Answer:
a)![V=\dfrac{5.3}{P}](https://tex.z-dn.net/?f=V%3D%5Cdfrac%7B5.3%7D%7BP%7D)
b)
.
Explanation:
Given that
Boyle's law
P V = Constant ,at constant temperature
a)
Given that
![P_1=50KPa](https://tex.z-dn.net/?f=P_1%3D50KPa)
![V_1=0.106m^3](https://tex.z-dn.net/?f=V_1%3D0.106m%5E3)
We know that for PV=C
![P_1V_1=P_2V_2=PV](https://tex.z-dn.net/?f=P_1V_1%3DP_2V_2%3DPV)
Now by putting the values
PV= 50 x 0.106
![V=\dfrac{5.3}{P}](https://tex.z-dn.net/?f=V%3D%5Cdfrac%7B5.3%7D%7BP%7D)
Where P is in KPa and V is in ![m^3](https://tex.z-dn.net/?f=m%5E3)
b)
PV= C
Take ln both sides
So ![\ln(PV)=\ln C](https://tex.z-dn.net/?f=%5Cln%28PV%29%3D%5Cln%20C)
lnP + lnV =lnC ( C is constant)
By differentiating
![\dfrac{dP}{P}+\dfrac{dV}{V}=0](https://tex.z-dn.net/?f=%5Cdfrac%7BdP%7D%7BP%7D%2B%5Cdfrac%7BdV%7D%7BV%7D%3D0)
So
![\dfrac{dP}{dV}=-\dfrac{P}{V}](https://tex.z-dn.net/?f=%5Cdfrac%7BdP%7D%7BdV%7D%3D-%5Cdfrac%7BP%7D%7BV%7D)
When P= 50 KPa
![\dfrac{dP}{dV}=-\dfrac{50}{V}\ \dfrac{KPa}{m^3}](https://tex.z-dn.net/?f=%5Cdfrac%7BdP%7D%7BdV%7D%3D-%5Cdfrac%7B50%7D%7BV%7D%5C%20%5Cdfrac%7BKPa%7D%7Bm%5E3%7D)
It indicates the slope of PV=C curve.
It unit is
.
Or we can say that
.
It is true that a physical change occurs when a material changes shape or size, but the composition of the material does not change. The correct answer is True.
Yes! Fossils, The outlines of the continents and geological features .
Answer:
D. Newton's first law
Explanation:
Newton's first law of inertia says that an object will remain how it is, unless affected by an outside force. In this case, the plates want to remain stationary(not moving). Therefore, if you pull the table cloth fast enough, the force of friction produced will be small enough so that the Inertia of the plates will overcome the force of friction.
Answer:
mb = 3.75 kg
Explanation:
System of forces in balance
ΣFx =0
ΣFy = 0
Forces acting on the box
T₁ : Tension in string 1 ,at angle of 50° with the horizontal on the left
T₂ = 40 N : Tension in string 2, at angle of 75° with the horizontal on the right.
Wb :Weightt of the box (vertical downward)
x-y T₁ and T₂ components
T₁x= T₁cos50°
T₁y= T₁sin50°
T₂x= 30*cos75° = 7.76 N
T₂y= 30*sin75° = 28.98 N
Calculation of the Wb
ΣFx = 0
T₂x-T₁x = 0
T₂x=T₁x
7.76 = T₁cos50°
T₁ = 7.76 /cos50° = 12.07 N
ΣFy = 0
T₂y+T₁y-Wb = 0
28.98 + 12.07(cos50°) = Wb
Wb = 36.74 N
Calculation of the mb ( mass of the box)
Wb = mb* g
g: acceleration due to gravity = 9.8 m/s²
mb = Wb/g
mb = 36.74 /9.8
mb = 3.75 kg