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xxTIMURxx [149]
3 years ago
14

Springfield's "classic rock" radio station broadcasts at a frequency of 102.1 mhz. what is the length of the radio wave in meter

s?
Physics
1 answer:
Mila [183]3 years ago
7 0
The frequency of the radio wave is:
f=102.1 MHz = 102.1 \cdot 10^6 Hz

The wavelength of an electromagnetic wave is related to its frequency by the relationship
\lambda= \frac{c}{f}
where c is the speed of light and f the frequency. Plugging numbers into the equation, we find
\lambda= \frac{3 \cdot 10^8 m/s}{102.1 \cdot 10^6 Hz}= 2.94 m
and this is the wavelength of the radio waves in the problem.
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How can a soccer ball and a small tennis ball bouncing relates to newtons first law interita?
Anna35 [415]

Answer:

I think it has something to do with both inertia and gravity?

7 0
3 years ago
Which best describes how the current scientific model of the atom was developed?
Lelu [443]

Answer:

A. The model was the result of hundreds of years of experiments.

Explanation:

Since it is not possible to visualize an atom in isolation, scientists have spent hundreds of years experimenting and creating atomic models, that is, images that serve to explain the constitution, properties and behavior of atoms.

The earliest who imagined the existence of the atoms were the Greek philosophers Leucippus and Democritus in about 450 BCE. According to them, everything would be formed by tiny indivisible particles. Hence the origin of the name "atom", which comes from the Greek a (no) and tome (parts).

But in the nineteenth century, some scientists began to conduct experimental tests increasingly accurate thanks to technological advances. Not only was it discovered that everything was actually made up of tiny particles, but it was also possible to understand more and more about the atomic structure.

Scientists used the information discovered by other scholars to develop the atomic model. In this way, the discoveries of one scientist were replaced by those of others. The concepts that were correct remained, but those that proved to be non-real were now abandoned. Thus, new atomic models were created. This series of discoveries of the atomic structure until arriving at the accepted models today was known like the evolution of the atomic model.

8 0
3 years ago
An orchestra is practicing a piece that is to be played at an allegro tempo. The conductor sets a metronome at 160 beats per min
omeli [17]

Answer:

375 ms

Explanation:

the frequency of metronome , f = 160 beats per minute

f = 160 /60 beats per sec

f = 2.67 beats /s

the period of a single beat , T = 1/f

T = 1/2.67 s

T = 0.375 s = 375 ms

the period of a single beat is 375 ms

3 0
3 years ago
A space station has a large ring-like component that rotates to simulategravity for the crew. This ring has a massM= 2.1×105kg a
ivann1987 [24]

Answer:

Each thruster has to applied a force of 294.5N in tangential direction

Explanation:

Mass of the ring ,M =2.1×105kg

Mass of the ship ,m = 3.5×104kg

Radius of the ring R = 86.0 m

distance of ship from center of the ring, r =31.0 m

Let force applied by each thruster be F

Time taken to reach  gravity ,t = 3hrs = 3600× 3 =10800sec

The movement of ring make the object kept at the edge feel a force of centrifuge in outward direction.

Centrifugal force = weight of the object on earth

Assume the ring is moving with angular speed ω

Centripetalforce of the object kept at ring

m₁R ω²=m₁g  (m₁=mas of object)

⇒Rω² = g

⇒ω = √g/R

The ring start from 0 angular speed with constant angular acceleration

Let the constant angular acceleration be ∝

∝ = ω  / t

(ω = √g/R)

∝ = \frac{1}{t } \sqrt{\frac{g}{R} } }

Consider Torque on the ring and ship system

T = FR + FR = 2FR

Moment of inertia of ring ship system

I = I(ring)+I(ship)+I(ship)

= MR² + mr² + mr² = MR² + 2mr²

angular acceleration of the ring ship system

∝ = \frac{2FR}{MR^2 + 2mr^2}

Now we have ,

∝ = \frac{1}{t} \sqrt{\frac{g}{R} }  , ∝ = \frac{2FR}{MR^2+2mr^2}

equating both values

We have,

F = \frac{MR^2+2mr^2}{2Rt} \sqrt{\frac{g}{R} }

where,

m = 2.1 × 10⁵kg, R = 86m, m = 3.5 × 10₄kg,

r = 31m, g = 9.8m/s² , t = 10800sec

F = \frac{(2.1 \times 10^5 \times86^2)+(2\times3.5v10^4\times31^2) }{2\times86\times10800} \times\sqrt{\frac{9.8}{86} } \\\\F = 294.47N\\\approx294.5N

Each thruster has to applied a force of 294.5N in tangential direction

3 0
3 years ago
A 65 kg bungee jumper leaps from a bridge. She is tied to a bungee cord that is 12 m long when unstretched, and falls a total of
Nimfa-mama [501]

Answer:

(a) k = 30.33 N/m

(b) a = 9.8 m/s²

Explanation:

First, we need to find the force acting on the bungee jumper. Since, this is a free fall motion. Therefore, the force must be equal to the weight of jumper:

F = W = mg

F = (65 kg)(9.8 m/s²)

F = 637 N

(a)

Now applying Hooke's Law:

F = k Δx

where,

k = spring constant = ?

Δx = change in length of bungee cord = 33 m - 12 m = 21 m

Therefore,

637 N = k(21 m)

k = 637 N/21 m

<u>k = 30.33 N/m</u>

<u></u>

(b)

Since, this is free fall motion. Thus, the maximum acceleration will be the acceleration due to gravity.

a = g

<u>a = 9.8 m/s²</u>

7 0
3 years ago
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