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ioda
3 years ago
9

The mass of 25cm of ivory was found to be 0.045kg. caculate the density of ivory in SI units​

Physics
1 answer:
solmaris [256]3 years ago
7 0

Answer:

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Explanation:

,ajsgfgsdfubcygduhdsbcdshyudbckjdshfyendhdndbhfdbuehundweugdjnbfcjdfhewjdnbewqwioehsajcnbsfhhgjhbgjhhbjdhgwbvdwnjjguhbwqhjbsjwqhshwbdjagdasndhsajuwqbdjasnjashdjaskxnjgjjsklx ms,dhwshjksnjdbhjssnj,

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If a 10. m3 volume of air (acting as an ideal gas) is at a pressure of 760 mm and a temperature of 27 degrees Celsius is taken t
kow [346]
We know, the ideal gas equation, 
P1V1 / T1 = P2V2 / T2

Here, P1 = 760 mm
V1 = 10 m3
T1 = 27 + 273 = 300 K

P2 = 400 mm Hg
T2 = -23 + 273 = 250 K

Substitute their values, 
760*10 / 300 = 400 * V2 / 250
25.33 * 250 = 400 * V2
V2 = 6333.333/ 400
V2 = 15.83

In short, Your Answer would be approx. 15.83 m3

Hope this helps!
7 0
3 years ago
What is the car's acceleration from 0 to 1 second?
dusya [7]
10 mph/s because there is 60 seconds in a minute then divide by 6 which is 10.
3 0
3 years ago
A motorist is driving at 20m/s when she sees that a traffic light 200m ahead has just turned red. She knows that this light stay
yuradex [85]

Answer:

5.71428571422 m/s

Explanation:

u = Initial velocity = 20 m/s

v = Final velocity

s = Displacement

a = Acceleration

Time taken = 15-1 = 14 s

Distance traveled in 1 second = 20\times 1=20\ m

s=ut+\frac{1}{2}at^2\\\Rightarrow 200-20=20\times 14+\frac{1}{2}\times a\times 14^2\\\Rightarrow a=\frac{2(180-20\times14)}{14^2}\\\Rightarrow a=-1.02040816327\ m/s^2

v=u+at\\\Rightarrow v=20-1.02040816327\times 14\\\Rightarrow v=5.71428571422\ m/s

The speed as she reaches the light at the instant it turns green is 5.71428571422 m/s

4 0
4 years ago
Bella makes the 2.5m distance to her food bowl in 9.1 seconds. What is her average velocity?
e-lub [12.9K]
  • Distance=2.5m
  • Time=9.1s

Average Velocity=Total Distance/Total Time

\\ \sf\longmapsto \dfrac{2.5}{9.1}

\\ \sf\longmapsto 0.3m/s

7 0
3 years ago
Please help me with the equations for this! Three uniform spheres are fixed at the positions shown in the diagram. ( there is a
lora16 [44]
The change in gravitational potential energy due to change in position must be the change in it's kinetic energy as the system is isolated! so find out the potential energies of the two different points!

<span>PE=−[G<span>M1</span><span>M2</span>]÷R

</span><span> Potential energy of a particle due to mass A is not affected by presence of any other mass B !</span>
7 0
3 years ago
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