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tino4ka555 [31]

3 years ago

13

Question 3 (Essay Worth 5 points)

1 answer:

Elis [28]3 years ago

8 0

Well, in order to figure out the answer is to divide until you figure out how many miles they went per second. If it takes 5 seconds to reach 50 miles per hour it took 10 seconds per every 10 miles meaning each mile took 1 second. (Not actually possible but the answer) So, If it finished a 100 mile trip in 2 hours it took an hour for 50 miles. If it took 1 hour for 50 miles divide 60/50 which gets you 1.2 so it took 1.2 miles per minute meaning the car went 120 miles per hour I believe. I hope this helps :)

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What is the wavelength of the radio waves from an FM station operating at a frequency of 99.5 MHz

ser-zykov [4K]

Answer:

Wavelength, $\lambda=3.01\ m$

Explanation:

It is given that,

Frequency, f = 99.5 MHz = 99.5 × 10⁶ Hz

We need to find the wavelength of the radio waves from an FM station operating at above frequency. The relationship between the frequency and the wavelength is given by :

$c=f\lambda$

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

c = speed of light

$\lambda=\dfrac{3\times 10^8\ m/s}{99.5\times 10^6\ Hz}$

$\lambda=3.01\ m$

So, the wavelength of the radio waves from an FM station is 3.01 m. Hence, this is the required solution.

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4 years ago

Compared to the mass and charge of a proton, the electron has

maxonik [38]

B. opposite charge and smaller mass

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4 years ago

Chromosomes are a form of blood cells.

Inessa05 [86]

Answer:

false im just trying to get it if you'd like to give it to me

8 0

3 years ago

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You are given a sample of metal and asked to determine its specific heat. You weight the sample and find that its weight is 28.4

lidiya [134]

Answer:

The samples specific heat is 14.8 J/kg.K

Explanation:

Given that,

Weight = 28.4 N

Suppose, heat energy $E=1.25\times10^{4}\ J$

Temperature = 18°C

We need to calculate the samples specific heat

Using formula of specific heat

$Q=mc\Delta T$

$c=\dfrac{Q}{m\Delta T}$

Where, m = mass

c = specific heat

$\Delta T$ = temperature

Q = heat

Put the value into the formula

$c=\dfrac{1.25\times10^{4}}{\dfrac{28.4}{9.8}\times(18+273)}$

$c=14.8\ J/kg. K$

Hence, The samples specific heat is 14.8 J/kg.K

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3 years ago

Use the following equation to help you answer the question. The peak intensity of radiation from a star named Sigma is 2 x 10 6

puteri [66]

Answer:

1449 K

Explanation:

The surface temperature of a star is related to its peak wavelength by Wien's displacement law:

$T=\frac{b}{\lambda}$

where

T is the surface temperature

b is Wien's displacement constant

$\lambda=2\cdot 10^{-6} m$

So the surface temperature of the star is

$T=\frac{2.898 \cdot 10^{-3} Km}{2\cdot 10^{-6} m}=1449 K$

8 0

3 years ago

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