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Amanda [17]
2 years ago
10

Calculate the amount of air in a room 6m long, 5m wide and 3mm high.​

Physics
2 answers:
sattari [20]2 years ago
6 0

Answer:

0.09kg of air

Explanation:

The dimensions of the room are given

change the height to meters by dividing it by thousand.

For the volume multiplying the length,width and height (all should be in the same unit most suitable being meters).

Volume refers to the amount of space inside a box or a object.

The amount of air is equal to the volume.

AlexFokin [52]2 years ago
3 0

Answer:

90 m^3

Explanation:

Volume of the room:

    6 m * 5 m * 3 m         =  90 m^3   <=====( I changed 3mm to 3 m)

<u>if  3mm is not a typo mistake </u>

<u>  volume becomes     ( 3 mm = .003 m) </u>

<u>       6 m * 5 m * .003 m   = .09 m^3   ( though unlikely )</u>

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When the daughter nucleus produced in a radioactive decay is itself unstable, it will eventually decay and form its own daughter
Dmitrij [34]

Answer:

4.981 MeV

Explanation:

The quantity of energy Q can be calculated using the formula

Q = (mass before - mass after) × c²

Atomic Mass of thorium = 232.038054 u, atomic of Radium = 228.0301069 u and mass of Helium = 4.00260. The difference of atomic number and atomic mass  between the thorium and radium ( 232 - 228)  and ( 90 - 88)  show α particle was emitted.

1 u = 931.494 Mev/c²

Q = (mass before - mass after) × c²

Q = ( mass of thorium - ( mass of Radium + mass of Helium ) )× c²

Q = 232.038054 u - ( 228.0301069 + 4.00260) × c²

Q = 0.0053471 u × c²

replace 1 u = 931.494 MeV/ c²

Q = 0.0053471 × c² × (931.494 MeV / c²)

cancel c²  from the equation

Q = 0.0053471 × 931.494 MeV = 4.981 MeV

3 0
3 years ago
4. Which of the following would be a good reference point to describe the motion of a dog?
saul85 [17]

ANOTHER RUNNING DOG

Explanation:

In the given question it is to find a suitable reference point to describe  the motion of dog. Here I could suggest that it is better to compare the dog with  another running dog to create the relative speed difference to get a reliable motion variation.

Because the motion of dog is in the linear with respect to the another dog and to the acceleration produced by the dog in the required interval is easy to calculate with respect to  another dog which is already in motion.

Hence, I suggest that Motion of dog can be analysed better by analyse the motion variation of dog with  another dog running.

4 0
3 years ago
How heavy is an object that displaces 400N of water in a pool?
julia-pushkina [17]
All we can say is that the object's volume is about 41 liters. That's the same as the volume of water displaced.

We can't say anything about the object's weight. There is no direct connection between the weight of the object and the weight of the water it displaces.
7 0
3 years ago
If one object is 103 km away and a second object is 106 km away, one could say that the second object is _____ times further awa
vaieri [72.5K]

Answer:

1.03

Explanation:

\frac{object_{second}}{object_{first}} = \frac{106}{103} = 1.02912621359

Round to three significant digits

1.03

7 0
3 years ago
The drawing shows two situations in which charges are placed on the x and y axes. They are all located at the same distance of 5
ra1l [238]

Answer:

For situation (a)

net charge E = E₊₂ + E₋₅ + E₋₃

E =  K(q/d²)

where K = 8.99e9

d = 5.7cm = 5.7e-2m

Therefore,

E₊₂(x) = K(q/d²) = (8.99e9)× ((2.0e-6)÷(5.7e-2)) = 3.15e5(+x)

E₋₅(y) = K(q/d²) = (8.99e9)× ((5.0e-6)÷(5.7e-2)) =  7.88e5(+y)

E₋₃(x) = K(q/d²) = (8.99e9)× ((3.0e6)÷(5.7e-2)) =  4.73e5(+x)

thus

E = E₊₂ + E₋₅ + E₋₃

= 3.15e5(x) + 7.88e5(y) + 4.73e6(x)

= 7.88e6(x) + 7.88e6(y)

use Pythagorean theorem

I <em>E </em>I  = \sqrt{(7.89e5)^{2}  + (7.89e5)^{2}} =  1.242e6\frac{N}{C}

∅ = tan^{-1}(\frac{7.88e5}{7.88e5} ) = tan^{-1}(1) = 45°

Thus for (a) net magnitude =  1.115e6\frac{N}{C} @ 45° above +x axis

for situation (b)

net charge E = E₊₄ + E₊₁ + E₋₁ + E₊₆

E₊₄(x) = K(q/d²) = (8.99e9)× ((4.0e-6)÷(5.7e-2)) = 6.30e5(+x)

 E₊₁(y) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(-y)

E₋₁(x) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(+x)

E₊₆(y) = K(q/d²) = (8.99e9)× ((6.0e-6)÷(5.7e-2)) = 9.46e5(+y)

thus,

E = E₊₄ + E₊₁ + E₋₁ + E₊₆

= 6.30e5(x) - 1.58e5(y) + 1.58e5(x) + 9.46e5(y)

= 7.88e5(x) + 7.88e5(y)

use Pythagorean theorem

I <em>E </em>I  = \sqrt{(7.88e5)^{2}  + (7.88e5)^{2}} =  1.242e6\frac{N}{C}

∅ = tan^{-1}(\frac{7.88e5}{7.88e5} ) = tan^{-1}(1) = 45°

Thus for (a) and (b) the net magnitude =  1.242e6\frac{N}{C} @ 45° above +x axis

Explanation:

I attached a sample image, i hope that corresponds to your question

5 0
3 years ago
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