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MAXImum [283]
3 years ago
10

What is the common size for air mass?

Physics
1 answer:
Serggg [28]3 years ago
5 0
Thecorrect answer is B. 

hope this helps brainliest would be cool if not its fine

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What is newton first law​
andrew11 [14]

Answer:

Newtons first law states that:

<em>If</em><em> </em><em>a</em><em> </em><em>body</em><em> </em><em>i</em><em>s</em><em> </em><em>in</em><em> </em><em>rest</em><em> </em><em>or</em><em> </em><em>motion</em><em> </em><em>in</em><em> </em><em>a</em><em> </em><em>straight</em><em> </em><em>line</em><em>,</em><em> </em><em>it</em><em> </em><em>remains</em><em> </em><em>at</em><em> </em><em>rest</em><em> </em><em>or</em><em> </em><em>at</em><em> </em><em>motion</em><em> </em><em>in</em><em> </em><em>a</em><em> </em><em>straight</em><em> </em><em>line</em><em> </em><em>with</em><em> </em><em>constant</em><em> </em><em>speed</em><em> </em><em>until</em><em> </em><em>and</em><em> </em><em>unless</em><em> </em><em>and</em><em> </em><em>external</em><em> </em><em>unbalanced</em><em> </em><em>force</em><em> </em><em>acts</em><em> </em><em>on</em><em> </em><em>it</em><em>.</em>

<em>'</em><em>This</em><em> </em><em>law</em><em> </em><em>i</em><em>s</em><em> </em><em>also</em><em> </em><em>known</em><em> </em><em>as</em><em> </em><em>the</em><em> </em><em>law</em><em> </em><em>of</em><em> </em><em>Inertia</em><em>.</em><em>'</em>

5 0
2 years ago
If a car increases its velocity from zero to 60 m/s in 10 seconds, its acceleration is A. 600 m/s2 B. 60 m/s2 C. 3 m/s2 D. 6 m/s
enot [183]

Answer:

a

Explanation:

7 0
3 years ago
Read 2 more answers
A 0.0780 kg lemming runs off a
kotegsom [21]

Answer:

5.01 J

Explanation:

Info given:

mass (m) = 0.0780kg

height (h) = 5.36m

velocity (v) = 4.84 m/s

gravity (g) = 9.81m/s^2

1. First, solve for Kinetic energy (KE)

KE = 1/2mv^2

1/2(0.0780kg)(4.84m/s)^2 = 0.91 J

so KE = 0.91 J

2. Next, solve for Potential energy (PE)

PE = mgh

(0.0780kg)(9.81m/s^2)(5.36m) = 4.10 J

so PE = 4.10 J

3. Mechanical Energy , E = KE + PE

Plug in values for KE and PE

KE + PE = 0.91J + 4.10 J = 5.01 J

4 0
3 years ago
T=2pi square root 1/g solve for g.<br> Explanation would be really helpful.
Natalija [7]

I added individual steps for clarity. Note that g must be positive if the solution is to be real.

T=2\pi \sqrt{\frac{1}{g}}=2\pi g^{-\frac{1}{2}}\\g^{-\frac{1}{2}} = \frac{T}{2\pi}\\(g^{-\frac{1}{2}})^{-2} = (\frac{T}{2\pi})^{-2}\\g = \frac{4\pi^2}{T^2}\,\,\,, g>0}

Let me know if you have any questions.

7 0
3 years ago
An 800-g block of ice at 0.00°C is resting in a large bath of water at 0.00°C insulated from the environment. After an entropy c
Allisa [31]

Answer:

Unmeltedd ice = 308.109 g

Explanation:

Gibbs Free energy:

A systems Gibbs Free Energy is defined as the free energy of the product of the absolute temperature and the entropy change less than the enthalpy change.

Therefore, G = ΔH-TΔS

where G is Gibbs Free Energy

          ΔH is enthalpy change

          T is absolute temperature

          ΔS is entropy change

Here since there is a phase change, therefore G will be 0.

∴ΔH = TΔS

Given: Temperature, T = 0°C = 273 K

           Entropy change,ΔS = 600 J/K

           Latent heat of fusion of water = 333 J/g

∴ΔH = TΔS

  ∴ΔH = 273 x 600

           = 163800 J

So this is the amount of enthalpy that will be used into melting of ice.

  ∴ΔH = mass of ice melted x latent heat of fusion of water

    Mass of ice melted = ΔH / latent heat of fusion of water

                                     = 163800 / 333

                                     = 491.891 g

This is the mass of ice melted.

And initial amount of ice is 800 g

Amount of ice left after melting = Initial amount of ice - amount of ice melted

                                                   = 800-491.891

                                                  = 308.109 g

Amount of ice remained after melting = 308.109 g

8 0
3 years ago
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