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soldier1979 [14.2K]
2 years ago
10

Our data for how much force is needed can be qualitative. This means we don’t have to know exact values for the force in Newtons

. Let’s use a scale of 1-5 to rank how much force we apply each time.
6. How many different masses should we use?

How many trials should we do for each mass?
Physics
2 answers:
kirill [66]2 years ago
8 0

Answer:

For macroscopic objects, we use units such as grams and kilograms to state their masses, but. There will not be much error if you estimate the mass of an atom by simply. to what we did for our hypothetical example, but the percentages are different: Add two atomic masses of carbon and six atomic masses of hydrogen:

Explanation:

schepotkina [342]2 years ago
3 0

Answer:We use units such as grams and kilograms to denote their masses for macroscopical objects. However. If you actually measure an atom's mass, there would be not much error. In our theory, however, percentages vary: add two carbon atomic masses and six hydrogen atomic masses:

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An automobile tire is inflated with air originally at 10.0°C and normal atmospheric pressure. During the process, the air is com
solong [7]

Answer:

(a) 3.81\times 10^5\ Pa

(b) 4.19\times 1065\ Pa

Explanation:

<u>Given:</u>

  • T_1 = The first temperature of air inside the tire = 10^\circ C =(273+10)\ K =283\ K
  • T_2 = The second temperature of air inside the tire = 46^\circ C =(273+46)\ K= 319\ K
  • T_3 = The third temperature of air inside the tire = 85^\circ C =(273+85)\ K=358 \ K
  • V_1 = The first volume of air inside the tire
  • V_2 = The second volume of air inside the tire = 30\% V_1 = 0.3V_1
  • V_3 = The third volume of air inside the tire = 2\%V_2+V_2= 102\%V_2=1.02V_2
  • P_1 = The first pressure of air inside the tire = 1.01325\times 10^5\ Pa

<u>Assume:</u>

  • P_2 = The second pressure of air inside the tire
  • P_3 = The third pressure of air inside the tire
  • n = number of moles of air

Since the amount pof air inside the tire remains the same, this means the number of moles of air in the tire will remain constant.

Using ideal gas equation, we have

PV = nRT\\\Rightarrow \dfrac{PV}{T}=nR = constant\,\,\,(\because n,\ R\ are\ constants)

Part (a):

Using the above equation for this part of compression in the air, we have

\therefore \dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow P_2 = \dfrac{V_1}{V_2}\times \dfrac{T_2}{T_1}\times P_1\\\Rightarrow P_2 = \dfrac{V_1}{0.3V_1}\times \dfrac{319}{283}\times 1.01325\times 10^5\\\Rightarrow P_2 =3.81\times 10^5\ Pa

Hence, the pressure in the tire after the compression is 3.81\times 10^5\ Pa.

Part (b):

Again using the equation for this part for the air, we have

\therefore \dfrac{P_2V_2}{T_2}=\dfrac{P_3V_3}{T_3}\\\Rightarrow P_3 = \dfrac{V_2}{V_3}\times \dfrac{T_3}{T_2}\times P_2\\\Rightarrow P_3 = \dfrac{V_2}{1.02V_2}\times \dfrac{358}{319}\times 3.81\times 10^5\\\Rightarrow P_3 =4.19\times 10^5\ Pa

Hence, the pressure in the tire after the car i driven at high speed is 4.19\times 10^5\ Pa.

8 0
2 years ago
A kangaroo jumps to a vertical height of 2.8 m. How long was it in the air before returning to earth
BaLLatris [955]
The answer would be 2.8m height on earth takes 
2.8=1/2*9.8*t^2 => <span>s = ut +1/2at^2 </span>
8 0
3 years ago
A piston of volume 0.1 m3 contains two moles of a monatomic ideal gas at 300K. If it undergoes an isothermal process and expands
seropon [69]

Answer:

the work is done by the gas on the environment -is W= - 3534.94 J (since the initial pressure is lower than the atmospheric pressure , it needs external work to expand)

Explanation:

assuming ideal gas behaviour of the gas , the equation for ideal gas is

P*V=n*R*T

where

P = absolute pressure

V= volume

T= absolute temperature

n= number of moles of gas

R= ideal gas constant = 8.314 J/mol K

P=n*R*T/V

the work that is done by the gas is calculated through

W=∫pdV=  ∫ (n*R*T/V) dV

for an isothermal process T=constant and since the piston is closed vessel also n=constant during the process then denoting 1 and 2 for initial and final state respectively:

W=∫pdV=  ∫ (n*R*T/V) dV =  n*R*T  ∫(1/V) dV = n*R*T * ln (V₂/V₁)

since

P₁=n*R*T/V₁

P₂=n*R*T/V₂

dividing both equations

V₂/V₁ = P₁/P₂

W= n*R*T * ln (V₂/V₁)  = n*R*T * ln (P₁/P₂ )

replacing values

P₁=n*R*T/V₁ = 2 moles* 8.314 J/mol K* 300K / 0.1 m3= 49884 Pa

since P₂ = 1 atm = 101325 Pa

W= n*R*T * ln (P₁/P₂ ) = 2 mol * 8.314 J/mol K * 300K * (49884 Pa/101325 Pa) = -3534.94 J

5 0
3 years ago
1. A car travels 36 miles North and then 45 miles East. How far does it
salantis [7]

Answer:

Explanation:

Displacement is the shortest distance or path between two points.

1) Displacement = √(36² + 45²) = 57.63 miles

2) Displacement = √(100² + 500²) = 509.9 meters

3) Displacement = √(60² + 40²) = 72.11 miles

4) Displacement = √(700² + 500²) = 860.23 miles

5) Displacement = 300 - 300 = 0 miles

6) Displacement = 200 + 100 = 300 miles

7) Displacement = √(650² + 650²) = 919.24 miles

8) Yes, since a distance is moved in a direction

6 0
3 years ago
PLEASE HELP ME I JUST CAN'T FIGURE THIS ONE OUT
kupik [55]
Indeed the answer is c!!
4 0
3 years ago
Read 2 more answers
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