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miv72 [106K]
3 years ago
15

The technological challenges of installing cellphones across the globe

Physics
1 answer:
Leona [35]3 years ago
3 0

Answer:

I don't know sorry For this question

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Wave traveling at 330 m/sec has a wavelength of 4.3 meters. What is the frequency of this wave?
Rasek [7]

Answer:

76.74 Hz

Explanation:

Given:

Wave velocity ( v ) = 330 m / sec

wavelength ( λ ) = 4.3 m

We have to calculate Frequency ( f ):

We know:

v = λ / t [ f = 1 / t ]

v = λ f

= > f = v / λ

Putting values here we get:

= > f = 330 / 4.3 Hz

= > f = 3300 / 43 Hz

= > f = 76.74 Hz

Hence, frequency of sound is 76.74 Hz.

6 0
2 years ago
2/25/20 or 2/28/20 Dispatch #53
mixer [17]

Answer:

Power = 21[W]

Explanation:

Initial data:

F = 35[N]

d = 18[m]

In order to solve this problem we must remember the definition of work, which tells us that it is equal to the product of a force for a distance.

Therefore:

Work = W = F*d = 35*18 = 630 [J]

And power is defined as the amount of work performed in a time interval.

Power = Work / time

Time = t = 30[s]

Power = 630/30

Power = 21 [W]

3 0
2 years ago
A force of 10N is required to stretch a spring from 20cm to 25cm. What is the spring constant in N/m2 Be careful of unit
kupik [55]

Answer:

C) 40 N/m

Explanation:

If we ASSUME that the spring is un-stretched at the zero cm position

k = F/Δx = 10/0.25 = 40 N/m

5 0
2 years ago
The ozone layer protects us from the harmful effects of which type of radiation?
leonid [27]

Answer:

Hey there

The ozone layer or ozone shield is a region of Earth's stratosphere that absorbs most of the Sun's ultraviolet radiation

Can u have brainly

3 0
2 years ago
The period of the earth around the sun is 1 year and its distance is 150 million km from the sun. An asteroid in a circular orbi
SOVA2 [1]

Answer:

5.024 years

Explanation:

T1 = 1 year

r1 = 150 million km

r2 = 440 million km

let the period of asteroid orbit is T2.

Use Kepler's third law

T² ∝ r³

So,

\left ( \frac{T_{2}}{T_{1}} \right )^2=\left ( \frac{r_{2}}{r_{1}} \right )^3

\left ( \frac{T_{2}}{1} \right )^2=\left ( \frac{440}{150} \right )^3

T2 = 5.024 years

Thus, the period of the asteroid's orbit is 5.024 years.

4 0
3 years ago
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