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Delicious77 [7]

3 years ago

14

A car left point a at 7:30 am and arrived at point b, 162 miles away at 10:30 am. what was its average speed in miles per hour?5

3545556skip

1 answer:

Reika [66]3 years ago

3 0

Dddddddddddddddddddddddddddddddddddddddddddddddddddddd

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laila [671]

3b. No

4a. 50N

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8 0

2 years ago

The intensity of the radiation from the Sun measured on Earth is 1360 W/m2 and frequency is f = 60 MHz. The distance between the

Mama L [17]

Answer: (a) power output = 3.85×10²⁶W

(b). There is no relative change in power as it is independent from frequency

(c). 590 W/m²

Explanation:

given Radius between earth and sun to be = 1.50 × 10¹¹m

Intensity of the radiation from the sun measured on earth to be = 1360 W/m²

Frequency = 60 MHz

(a). surface area A of the sun on earth is = 4πR²

substituting value of R;

A = 4π(.50 × 10¹¹)² = 2.863 10²³×m²

A = 2.863 10²³×m²

now to get the power output of the sun we have;

<em>P </em>sun = <em>I </em><em>sun-earth </em><em>A </em><em>sun-earth</em>

where A = 2.863 10²³×m², and <em>I </em> is 1360 W/m²

<em>P </em>sun = 2.863 10²³ × 1360

<em>P </em>sun = 3.85×10²⁶W

(c). surface area A of the sun on mars is = 4πR²

now we substitute value of 2.28 ×10¹¹ for R sun-mars, we have

A sun-mars = 4π(2.28× 10¹¹)²

A sun-mars = 6.53 × 10²³m²

now to calculate the intensity of the sun;

<em>I </em><em>sun-mars = </em><em>P </em>sun / A sun-mars

where <em>P </em>sun = 3.85×10²⁶W and A sun-mars = 6.53 × 10²³m²

<em>I </em><em>sun-mars = </em>3.85×10²⁶W / 6.53 × 10²³m²

<em>I </em><em>sun-mars = </em>589.6 ≈ 590 W/m²

<em>I </em><em>sun-mars = </em>590 W/m²

6 0

3 years ago

How did France and Spain use what they learned during the American revolution to benefit their own countries

alexgriva [62]

Spain declared war on Britain as an ally of France, itself an ally of the American colonies, and provided supplies and munitions to the American forces. Beginning in 1776, it jointly funded Roderigue Hortalez and Company, a trading company that provided critical military supplies.

4 0

3 years ago

Raindrops acquire an electric charge as they fall. Suppose a 2.4-mm-diameter drop has a charge of +18 pC, fairly typical values.

muminat

To solve this problem we will apply the concepts related to the potential, defined from the Coulomb laws for which it is defined as the product between the Coulomb constant and the load, over the distance that separates the two objects. Mathematically this is

$V = \frac{kq}{r}$

k = Coulomb's constant

q = Charge

r = Distance between them

$q = 18 pC \rightarrow q = 1.8*10^-11 C$

$d = 2.4mm \rightarrow r = 1.2 mm = 1.2*10^-3 m$

Replacing,

$V = \frac{kq}{r}$

$V = \frac{ (9*10^9)*(1.8*10^{-11})}{(1.2*10^{-3})}$

$V = 135 V$

Therefore the potential at the surface of the raindrop is 135 V

3 0

3 years ago

When a rubber ball is thrown against a wall, energy is transferred and transformed. Select the

Nikolay [14]

Answer:

B. When the ball is released, the thrower's arm transfers its energy to the ball.

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3 years ago