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3 years ago

11

Look at the picture to answer :)

1 answer:

cricket20 [7]3 years ago

3 0

Answer:

The stress of the movement of tectonic plates caused this fold.

Sorry I don't know the answer to the first part.

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un futbolista patea una pelota que se encuentra en el pasto con un angulo de 30° (medido desde la horizontal) con la intención d

Rus_ich [418]

Answer:

i dont really know what it is

8 0

3 years ago

What is the critical angle of light traveling from benzene (n=1.501) into air?

Taya2010 [7]

The critical angle formula should be: sin^-1(1/n)

where "n" is 1.501 into the air

<span>The critical angle of light travelling from benzene, happens because the larger angles of incidence from the inside of the benzene has experienced the total internal reflection. </span>

4 0

3 years ago

Vitamins and iron are nutrients that provide energy to the human body.<br> A true<br> B false

SIZIF [17.4K]

I think it is A. but then you can also produce your own energy

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2 years ago

How is frequency related to the sound we hear?

Leviafan [203]

Frequency is the vibration of noise and the vibration determines the pitch, which we depend on to be a pitch or frequency we can hear. If it's too high or too low our ears can't hear it

8 0

3 years ago

The force exerted by the wind on the sails of a sailboat is Fsail = 330 N north. The water exerts a force of Fkeel = 210 N east.

Elena L [17]

Answer:

The magnitude of the acceleration is $a_r = 1.50 \ m/s^2$

The direction is $\theta = 32.5 6^o$ north of east

Explanation:

From the question we are told that

The force exerted by the wind is $F_{sail} = (330 ) \ N \ north$

The force exerted by water is $F_{keel} = (210 ) \ N \ east$

The mass of the boat(+ crew) is $m_b = 260 \ kg$

Now Force is mathematically represented as

$F = ma$

Now the acceleration towards the north is mathematically represented as

$a_n = \frac{F_{sail}}{m_b}$

substituting values

$a_n = \frac{330 }{260}$

$a_n = 1.269 \ m/s^2$

Now the acceleration towards the east is mathematically represented as

$a_e = \frac{F_{keel}}{m_b }$

substituting values

$a_e = \frac{210}{260}$

$a_e =0.808 \ m/s^2$

The resultant acceleration is

$a_r = \sqrt{a_e^2 + a_n^2}$

substituting values

$a_r = \sqrt{(0.808)^2 + (1.269)^2}$

$a_r = 1.50 \ m/s^2$

The direction with reference from the north is evaluated as

Apply SOHCAHTOA

$tan \theta = \frac{a_e}{a_n}$

$\theta = tan ^{-1} [\frac{a_e}{a_n } ]$

substituting values

$\theta = tan ^{-1} [\frac{0.808}{1.269 } ]$

$\theta = tan ^{-1} [0.636 ]$

$\theta = 32.5 6^o$

5 0

3 years ago