency-dependent bending of light at a surface between media.

Explanation: Please give me brainliest! :)

Dafna11 [192]2 years ago

5 0

The amplitude of a wave tells us about the intensity or brightness of the light relative to other light waves of the same wavelength.

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To measure the height of a building without a ruler or tape measure, an engineer drops a rock off the top of the building and fi

serg [7]

The relevant equation we can use in this problem is:

h = v0 t + 0.5 g t^2

where h is height, v0 is initial velocity, t is time, g is gravity

Since it was stated that the rock was drop, so it was free fall and v0 = 0, therefore:

h = 0 + 0.5 * 9.81 m/s^2 * (4.9 s)^2

6 0

3 years ago

Write a short description of how the motion of the racers might change from the start of the race to the finish line

Ad libitum [116K]

The motion of the racers might change from the start because the pressure goes up so all the racer wants is to speed up and win, so when the racer first starts he or she is calm because he's not driving yet and when he or she is on his/hers way to he finish line he/she just wants to win and gets under pressure so he speeds up even more and drifts. Your welcome

6 0

3 years ago

A bicycle tire has a pressure of 7.00×105 N/m2 at a temperature of 18.0ºC and contains 2.00 L of gas. What will its pressure be

stepan [7]

Answer:

$p_2 = 664081 N/m^{2}$

Explanation:

from the ideal gas law we have

PV = mRT

$P = \rho RT$

$\rho = \frac{P}{RT}$

HERE R is gas constant for dry air = 287 J K^{-1} kg^{-1}

$\rho = \frac{7.00 10^{5}}{287(18+273)}$

$\rho = 8.38 kg/m^{3}$

We know by ideal gas law

$\rho = \frac{m_1}{V_1}$

$m_1 = \rho V_1 = 8.38 *2*10^{-3}$

$m_1 = 0.0167 kg$

for m_2

$m_2 = \rho V_i - V_removed$

$m_2 = 8.38*(.002 - 10^{-4})$

$m_2 = 0.0159 kg$

WE KNOW

PV = mRT

V, R and T are constant therefore we have

P is directly proportional to mass

$\frac{p_2}{p_1}=\frac{m_2}{m_1}$

$p_2 = p_1 * \frac{m_2}{m_1}$

$p_2 =7*10^{5} * \frac {.0159}{0.0167}$

$p_2 = 664081 N/m^{2}$

8 0

3 years ago

Help please 20 points ? ASAP

KengaRu [80]

Answer:

21: Mass

22: Unbalanced Force

23: Friction

24: Balanced

Explanation:

3 0

3 years ago

Two students make the following claims:

antiseptic1488 [7]

Answer:

E. Student 1 is correct, because as θ is increased, h is the same.

Explanation:

Here we have the object of a certain mass falling under gravity so the force acting on the it will depend on mass of the object and the acceleration due to gravity.

Mathematically:

$F=m.g$

As we know that the work done is evaluated as the force applied on a body and the displacement of the body in the direction of the force.

And for work we have:

$W=F.s\cos\theta$

where:

$s=$ displacement of the object

$\theta=$ angle between the force and displacement vectors

Given that the height of the object is same in each trail of falling object under the gravity be it a free-fall or the incline plane.

In case of free-fall the angle between the force is and the displacement is zero.
In case when the body moves along the inclined plane the force applied by the gravity is same because it depends upon the mass of the object. And the net displacement in the direction of the gravitational force is the height of the object which is constant in both the cases.

So, the work done by the gravitational force is same in the two cases.

6 0

3 years ago