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MAXImum [283]
3 years ago
11

How many atoms are present in 3 molecules of chromium

Physics
1 answer:
iren2701 [21]3 years ago
5 0
Answer:

1,8066 x 1024 atoms

Explanation:

1 mole of chromium contains 6,022 x 1023 atoms

so

3 moles of chromium contains 3 times as many atoms

3 x 6,022 x 1023 = 1,8 066 x 1024

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During nuclear fission and fusion, matter that seems to disappear is actually converted into
iris [78.8K]

Answer:

Energy (I need one more brainlist can i has?)

Explanation:

- Nuclear fusion occurs when two light nuclei fuse together into a heavier nucleus

- Nuclear fission occurs when a heavy, unstable nucleus breaks apart into two or more lighter nuclei

In both processes, the mass of the products is always smaller than the mass of the initial nuclei. This means that part of the initial mass has been converted into something else: into energy, which is released in the process.

The amount of energy released in the process can be calculated by using the famous Einstein's equivalence:

where m is the difference between the mass of the product and the initial mass of the nuclei, and c is the speed of light.

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When the dry-bulb reading of a thermometer is 20°C and the wet-bulb reading is 11°C, the relative humidity is approximately
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Explanation:

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2 years ago
A skydiver has jumped out of a plane and is falling faster and faster. what forces are present in this situation
kompoz [17]
Gravity and air resistance 

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3 years ago
NASA is designing a Mars-lander that will enter the Martian atmosphere at high speed. To land safely it must slow to a constant
Viktor [21]

Answer:

a) maximum mass of the Mars lander to ensure it can land safely is 200 kg

b) area of the parachute required is 480 m² which is larger than 400 m²

c) area of the parachute should be 12.68 m²

Explanation:

Given the data in the question;

V = 20 m/s

A = 200 m²

drag co-efficient CD = 1.855

g = 3.71 m/s²

density of the atmospheric pressure β = 0.01 kg/m³

a. Calculate the maximum mass of the Mars lander to ensure it can land safely?

Drag force FD = 1/2 × CD × β × A × V²

we substitute

FD = 1/2 × 1.855 × 0.01 kg/m × 200 m² × ( 20 m/s )²

FD = 742 N

we know that;

FD = Fg

Fg = gravity force

Fg = mg

so

FD = mg

m = FD/g

we substitute

m = 742 N / 3.71 m/s²

m = 200 kg

Therefore, the maximum mass of the Mars lander to ensure it can land safely is 200 kg

b. The mission designers consider a larger lander with a mass of 480 kg. Show that the parachute required would be larger than 400 m²;

Given that;

M = 480 kg

Show that the parachute required would be larger than 400 m²

we know that;

FD = Fg = Mg = 480 kg × 3.71 m/s²

FD = 1780.8 N

Now, FD = 1/2 × CD × β × A × V², we solve for A

A = FD / 0.5 × CD × β × V²

we substitute

A = 1780.8  / 0.5 × 1.855 × 0.1 × (20)²

A = 1780.8 / 3.71

A = 480 m²

Therefore, area of the parachute required 480 m² which is larger than 400 m²

c. To test the lander before launching it to Mars, it is tested on Earth where g = 9.8 m/s^2 and the atmospheric density is 1.0 kg m-3. How big should the parachute be for the terminal speed to be 20 m/s, if the mass of the lander is 480 kg?

Given that;

g = 9.8 m/s²,

β" = 1 kg/m³

v" = 20 m/s

M" = 480 kg

we know that;

FD = Fg = M"g

FD = 480 kg × 9.8 m/s² = 4704 N

from the expression; FD = 1/2 × CD × β × A × V²

A = FD / 0.5 × CD × β" × V"²

we substitute

A = 4704 / 0.5 × 1.855 × 1 × (20)²

A = 4704 / 371

A = 12.68 m²

Therefore area of the parachute should be 12.68 m²

3 0
3 years ago
A 4000kg truck has a head-on inelastic collision with a 2500kg truck.
iogann1982 [59]

Answer:it could be B

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im not sure

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