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2 years ago

13

A student walks 12 blocks to a coffee shop. (Assume all the blocks are equal length.) How long will it take him to reach the cof

fee shop if he walks 3 blocks in 2 min?

2 answers:

horrorfan [7]2 years ago

7 0

It will take 8 minutes

Nina [5.8K]2 years ago

6 0

Answer: 18

Explanation:

3 divided by 2 is 1.5

1.5 x 12 is 18

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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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3 years ago

Ocean waves pass through two small openings, 20.0 m apart, in a breakwater. You're in a boat 70.0 m from the breakwater and init

Klio2033 [76]

Answer:

λ = 5.65m

Explanation:

The Path Difference Condition is given as:

δ= $(m+\frac{1}{2})\frac{lamda}{n}$ ;

where lamda is represent by the symbol (λ) and is the wavelength we are meant to calculate.

m = no of openings which is 2

∴δ= $\frac{3*lamda}{2}$

n is the index of refraction of the medium in which the wave is traveling

To find δ we have;

δ= $\sqrt{70^2+(33+\frac{20}{2})^2 }-\sqrt{70^2+(33-\frac{20}{2})^2 }$

δ= $\sqrt{4900+(\frac{66+20}{2})^2}-\sqrt{4900+(\frac{66-20}{2})^2}$

δ= $\sqrt{4900+(\frac{86}{2})^2 }-\sqrt{4900+(\frac{46}{2})^2 }$

δ= $\sqrt{4900+43^2}-\sqrt{4900+23^2}$

δ= $\sqrt{4900+1849}-\sqrt{4900+529}$

δ= $\sqrt{6749}-\sqrt{5429}$

δ= 82.15 -73.68

δ= 8.47

Again remember; to calculate the wavelength of the ocean waves; we have:

δ= $\frac{3*lamda}{2}$

δ= 8.47

8.47 = $\frac{3*lamda}{2}$

λ = $\frac{8.47*2}{3}$

λ = 5.65m

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a load of 800 newton is lifted by an effort of 200 Newton. if the load is placed at a distance of 10 cm from the fulcrum. what w

nataly862011 [7]

Answer:

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Explanation:

We are given that

Load=800 N

Effort=200 N

Load distance=10 cm

We have to find the effort distance.

We know that

$load\times load\;distance=Effort\times effort\;distance$

Using the formula

$800\times 10=200\times effort\;distance$

Effort distance= $\frac{800\times 10}{200}$

Effort distance= $\frac{8000}{200}$

Effort distance=40 cm

Hence, the effort distance will be 40 cm.

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