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3 years ago

What season is it when the earth axis is tilted so that it is facing toward the sun?

Physics

2 answers:

iVinArrow [24]3 years ago

5 0

Summer (but only in the northern hemisphere)

Tpy6a [65]3 years ago

3 0

C summer because of the tilt but it could be spring due to the rotation.

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Oxygen enters the blood through the.... I do not know the answer. Can someone please help me?!?!

frez [133]

Oxygen enters the blood Through your lungs

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3 years ago

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A 4.60 gg coin is placed 19.0 cmcm from the center of a turntable. The coin has static and kinetic coefficients of friction with

Komok [63]

Answer:

62.64 RPM.

Explanation:

Given that

m= 4.6 g

r= 19 cm

μs = 0.820

μk = 0.440.

The angular speed of the turntable = ω rad/s

Condition just before the slipping starts

The maximum value of the static friction force =Centripetal force

$\mu_s\ m g=m\ \omega^2\ r\\ \omega^2=\dfrac{\mu_s\ m g}{m r}\\ \omega=\sqrt{\dfrac{\mu_s\ g}{ r}}\\ \omega=\sqrt{\dfrac{0.82\times 10}{ 0.19}}\ rad/s\\\omega= 6.56\ rad/s$

$\omega=\dfrac{2\pi N}{60}\\N=\dfrac{60\times \omega}{2\pi }\\N=\dfrac{60\times 6.56}{2\pi }\ RPM\\N=62.64\ RPM$

Therefore the speed in RPM will be 62.64 RPM.

5 0

2 years ago

Could someone ples help

natulia [17]

Yes because it helps them locate their position and direction

5 0

3 years ago

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At the surface of Jupiter's moon Io, the acceleration due to gravity is 1.81 m/s2 . A watermelon has a weight of 40.0 N at the s

Mumz [18]

Answer

acceleration due to gravity on Jupiter's moon,g' = 1.81 m/s²

weight of water melon on earth, W = 40 N

acceleration due to gravity on earth, g = 9.8 m/s²

a) Mass on the earth surface

$M = \dfrac{W}{g}$

$M = \dfrac{40}{9.8}$

M = 4.08 Kg

b) Mass on the surface of Lo

Mass of an object remain same.

Hence, mass of object at the surface of Lo = 4.08 Kg.

c) Weight at the surface of Lo

W' = m g'

W' =4.08 x 1.81

W' = 7.38 N

8 0

2 years ago

Charge is uniformly distributed around a ring of radius R and the resulting electric field magnitude E is measured along the rin

Arte-miy333 [17]

Answer:

$x=\dfrac{r}{\sqrt2}$

Explanation:

Given that

Radius =r

Electric filed =E

Q=Charge on the ring

The electric filed at distance x given as

$E=K\dfrac{Q}{(r^2+x^2)^{3/2}}$

For maximum condition

$\dfrac{dE}{dx}=0$

$E=K{Q}{(r^2+x^2)^{-3/2}}$

$\dfrac{dE}{dx}=K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}$

For maximum condition

$\dfrac{dE}{dx}=0$

$K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}=0$

$r^2+x^2-3x^2=0$

$x=\dfrac{r}{\sqrt2}$

At $x=\dfrac{r}{\sqrt2}$ the electric field will be maximum.

3 0

2 years ago