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Korolek [52]
3 years ago
7

Light can be made to have a higher intensity by raising its

Physics
1 answer:
e-lub [12.9K]3 years ago
3 0

Answer:

Frequency

Explanation:

The intensity of light is increased by increasing the number of photon per second, and the number of photon per second is referred to as frequency. A photon the smallest unit of light. If the frequency is increased, the number of photons per second increases hence an increase in intensity.

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The angular velocity of a 755-g wheel 15.0 cm in diameter is given by the equation ω(t) = (2.00 rad/s2)t + (1.00 rad/s4)t 3. (a)
Lunna [17]

Answer:

a

The number of radians turned by the wheel in 2s is   \theta= 8\ radians

b

The angular acceleration is  \alpha =14 rad/s^2

Explanation:

        The angular velocity  is given as

                 w(t) = (2.00 \ rda/s^2)t + (1.00 rad /s^4)t^3

Now generally the integral of angular velocity gives angular displacement

           So integrating the equation of angular velocity through the limit 0 to 2 will gives us the angular displacement for 2 sec

    This is mathematically evaluated as

            \theta(t ) = \int\limits^2_0 {2t + t^3} \, dt

                  = [\frac{2t^2}{2} + \frac{t^4}{4}] \left\{ 2} \atop {0}} \right.

                  = [\frac{2(2^2)}{2} + \frac{2^4}{4}] - 0

                  = 4 +4

                 \theta= 8\ radians

Now generally the derivative  of angular velocity gives angular acceleration

      So the value of the derivative of angular velocity equation at t= 2 gives us the angular acceleration

    This is mathematically evaluated as          

           \frac{dw}{dt}  = \alpha (t) = 2 + 3t^2

so at t=2

            \alpha (2) = 2 +3(2)^2

                   \alpha =14 rad/s^2

7 0
2 years ago
I really need help please just answer at least one
yuradex [85]

Answer:

9) a = 25 [m/s^2], t = 4 [s]

10) a = 0.0875 [m/s^2], t = 34.3 [s]

11) t = 32 [s]

Explanation:

To solve this problem we must use kinematics equations. In this way we have:

9)

a)

v_{f}^{2} = v_{i}^{2}-(2*a*x)\\

where:

Vf = final velocity = 0

Vi = initial velocity = 100 [m/s]

a = acceleration [m/s^2]

x = distance = 200 [m]

Note: the final speed is zero, as the car stops completely when it stops. The negative sign of the equation means that the car loses speed or slows down as it stops.

0 = (100)^2 - (2*a*200)

a = 25 [m/s^2]

b)

Now using the following equation:

v_{f} =v_{i} - (a*t)

0 = 100 - (25*t)

t = 4 [s]

10)

a)

To solve this problem we must use kinematics equations. In this way we have:

v_{f} ^{2} =  v_{i} ^{2} + 2*a*(x-x_{o})

Note:  The positive sign of the equation means that the car increases his speed.

5^2 = 2^2 + 2*a*(125 - 5)

25 - 4 = 2*a* (120)

a = 0.0875 [m/s^2]

b)

Now using the following equation:

v_{f}= v_{i}+a*t\\

5 = 2 + 0.0875*t

3 = 0.0875*t

t = 34.3 [s]

11)

To solve this problem we must use kinematics equations. In this way we have:

v_{f} ^{2} =  v_{i} ^{2} + 2*a*(x-x_{o})

10^2 = 2^2 + 2*a*(200 - 10)

100 - 4 = 2*a* (190)

a = 0.25 [m/s^2]

Now using the following equation:

v_{f}= v_{i}+a*t\\

10 = 2 + 0.25*t

8 = 0.25*t

t = 32 [s]

4 0
3 years ago
Define the types of friction and give FOUR examples of each
Verdich [7]

Answer:

Static Friction - acts on objects when they are resting on a surface

Sliding Friction -  friction that acts on objects when they are sliding over a surface

Rolling Friction - friction that acts on objects when they are rolling over a surface

Fluid Friction - friction that acts on objects that are moving through a fluid

Explanation:

Examples of static include papers on a tabletop, towel hanging on a rack, bookmark in a book , car parked on a hill.

Example of sliding include sledding, pushing an object across a surface, rubbing one's hands together, a car sliding on ice.

Examples of rolling include truck tires, ball bearings, bike wheels, and car tires.

Examples of fluid include water pushing against a swimmer's body as they move through it , the movement of your coffee as you stir it with a spoon,  sucking water through a straw, submarine moving through water.

4 0
3 years ago
What is the equation that defines work
mina [271]
Work=force*displacment
3 0
3 years ago
The MAC is 58 inches, The CG limits are from 26% to 43% MAC. If the CG is found to be
eimsori [14]

By working with percentages, we want to see how many inches is the center of gravity out of the limits. We will find that the CG is 1.45 inches out of limits.

<h3>What are the limits?</h3>

First, we need to find the limits.

We know that the MAC is 58 inches, and the limits are from 26% to 43% MAC.

So if 58 in is the 100%, the 26% and 43% of that are:

  • 26% → (26%/100%)*58in = 0.26*58 in = 15.08 in
  • 43% → (43%/100%)*58in = 0.43*58 in = 24.94 in.

But we know that the CG is found to be 45.5% MAC, then it measures:

(45.5%/100%)*58in = 0.455*58in = 26.39 in

We need to compare it with the largest limit, so we get:

26.39 in - 24.94 in = 1.45 in

This means that the CG is 1.45 inches out of limits.

If you want to learn more about percentages, you can read:

brainly.com/question/14345924

6 0
2 years ago
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