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Juliette [100K]
3 years ago
6

What does the word invulnerable mean​

Physics
1 answer:
nikitadnepr [17]3 years ago
7 0
Can’t be harmed/hurt.
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Somebody please explain how to solve this. Thanks in advance!
Jobisdone [24]

work done is product of force and displacement of point of application of force

so here we have to check the product of force and displacement both

Now we will put the least to maximum work in the following order

1. -A man exerts strenuous effort in pushing a stationary wall

2. -A flea pushes a speck of dirt 1 cm

3. -A farmer pushes a 2 kg wheelbarrow 20 m

4. -A cannon launches a 3 kg cannonball a distance of 200

5. -A 2000 kg car travels 400 m down a road

6. -Space shuttle Atlantis launches from the ground into near-Earth orbit

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3 years ago
In which year was the first Badminton game played in the Olympics
Kisachek [45]
Year 1972
if I'm not wrong :)
4 0
3 years ago
Read 2 more answers
A 5.45-g combustible sample is burned in a calorimeter. the heat generated changes the temperature of 555 g of water from 20.5°c
Y_Kistochka [10]
Given:
m = 555 g, the mass of water in the calorimeter
ΔT = 39.5 - 20.5 = 19 °C, temperature change
c = 4.18 J/(°C-g), specific heat of  water

Assume that all generated heat goes into heating the water.
Then the energy released is
Q = mcΔT
    = (555 g)*(4.18 J/(°C-g)*(19 °C)
    = 44,078.1 J
    = 44,100 J (approximately)

Answer:  44,100 J

3 0
3 years ago
A sealed container holding 0.0255 L of an ideal gas at 0.981 atm and 65 ∘ C is placed into a refrigerator and cooled to 41 ∘ C w
user100 [1]

Answer:

0.911 atm

Explanation:

In this problem, there is no change in volume of the gas, since the container is sealed.

Therefore, we can apply Gay-Lussac's law, which states that:

"For a fixed mass of an ideal gas kept at constant volume, the pressure of the gas is proportional to its absolute temperature"

Mathematically:

p\propto T

where

p is the gas pressure

T is the absolute temperature

For a gas undergoing a transformation, the law can be rewritten as:

\frac{p_1}{T_1}=\frac{p_2}{T_2}

where in this problem:

p_1=0.981 atm is the initial pressure of the gas

T_1=65^{\circ}+273=338 K is the initial absolute temperature of the gas

T_2=41^{\circ}+273=314 K is the final temperature of the gas

Solving for p2, we find the final pressure of the gas:

p_2=\frac{p_1 T_2}{T_1}=\frac{(0.981)(314)}{338}=0.911 atm

3 0
4 years ago
How physical science can be truth?
JulsSmile [24]
Because it is physically true day really know that by just knowing it very well
4 0
3 years ago
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