2 years ago

Why is it possible to see the moon during the day?

1 answer:

6 0

It depends on the brightness of its light—which is really just the sun's light reflecting off of the moon's surface—and the earth's rotation. In order for the moon to be visible in the sky, it needs to be above the horizon.

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If y^2= 3.249 x 10^-11, y = ?

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Answer:

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E Notation: 3.45e5

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A train increases its speed steadily from 10 m/s to 20 m/s in

rewona [7]

Answer:

15m/s

Explanation:

add the two speeds and divide by 2

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Consider the three dip1acement vectors A = (3i - 3j) m, B = (i-4j) m, and C = (-2i + 5j) m. Use the component method to determin

tia_tia [17]

Answer with Explanation:

We are given that

A=3i-3j m

B=i-4 j m

C=-2i+5j m

a. $D=A+B+C$

$D=3i-3j+i-4j-2i+5j$

$D=2i-2j$

Compare with the vector r=xi+yj

We get x=2 and y=-2

Magnitude= $\mid D\mid=\sqrt{x^2+y^2}=\sqrt{(2)^2+(-2)^2}=2\sqrt 2$ units

By using the formula $\mid r\mid=\sqrt{x^2+y^2}$

Direction: $\theta=tan^{-1}\frac{y}{x}$

By using the formula

Direction of D: $\theta=tan^{-1}(\frac{-2}{2})=tan^{-1}(-1)=tan^{-1}(-tan45^{\circ})=-45^{\circ}$

b.E=-A-B+C

$E=-3i+3j-i+4j-2i+5j$

$E=-6i+12j$

$\mid E\mid=\sqrt{(-6)^2+(12)^2}=13.4$ units

Direction of E= $\theta=tan^{-1}(\frac{12}{-6}=tan^{-1}(-2)=-63.4^{\circ}$

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150J of heat energy

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Btu/(lb-°F) J/(g-°C i mean this is the correct answer

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The rate of a reaction is the speed at which products form or reactants disappear. T/F

Natali [406]

Answer: TRUE (:

Explanation:

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