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3 years ago

Please choose the answer that describes the scientific notation 5098000

1 answer:

3 0

So there is a decimal after the last zero and it looks like this 5098000. You have to move the decimal point six back to get in between the five and the zero which looks like this 5.098000

Scientific notation is the way that scientists easily handle very large numbers or very small numbers. For example, instead of writing 0.0000000056, we write 5.6 x 10^9.

Being that we moved the decimal six places back the answer is 5.098 x 10^6

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I believe this is right but double check to make sure :)

7 0

2 years ago

Which component of a loudspeaker produces sound waves? a solenoid coil a thin membrane a permanent magnet

zimovet [89]

Answer: A thin membrane.

Explanation:

When a sound signal is allowed to pass through the voice coil suspended between permanent magnet, magnetic field will be induced which will cause vibration in the diaphragm - a thin membrane causing disturbance of air in the surrounding of membranes which eventually produce sound waves.

Therefore, we can conclude that a thin membrane of the loud speaker produces sound waves

4 0

2 years ago

Infrared waves from the sun are what make our skin feel warm on a sunny day. If an infrared wave has a frequency of 3.0 x 1012 H

djyliett [7]

Answer:

The wavelength of the infrared wave is 0.0001 m.

Explanation:

Given:

Frequency of an infrared wave is, $f=3.0\times 10^{12}\ Hz$

We know that, infrared waves are electromagnetic waves. All electromagnetic waves travel with the same speed and their magnitude is equal to the speed of light in air.

So, speed of infrared waves coming from the Sun travels with the speed of light and thus its magnitude is given as:

$v=c=3.0\times 10^8\ m/s$

Where, 'v' is the speed of infrared waves and 'c' is the speed of light.

Now, we have a formula for the speed of any wave and is given as:

$v=f\lambda$

Where, $\lambda \to \textrm{Wavelength of infrared wave}$

Now, rewriting the above formula in terms of wavelength, $\lambda$ , we get:

$\lambda=\dfrac{v}{f}$

Now, plug in $3.0\times 10^8$ for 'v', $3.0\times 10^{12}$ for 'f' and solve for $\lambda$ . This gives,

$\lambda=\frac{3.0\times 10^8}{3.0\times 10^{12}}\\\\\lambda=0.0001\ m$

Therefore, the wavelength of the infrared wave is 0.0001 m.

5 0

3 years ago

A seagull flies at a velocity of 9.00 m/s straight into the wind. (a) if it takes the bird 20.0 min to travel 6.00 km relative t

enot [183]

Here we will the speed of seagull which is v = 9 m/s

this is the speed of seagull when there is no effect of wind on it

now in part a)

if effect of wind is in opposite direction then it travels 6 km in 20 min

so the average speed is given by the ratio of total distance and total time

$v_{avg} = \frac{6000}{20*60}$

$v_{avg} = 5m/s$

now since effect of wind is in opposite direction then we can say

$V_{net} = v_{bird} - v_{wind}$

$5 = 9 - v_{wind}$

$v_{wind}= 4 m/s$

Part b)

now if bird travels in the same direction of wind then we will have

$v_{net}= v_{bird} + v_{wind}$

$v_{net} = 9 + 4 = 13 m/s$

now we can find the time to go back

$time = \frac{distance}{speed}$

$time = \frac{6000}{13}$

$time = 7.7 minutes$

Part c)

Total time of round trip when wind is present

$T = t_1 + t_2$

$T = 20 + 7.7 = 27.7 min$

now when there is no wind total time is given by

$T = \frac{6000}{9} + \frac{6000}{9}$

$T = 22.22 min$

So due to wind time will be more

4 0

3 years ago

Use the following half-life graph to answer the following question:

Temka [501]

Answer:

A 1.0 min

Explanation:

The half-life of a radioisotope is defined as the time it takes for the mass of the isotope to halve compared to the initial value.

From the graph in the problem, we see that the initial mass of the isotope at time t=0 is

$m_0 = 50.0 g$

The half-life of the isotope is the time it takes for half the mass of the sample to decay, so it is the time t at which the mass will be halved:

$m'=\frac{50.0 g}{2}=25.0 g$

We see that this occurs at t = 1.0 min, so the half-life of the isotope is exactly 1.0 min.

3 0

3 years ago