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Lady_Fox [76]
3 years ago
11

Given the following frequencies, calculate the corresponding periods. a. 60 Hz b. 8 MHz c. 140 kHz d. 2.4 GHz

Physics
1 answer:
Ipatiy [6.2K]3 years ago
6 0

The frequency can be defined as the inverse of the period, that is, it can be expressed as

T = \frac{1}{f}

Here,

T = Period

f = Frequency

For each value we only need to replace the value and do the calculation:

PART A)

T = \frac{1}{f}

T = \frac{1}{60Hz}

T = 0.0166s

PART B)

T = \frac{1}{f}

T = \frac{1}{8*10^6}

T = 1.25*10^{-7} s

PART C)

T = \frac{1}{f}

T = \frac{1}{140*10^{3}}

T = 7.14*10^{-6}s

PART D)

T = \frac{1}{f}

T = \frac{1}{2.4*10^{9}}

T = 4.166*10^{-10}s

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a hawk flies in a horizontal arc of radius 10.3 m at a constant speed of 4.8 m/s. find its centripetal acceleration. answer in u
n200080 [17]

The hawk’s centripetal acceleration is 2.23 m/s²

The magnitude of the acceleration under new conditions is 2.316 m/s²

radius of the horizontal arc = 10.3 m

the initial constant speed = 4.8 m/s

we know that the centripetal acceleration is given by

    a_{c}  = \frac{v^{2} }{r}

   a_{c}  = 23.04/10.3

    a_{c}  = 2.23 m/s²

It continues to fly but now with some tangential acceleration

a_{t} = 0.63 m/s²

therefore the net value of acceleration is given by the resultant of the centripetal acceleration and the tangential acceleration

so

a_{net}  =  \sqrt{a_{c} ^{2} +a_{t} ^{2}   }

a_{net}  =  \sqrt{4.97 + 0.396}

a_{net}  =  2.316 m/s²

So the magnitude of  net acceleration will become 2.316 m/s².

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brainly.com/question/11560829

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8 0
1 year ago
Which of these is a type of matter that rarely interacts with other matter and is created in some nuclear fusion reactions withi
marysya [2.9K]

Answer:

the answer is helium you already know that because im in college on my way to the military next week bless me

Explanation:

7 0
3 years ago
1) Subatomic particles called muons can be created in the upper atmosphere by collisions of cosmic rays (energetic particles com
Vsevolod [243]

Answer and explanation:

A.

Muon travelled straight down towards the earth. Therefore the tree moves up in the rest frame of muon (option a)

B.

In muon rest frame it travels Zero meters

C.

Distance, d = Velocity, v * Time, s

where, v = 0.9c = 0.9 \times 8 \times 10^8 , s = 2.2 \mu s

d = 0.9 \times 3 \times 10^8 \times 2.2 \times 10^{-6}\\\\d = 594m

D.

Distance from the top of the mountain to the tree is the same as the distance travelled by the tree in the muons rest frame

that is same as in part C which is 594m

E.

Using lorentz contraction

In the rest frame of someone standing on the mountain

the distance is given by

d' = \frac{d}{\gamma} = d\sqrt{1 - \frac{v^2}{c^2}}, where, \frac{1}{\gamma}= \sqrt{1 - \frac{v^2}{c^2}}

d' = 594\sqrt{1 - \frac{(0.9c)^2}{c^2}}

d' = 594\sqrt{1 - 0.81}

d' = 594 \times 0.4359

d' = 258.92m

F.

in the rest frame of someone standing on the mountain,

muon moves straight down

3 0
3 years ago
Car B is following Car A and has a greater speed than Car A. the two cars are moving in a straight line and in the same directio
ElenaW [278]

Answer:

Collision force will be same in both the cases.

Explanation:

A perfectly inelastic collision is said to take place  when a system loses the amount of its Kinetic Energy at its maximum. In a perfectly inelastic collision,  the colliding particles stick to each other. In such a collision, kinetic energy is lost by combining the two bodies with each other.

In situation 1:

Speed of Car A, v_{A} = 10 mph

Speed of Car B, v_{B} = 20 mph

Relative speed of car A and car B, v = v_{b} - v_{a} = 10 m/s

Now, in the situation 2:

Speed of car A, v_{A} = 30 mph

Speed of car B, v_{B} = 40 mph

Relative speed of car A and car B, v = v_{b} - v_{a} = 10 m/s

Therefore, Car A and Car B both have the same relative speed, v = 10 m/s

7 0
3 years ago
A mobile phone is 35% efficient. Over half an hour 11 kJ of energy is transferred to the phone.
kramer
<h3><u>A</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u><u>:</u><u>-</u></h3>

  • Energy Transferre=11KJ
  • Efficiency=35%
<h3>☆Usefully transferred energy:-</h3>

\\ \sf\longmapsto 35\%\:of 11

\\ \sf\longmapsto 35\%\times 11

\\ \sf\longmapsto \dfrac{35}{100}\times 11

\\ \sf\longmapsto \dfrac{385}{100}

\\ \sf\longmapsto 3.85KJ

\\ \sf\longmapsto 3850J

6 0
3 years ago
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