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4 years ago

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1 answer:

Agata [3.3K]4 years ago

5 0

Answer:

2.6 then 1.5 last 0.8

Explanation:

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Help me label the respiratory system

lutik1710 [3]

8. epiglottis
9. trachea
10. lugs
11.diaphragm

5 0

3 years ago

Clara's grandfather is a captain of a cargo ship. He prefers to navigate his ship through the moderate neap tides. At these time

spayn [35]

Answer:

The phases are the first and third quarters.

Explanation:

During the first quarter of the moon phases, the line between Earth and the Sun is at a right angle to the line between Earth and the moon. At this time the moon has completed the first quarter of its orbit around the earth. At this point, half of the moon is fully reflecting light from the sun.

Also during the third quarter which is the last quarter, the line between the earth and the sun is at a right angle to the line between the earth and the moon. This is the time when the second half of the moon which was not illuminated during the first quarter becomes completely illuminated.

6 0

3 years ago

In a particular case of Compton scattering, a photon collides with a free electron and scatters backwards. The wavelength after

mafiozo [28]

Answer:

hence initial wavelength is $\lambda =4.86\times10^{-12}m$

Explanation:

shift in wavelength due to compton effect is given by

$\lambda ^{'}-\lambda =\frac{h}{m_{e}c}\times(1-cos\theta )$

λ' = the wavelength after scattering

λ= initial wave length

h= planks constant

m_{e}= electron rest mass

c= speed of light

θ= scattering angle = 180°

compton wavelength is

$\frac{h}{m_{e}c}= 2.43\times10^{-12}m$

$\lambda '-\lambda =2.43\times10^{-12}\times(1-cos\theta )$

$\lambda '-\lambda =2.43\times10^{-12}\times(1+1 )$ ( put cos 180°=-1)

also given λ'=2λ

putting values and solving we get

$\lambda =4.86\times10^{-12}m$

hence initial wavelength is $\lambda =4.86\times10^{-12}m$

4 0

4 years ago

Two particles having charges of 0.440 nC and 11.0 nC are separated by a distance of 1.80 m . 1) At what point along the line con

Harlamova29_29 [7]

Answer:

a) 0.3 m

b) r = 0.45 m

Explanation:

given,

q₁ = 0.44 n C and q₂ = 11.0 n C

assume the distance be r from q₁ where the electric field is zero.

distance of point from q₂ be equal to 1.8 -r

now,

E₁ = E₂

$\dfrac{K q_2}{(1.8-r)^2} = \dfrac{K q_1}{r^2}$

$(\dfrac{1.8-r}{r})^2= \dfrac{q_2}{q_1}$

$\dfrac{1.8-r}{r}= \sqrt{\dfrac{11}{0.44}}$

1.8 = 6 r

r = 0.3 m

<h3>b) zero when one charge is negative.</h3>

let us assume q₁ be negative so, distance from q₁ be r

from charge q₂ the distance of the point be 1.8 +r

now,

E₁ = E₂

$\dfrac{K q_2}{(1.8+r)^2} = \dfrac{K q_1}{r^2}$

$(\dfrac{1.8+r}{r})^2= \dfrac{q_2}{q_1}$

$\dfrac{1.8+r}{r}= \sqrt{\dfrac{11}{0.44}}$

1.8 =4 r

r = 0.45 m

4 0

4 years ago

This type of bike tire is thinner, lighter, more expensive, and punctures easily.

Dmitriy789 [7]

Answer:tublar

Explanation:tublar bikes usaly cost 30-40% more

6 0

3 years ago

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