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3 years ago

5

If the diameter of each of a car's wheels is 0.4 meters, approximately, what was the average speed of the car for its three-hour

trip in which each wheel made 100,000 revolutions?

2 answers:

Reptile [31]3 years ago

8 0

Answer:

42 km/h was the average speed of the car for its three-hour trip.

Explanation:

Diameter of the car wheel = d = 0.4 m

Radius of the car wheel = r = 0.5 d = 0.5 × 0.4 m =0.2 m

Distance covered in 1 complete revolution = $2\pi r$

Distance covered in 100,000 complete revolution,D = $2\pi r\times 100,000$

$D=2\times 3.14\times 0.2 m\times 100,000=125,663.71 m$

1 m = 0.001 km

125,663.71 m = 125,663.71 × 0.001 km = 125.66371 km

Time taken to complete 100,000 revolution by a wheel = t = 3 hours

Speed of car : S

$S=\frac{125.66371 km}{3hours}=41.89 km/h\approx 42 km/h$

42 km/h was the average speed of the car for its three-hour trip.

bogdanovich [222]3 years ago

3 0

Answer:

Explanation:

diameter = 0.4 m

Radius, r = 0.2 m

Time, t = 3 hour

number of revolutions, = 100,000

Distance traveled in one revolution = 2 x π x r = 2 x 3.14 x 0.2 = 1.256 m

Total distance in 100,000 revolutions,

d = 100,000 x 1.256 = 125600 m

d = 125.6 km

Average speed = distance / time

Average speed = 125.6 / 3 = 41.87 km/h

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3 years ago

An ancient club is found that contains 100 g of pure carbon and has an activity of 6.5 decays per second. Determine its age assu

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Answer:

The age of living tree is 11104 years.

Explanation:

Given that,

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Activity of this carbon is = 6.5 decays per second = 6.5 x60 decays/min =390 decays/m

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$R=\dfrac{-dN}{dt}=\lambda N=\dfrac{0.693}{t_{\frac{1}{2}}}N$ ....(I)

Where, N = number of radio active atoms

$t_{\frac{1}{2}}$ =half life

We need to calculate the number of radio active atoms

For $N_{12_{c}}$

$N_{12_{c}}=\dfrac{N_{A}}{M}$

Where, $N_{A}$ =Avogadro number

$N_{12_{c}}=\dfrac{6.02\times10^{23}}{12}$

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$N_{c_{14}}=1.30\times10^{-12}N_{12_{c}}$

$N_{c_{14}}=1.30\times10^{-12}\times5.02\times10^{22}$

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Put the value in the equation (I)

$R=\dfrac{0.693\times6.526\times10^{10}\times60}{5700\times3.16\times10^{7}}$

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$2^n=\dfrac{1507}{390}$

$2^n=3.86$

$n ln 2=ln 3.86$

$n=\dfrac{ln 3.86}{ln 2}$

$n =1.948$

We need to calculate the age of living tree

Using formula of age

$t=n\times t_{\frac{1}{2}}$

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$t=11103.6 =11104\ years$

Hence, The age of living tree is 11104 years.

5 0

3 years ago

The 630-nm light from a helium-neon laser irradiates a grating. The light then falls on a screen where the first bright spot is

dimaraw [331]

Answer:

464.8 nm

Explanation:

The second wavelength of light can be calculated using the next equation:

$\lambda = \frac{x*d}{L}$

<u>Where:</u>

<em>λ : is the wavelength of light</em>

<em>x: is the distance from the central maximum</em>

<em>d: is the distance between the spots </em>

<em>L: is the lenght from the screen to the bright spot</em>

For the first wavelength of light we have:

$\lambda_{1} = \frac{x_{1}*d}{L}$

$630 \cdot 10^{-9} m = \frac{0.61 m*d}{L}$

$\frac{d}{L} = \frac{630 \cdot 10^{-9} m}{0.61 m} = 1.033 \cdot 10^{-6}$ (1)

For the second wavelength of light we have:

$\lambda_{2} = \frac{x_{2}*d}{L}$

$\lambda_{2} = 0.45 m*\frac{d}{L}$ (2)

By entering equation (1) into equation (2) we have:

$\lambda_{2} = 0.45 m* 1.033 \cdot 10^{-6} = 4.648 \cdot 10^{-7} m = 464.8 nm$

Therefore, the second wavelength is 464.8 nm

I hope it helps you!

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2 years ago

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