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oksian1 [2.3K]
3 years ago
14

Which of the following occurs in January when the earth is closest to the sun?

Physics
1 answer:
Vlada [557]3 years ago
4 0
Its January 4th 2003
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Helium

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Incident ray

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"To describe the reflection of light, we will use the following terminology. The incoming light ray is called the incident ray. The light ray moving away from the surface is the reflected ray. The most important characteristic of these rays is their angles in relation to the reflecting surface."

https://www.siyavula.com/read/science/grade-11/geometrical-optics/05-geometrical-optics-03

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Before the widespread use of computers, how was the epicenter of an earthquake determined?
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Web Spiders and Oscillations All spiders have special organs that make them exquisitely sensitive to vibrations. Web spiders det
chubhunter [2.5K]

Complete Question

The complete quetion is shown on the first uploaded image

Answer:

Explanation:

From the question we are told that

       The mass of the fly is  m_f =  11 mg =  11*10^{-3} g =  1.1*10^{-5} \ kg

        The extension of the web is  e=  4.00 \ mm = 0.004 \ m

       

The spring constant is mathematically evaluated as

          k = \frac{mg}{e}

substituting values

        k = \frac{1.1 *10^{-5} *9.8}{0.004}

         k = 0.027 \ N/m

The frequency of vibration is

         f =  \frac{1}{2 \pi} \sqrt{\frac{k}{m} }

substituting values

       f =  \frac{1}{2 * 3.142 } \sqrt{\frac{0.027}{1.1*10^{-5}} }

      f = 7.9 Hz

         

8 0
3 years ago
While driving, your car has an initial position of 3.2 m, an initial velocity of -8.4 m/s, and
KIM [24]

Answer:

The position of the car at t = 1.5 s is at -8.1625 meters

Explanation:

The initial position of the car is 3.2 meters

The initial velocity is -8.4 m/s

The constant acceleration is 1.1 m/s²

We need to find the final position of the car at the time t = 1.5 seconds

The displacement <em>s</em> = final position - initial position

s=ut+\frac{1}{2}at^{2}, where <em>u</em> is the initial velocity, <em>a</em> is the

constant acceleration and <em>t</em> is the time

So we can find the final velocity by using the rule:

final position - initial position = ut+\frac{1}{2}at^{2}

initial position = 3.2 meters , u = -8.4 m/s , a = 1.1 ²m/s , t = 1.5 s

Substitute these values in the rule

final position - 3.2 = (-8.4)(1.5)+\frac{1}{2}(1.1)(1.5)^{2}

final position - 3.2 = -12.6 + 1.2375

final position - 3.2 = -11.3625

add 3.2 for both sides

final position = -8.1625

<em>That means the car is at 8.1625 meters in opposite direction</em>

<em>The position of the car at t = 1.5 s is at -8.1625 meters </em>

4 0
3 years ago
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