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kipiarov [429]
3 years ago
10

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

urface of the earth. In this problem, use 9.80 m/s2 for the acceleration due to gravity on earth.Part A
What is its mass on the earth's surface?

Part B

What is its mass on the surface of Io?

Part C

What is its weight on the surface of Io?
Physics
1 answer:
Mumz [18]3 years ago
8 0

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

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Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a larg
valentina_108 [34]

Answer:

-time it takes for the sled to come to a stop after launch of rocket = 7.244 s

-distance sled has travelled from its starting point by the time it finally comes to rest is = 234.8655 m

Explanation:

From the question, looking at the motion while accelerating, we have;

Initial velocity; u = 0 m/s

Acceleration; a = 13.5 m/s²

Time; t = 3.3 s

Let's use first equation of motion to find final velocity (v).

v = u + at

v = 0 + (13.5 × 3.3)

v = 44.55 m/s

In this forward direction, let's calculate the displacement(d1) using newton's 3rd equation of motion.

d1 = ut + ½at²

d1 = 0(3.3) + ½(13.5 × 3.3²)

d1 = 73.5075 m

Now, let's consider the motion while slowing down and our final velocity will be 0 m/s while initial velocity will now be 44.55 m/s while acceleration is 6.15 m/s².

Thus, from v = u + at, we can find the time it take for the sled to come to a stop.

Now, since it's coming to rest acceleration will be negative. Thus;

0 = 44.55 + (-6.15t)

0 = 44.55 - 6.15t

t = 44.55/6.15

t = 7.244 s

Now we want to find out how far the sled has travelled from its starting point by the time it finally comes to rest.

Thus, we'll use the equation;

v² = u² + 2as

Where s will be the second displacement which we will call d2.

Thus;

0² = 44.55² + (-2 × 6.15 × s)

0 = 1984.7025 - 12.3s

12.3s = 1984.7025

s = 1984.7025/12.3

s = 161.358

Thus, d2 = s = 161.358 m

Thus, distance sled has travelled from its starting point by the time it finally comes to rest is ;

= d1 + d2 = 73.5075 + 161.358 = 234.8655 m

4 0
4 years ago
3. As an object’s temperature increases, the ____________________ at which it radiates energy increases.
Vladimir [108]

Answer:

As an object’s temperature increases, the Rate at which it radiates energy increases.

7 0
3 years ago
Read 2 more answers
calculate the work done in kilo joules in lifting a mass of 20kg at steady velocity through a vertical height of 20m
bulgar [2K]

Answer:

\huge\boxed{\sf Work\ done = 4 kJ}

Explanation:

Since work done is in the form of potential energy, we will use the formula of potential energy here.

We know that,

<h3>P.E. = mgh </h3>

Where,

m = mass = 20 kg

g = acceleration due to gravity = 10 m/s²

h = vertical height = 20 m

So,

<h3>Work done = mgh</h3>

Work done = (20)(10)(20)

Work done = 4000 joules

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\rule[225]{225}{2}

5 0
2 years ago
How does a metamorphic rock become an igneous rock?
Oksana_A [137]

Answer:

If the newly formed metamorphic rock continues to heat, it can eventually melt and become molten (magma). When the molten rock cools it forms an igneous rock. Metamorphic rocks can form from either sedimentary or igneous rocks.

7 0
3 years ago
How much energy is required to heat 70 g of water at 20°C to boiling
choli [55]

Answer:

Q=23,430J

Explanation:

Hello,

In this case, since we compute the required energy via:

Q=mC\Delta T

Whereas m is the mass which here is 70 g, C the specific heat which for water is 4.184 J/(g°C) and ΔT is the temperature difference which is:

\Delta T=100-20=80\°C

Therefore, the energy turns out:

Q=70g*4.184\frac{J}{g\°C}*80\°C\\ \\Q=23,430J

Best regards.

3 0
3 years ago
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