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3 years ago

A boy flies a kite with the string at a 30 degree angle to the horizontal. The tension in the string is 4.5N .

1 answer:

8 0

From our physics class, we know that Work is a product of Force and Displacement. The force must be the force in the same direction with the displacement. So force and displacement must be parallel.

A. How much work in J does the string do on the boy if the boy stands still?

Since we know that: Work = Force * Displacement

and in this case displacement is equal to zero, therefore Work must also be equal to zero.

Work = 0

B. How much work does the string do on the boy if the boy walks a horizontal distance of 11m away from the kite?

Now are given a displacement of 11 m which is a horizontal distance. Therefore we must first calculate the horizontal component of force:

Fx = 4.5 N * cos 30

Fx = 3.897 N

So work is:

Work = 3.897 N * 11 m

Work = 42.87 J ~ 43 J

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3 years ago

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3 years ago

What is the correct sequence for light traveling through a microscope?

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The answer would be B.

Since the light source comes from the bottom, you just need to find the sequence from the bottom going up.

It will start with the light source, then go through the slide, which will then go through the objective lens and lastly, the eyepiece lens. In summary, the sequence is:

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4 years ago

OSCilloscope shows wave starts cycle minimum 20 units reaches max of 100 units before completing 5 milliseconds what is relation

Vitek1552 [10]

Answer:

$T=0.02\ s$

$f=50\ Hz$

Explanation:

Given:

minimum amplitude at the start of oscillation cycle, $a_0=20\ unit$

the first maximum amplitude after the start of oscillation cycle, $a_{m1}=100\ units$

Time taken to reach from the first minima to the first maxima, $t=5\times 10^{-3}\ s$

As we know that an oscilloscope executes a wave cycle represented by a sine wave. So we can deduce that it has executed one-fourth of the cycle in going from the amplitude of 20 units to 100 units in 0.005 seconds.

So the time taken to complete one cycle of the oscillation:

$T=4\times 0.005$

$T=0.02\ s$ is the time period of the oscillation

We know frequency:

$f=\frac{1}{T}$

$f=\frac{1}{0.02}$

$f=50\ Hz$

6 0

4 years ago

A narrow beam of light containing orange (610 nm) and blue (470 nm) wavelengths goes from polystyrene to air, striking the surfa

Eva8 [605]

Answer: The angle between the two colours when they emerge is 0.4°

Explanation:

The refracting angle for different colors knowing its refractive index can be calculated using Snell's law.

Ni × sin α = Nr × sinβ

Where Ni is the refractive index for light in incident medium

α is the angle the incident makes with normal

Nr is the refractive index for light in the refractive medium

β is the angle the refracted makes with the normal

Making β the subject

β = arcSin ( Ni × sinα)/Nr

For Orange color

The value of refractive index in polystyrene medium is 1.490 ,that is Ni= 1.490

α is 30° and Nr is the refractive index of air = 1

β = arcSin(1.490×sin30°)/1

β= 48.15°

For Blue color

The value of refractive index in polystyrene medium is 1.499,that is Ni= 1.499

α is 30° and Nr is the refractive index of air = 1

β = arcSin(1.499×sin30°)/1

β= 48.55°

The angle between the two colours is the difference in the angles of their refracted rays

48.55-48.15=0.4°

6 0

3 years ago