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Savatey [412]
4 years ago
13

Light intensity is?

Physics
2 answers:
IRISSAK [1]4 years ago
6 0

Answer:

The rate at which a waves energy flows through a given unit of area

Fittoniya [83]4 years ago
5 0
The correct answer is:
<span>The rate at which a waves energy flows through a given unit of area

In fact, light intensity is defined as the light power per unit of area:
</span>I= \frac{P}{A}
<span>but the power is the energy carried by the light per unit of time:
</span>P= \frac{E}{t}
<span>this means that the intensity can be rewritten as
</span>I= \frac{E}{tA}<span>
So, it's basically the rate of energy (per unit of time) through a given surface.</span>
You might be interested in
The resultant of two forces is 250 N and the same are inclined at 30° and 45° with resultant one on either side calculate the ma
Varvara68 [4.7K]

Answer:

The two forces are;

1) Force 1 with magnitude of approximately 183.013 N, acting 30° to the left of the resultant force

2) Force 2 with magnitude of approximately 129.41 N acting at an inclination of 45° to the right of the resultant force

Explanation:

The given parameters are;

The (magnitude) of the resultant of two forces = 250 N

The angle of inclination of the two forces to the resultant = 30° and 45°

Let, F₁ and F₂ represent the two forces, we have;

F₁ is inclined 30° to the left of the resultant force and F₂ is inclined 45° to the right of the resultant force

The components of F₁ are \underset{F_1}{\rightarrow} = -F₁ × sin(30°)·i + F₁ × cos(30°)·j

The components of F₂ are \underset{F_2}{\rightarrow} = F₂ × sin(45°)·i + F₂ × cos(45°)·j

The sum of the forces = F₂ × sin(45°)·i + F₂ × cos(45°)·j + (-F₁ × sin(30°)·i + F₁ × cos(30°)·j) = 250·j

The resultant force, R = 250·j, which is in the y-direction, therefore, the component of the two forces in the x-direction cancel out

We have;

F₂ × sin(45°)·i = F₁ × sin(30°)·i

F₂ ·√2/2 = F₁/2

∴ F₁ = F₂ ·√2

∴ F₂ × cos(45°)·j  + F₁ × cos(30°)·j = 250·j

Which gives;

F₂ × cos(45°)·j  + F₂ ·√2 × cos(30°)·j = 250·j

F₂ × ((cos(45°) + √2 × cos(30°))·j = 250·j

F₂ × ((√2)/2 × (1 + √3))·j = 250·j

F₂ × ((√2)/2 × (1 + √3))·j = 250·j

F₂ = 250·j/(((√2)/2 × (1 + √3))·j) ≈ 129.41 N

F₂ ≈ 129.41 N

F₁  = √2 × F₂ = √2 × 129.41 N ≈ 183.013 N

F₁  ≈ 183.013 N

The two forces are;

A force with magnitude of approximately 183.013 N is inclined 30° to the left of the resultant force and a force with magnitude of approximately 129.41 N is inclined 45° to the right of the resultant force.

5 0
3 years ago
Two manned satellites approaching one another at a relative speed of 0.450 m/s intend to dock. The first has a mass of 4.50 ✕ 10
vovangra [49]
Ok, so adopting that the 2nd satellite is at rest and that we're not moving anywhere near the speed of light (so no special relativity considerations), we can just add the two speed together, and say the 1st satellite is moving at 0.9m/s at the 2nd satellite. We can then set up our conservation of momentum equation, m₁v₁+m₂v₂ = m₁v₃+m₂v₄, where I'm calling v 1 and 2 the initial velocities of satellite 1 and 2 and v 3 and 4 the final velocities of satellite 1 and 2 respectively. We know, based on our chosen frame, that v₂ = 0, so that falls out to leave m₁v₁ = m₁v₃+m₂v₄, but we don't know v₃ or v₄, so we need another equation. Let's set up conversation of energy (elastic collisions conserve energy), where we only have to worry about kinetic energy (K = 1/2mv²) for each satellite before and after the collision. So we get 1/2m₁v₁²+1/2m₂v₂² = 1/2m₁v₃²+1/2m₂v₄². Now we have 2 equations and two unknown variables so let's solve with substitution. Let's solve the momentum equation for v₃, v₃ = (m₁v₁ - m₂v₄)/m₁, sub that into the energy equation, cancel the 1/2's and let's drop the v₂ terms since it's zero and we get: m₁v₁² = m₁((m₁v₁ - m₂v₄)/m₁)²+m₂v₄², then after some algebra we get v₄ = sqrt(m₁v₁/((v₁ - m₂/m₁)²+m₂)), then we plug in numbers v₄ = sqrt((4.5*10³*0.9/((0.9-(7.5/4.5))²+7.5*10³) = 0.73 m/s for the 2nd satellite after the collision. Then go back to v₃ = (m₁v₁ - m₂v₄)/m₁ and plug in numbers now that we know v₄ and we get v₃ = (4.5*10³*0.9 - 7.5*10³*0.73)/(4.5*10³) = -0.3167 m/s for the 1st satellite.
6 0
3 years ago
a tennis ball is hit upward with a tennis racket an initial velocity of 12 m/s. What will the ball's speed be when it returns to
zaharov [31]

Answer: 12 m/s

Explanation:

If we ignore air resistance, gravity alone will reduce the upward velocity to zero at the top of the flight. As gravity is a conservative force, it will return exactly the same amount of energy to the tennis ball when it returns to the original elevation.

7 0
3 years ago
A string is vibrating between two posts as shown above. Students are to determine the speed of the wave within this string. They
kow [346]

Answer:

The distance between two posts

Explanation:

As the students have already measured the amount of time necessary for the wave to oscillate up and down that is on the string between two posts,They have to measure the speed of the wave.

        ie  v=fλ

<em>They have to measure distance between the posts. </em>

Because wavelength is equal to the distance between the posts ie full length of string.

The frequency of the wave is calculated by

               f= 1/T

Where is the time period which the students have calculated ie the amount of time taken to oscillate up and down,

Thus the wave speed is calculated using formula

            v=λ/T

6 0
3 years ago
An unbalanced force of 20 N is applied to a 10 kg mass. What is the acceleration of the mass?
NemiM [27]

Answer:

2 m/s2

Explanation:

Newton's 2nd law:

F=ma

20 N= (10 kg)a

a= 2 m/s2

8 0
3 years ago
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