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pickupchik [31]
3 years ago
7

Your iclicker operates at a frequency of approximately 900 mhz (900x106 hz). what is the approximate wavelength of the em wave p

roduced by your iclicker
Physics
1 answer:
Umnica [9.8K]3 years ago
5 0
The clicker emits EM (electromagnetic) wave which travels at the speed of light, that is
v = 3 x 10⁸ m/s

The frequency is
f = 900mHz = 9 x 10⁸ Hz

Because velocity = frequency * wavelength, the wavelength, λ,  is given by
fλ = v
λ = v/f
   = (3 x 10⁸ m/s) / (9 x 10⁸ 1/s)
   = 1/3 m

Answer:  1/3 m

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pickupchik [31]

interesting question.

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3 years ago
Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a larg
IRINA_888 [86]

Answer:

Time : <u>7.96 s</u>

Distance Traveled : <u>357.8 m</u>  

Explanation:

In order to solve this problem, we first consider the accelerated motion of rocket. We will be using the subscript 1 for accelerated motion.

So, for accelerated motion, we have:

Acceleration = a₁ = 14.5 m/s²

Time Period = t₁ = 3.1 s

Initial Velocity = Vi₁ = 0 m/s    (Since, it starts from rest)

Final Velocity = Vf₁

Distance covered by sled during acceleration motion = s₁

Now, using 1st equation of motion:

Vf₁ = Vi₁ + (a₁)(t₁)

Vf₁ = 0 m/s + (14.5 m/s²)(3.1 s)

Vf₁ = 44.95 m/s

Now, using 2nd equation of motion:

s₁ = (Vi₁)(t) + (0.5)(a₁)(t₁)

s₁ = (0 m/s)(3.1 s) + (0.5)(14.5 m/s²)(3.1 s)

s₁ = 22.5 m

Now, we first consider the decelerated motion of rocket. We will be using the subscript 2 for decelerated motion.

So, for accelerated motion, we have:

Deceleration = a₂ = - 5.65 m/s²

Time Period = t₂ = ?

Initial Velocity = Vi₂ = Vf₁ = 44.95 m/s    (Since, decelerate motion starts, where accelerated motion ends)

Final Velocity = Vf₂ = 0 m/s    (Since, rocket will eventually stop)

Distance covered by sled during deceleration motion = s₂

Now, using 1st equation of motion:

Vf₂ = Vi₂ + (a₂)(t₂)

0 m/s = 44.95 m/s + (- 5.65 m/s²)(t₂)

t₂ = (44.95 m/s)/(5.65 m/s²)

<u>t₂ = 7.96 s</u>

Now, using 2nd equation of motion:

s₂ = (Vi₂)(t₂) + (0.5)(a₂)(t₂)

s₂ = (44.95 m/s)(7.96 s) + (0.5)(- 5.65 m/s²)(7.96 s)

s₂ = 357.8 m - 22.5 m

s₂ = 335.3 m

Thus, the total distance covered by sled will be:

Total Dustance = S = s₁ + s₂

S = 22.5 m + 335.3 m

<u>S = 357.8 m</u>

7 0
3 years ago
Science: Physics AND Biology - Please help me with these, it's due tonight.
Ostrovityanka [42]
The answer to the first question is that plants make glucose and oxygen by taking in CO2 and H20.

The answer to the second question is having a heavier mass increases the force of a moving object. That is the Second Law of Motion, F = ma.
7 0
4 years ago
an airplane traveling 245 m/s east expericences turbulence, so the pilot slows down to 230 m/s. it takes the pilot 7 seconds to
lana66690 [7]

Answer:

a=v-u/t

a=245-230/7

a=2

8 0
3 years ago
The spaceship Intergalactica lands on the surface of the uninhabited Pink Planet, which orbits a rather average star in the dist
Sidana [21]

Complete Question

The spaceship Intergalactica lands on the surface of the uninhabited Pink Planet, which orbits a rather average star in the distant Garbanzo Galaxy. A scouting party sets out to explore. The party's leader–a physicist, naturally–immediately makes a determination of the acceleration due to gravity on the Pink Planet's surface by means of a simple pendulum of length 1.08m. She sets the pendulum swinging, and her collaborators carefully count 101 complete cycles of oscillation during 2.00×102 s. What is the result? acceleration due to gravity:acceleration due to gravity: m/s2

Answer:

The acceleration due to gravity is  g = 167.2 \ m/s^2  

Explanation:

From the question we are told that

     The length of the simple pendulum is L = 1.081.08 \ m

      The number of cycles is  N =  101

       The time take is  t =  2.00 *10^{2 \ }s

Generally the period of this oscillation is mathematically evaluated as

         T = \frac{N}{t }

substituting values

         T = \frac{101}{2.0*10^2 }

        T = 0.505 \  s

The period of this oscillation is mathematically represented  as

               T = 2 \pi \sqrt{\frac{l}{g} }

making g the subject of the formula we have

              g = \frac{L}{[\frac{T}{2 \pi } ]^2 }

              g = \frac{4 \pi ^2 L }{T^2 }

Substituting values

               g = \frac{4 * 3.142 ^2  * 1.08 }{505.505^2 }

               g = \frac{4 * 3.142 ^2  * 1.08 }{0.505^2 }  

              g = 167.2 \ m/s^2  

7 0
3 years ago
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