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3 years ago

How do you solve 0.004 dm + 0.12508 dm?

1 answer:

7 0

0.004 of something added to 0.12508 of the same thing
adds up to 0.12908 of it.

The thing could be a glass of water, a sheet of paper,
a pound of ground beef, a gallon of gas, or a snowball.
In this problem, it just happens to be a dm.

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Please, someone help me I'm begging yall 1. it's estimated that 1 kg of body fat will provide 3.8 * 10^7 J of energy. A 67 kg mo

Leto [7]

Answer:

0.24 kg used up

Explanation:

He has a mass of 67 kg

The gravitational constant is 9.81 m/s^2

The distance upward is 3500 m

W = m*g*h

W = 67 * 9.81 * 3500

Work = 2,300,445 Joules

Work = 2300 kj

work = 2.30 * 10^6 joules in scientific notation.

Part B

He needs 4 times this amount to climb the mountain because the body is only 25% efficient in converting energy.

4*2.30 * 10^6 = 9.20 * 10^6 Joules of energy are therefore required.

The total amount in a kg of fat = 3.8 * 10^7 joules

x kg of fat is needed to provide 9.20.*10^6 joules

1 kg / (3.8 * 10^7 J ) = x kg / (9.20 * 10^6 J)

9.20 * 10 ^6 * 1 = 3.8 * 10^7 *x

9.20 * 10 ^6 / 3.8 * 10^7 = x

x = 0.24 kg of fat are needed

7 0

3 years ago

A skier descends a mountain at an angle of 35.0º to the horizontal. If the mountain is 235 m long, what are the horizontal and v

kumpel [21]

Opposite from the angle is 134.7905

Adjacent to the angle is 192.5007

6 0

4 years ago

Read 2 more answers

Venus has an average distance to the sun of 0.723 AU. In two or more complete sentences, explain how to calculate the orbital pe

Mice21 [21]

As per the question the distance of venus from sun is given as 0.723 AU

We have been asked to calculate the time period of the planet venus.

As per kepler's laws of planetary motion the square of time period of planet is directly proportional to the cube of semi major axis. mathematically

$T^{2} \alpha R^{3}$

⇒ $T^{2} = KR^{3}$ where is k is the proportionality constant

We may solve this problem by comparing with the time period of the earth . We know that time period of earth is 365.5 days

Hence $T_{1} =365.5 days$

The distance of sun from earth is taken as 1 AU i.e the mean distance of earth from sun

Hence $R_{1} =1 AU$

The distance of venus from sun is 0.723 AU i.e $R_{2} =0.723$

From keplers law we know that- $\frac{T_{1} ^{2} }{T_{2} ^{2} } =\frac{R_{1} ^{3} }{R_{2} ^{3} }$

⇒ $T_{2} ^{2} =T_{1} ^{2} *\frac{R_{2} ^{3} }{R_{1} ^{3} }$

Putting the values mentioned above we get-

$T_{2} ^{2} =50,350.132851075$

⇒ $T_{2} =\sqrt{50,350.132851075}$

⇒ $T_{2} = 224.388352752710 days.$

Hence the time period of venus is 224.388352752710 days

7 0

4 years ago

Read 2 more answers

How to calculate depth of sea whan a ship receives sound produced by its fathometer after 3.5 second

HACTEHA [7]

Answer:

2600 m

Explanation:

A fathometer produces a sound wave and then detects the echo. It takes 3.5 seconds for the echo to reach the ship, so that means it takes half the time (1.75 seconds) to reach the ocean floor.

The speed of sound in seawater is approximately 1500 m/s, so the depth of the ocean at that point is:

d = 1500 m/s × 1.75 s

d = 2625 m

Rounding to two significant figures, the depth is approximately 2600 m.

5 0

4 years ago

Read 2 more answers

Please help I have no idea

DochEvi [55]

Answer:

Explanation:

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3 0

3 years ago