Answer:
The answer is A good luck :P
(5 mi/hr) x (1hr/60min) x (10min) = 5 x 10 / 60 = <em>5/6 mile</em>
(5/6 mile) x (1,760 yd/mile) = <em>1,466 and 2/3 yards</em>
^^^^^^^^^^^^^^^^^^^^^^^^^^^
Answer:
Ro = 7.8 [g/cm³]
Explanation:
According to the principle of Archimedes, the volume of a body immersed in a liquid is equal to the volume displaced by water. That is, in this problem The displacement volume is equal to the new volume minus the original volume.
![V_{n}=110[cm^{3} ]\\V_{o}=100[cm^{3} ]\\V_{d}=110-100 = 10 [cm^{3} ]](https://tex.z-dn.net/?f=V_%7Bn%7D%3D110%5Bcm%5E%7B3%7D%20%5D%5C%5CV_%7Bo%7D%3D100%5Bcm%5E%7B3%7D%20%5D%5C%5CV_%7Bd%7D%3D110-100%20%3D%2010%20%5Bcm%5E%7B3%7D%20%5D)
We now know that density is defined as the relationship between mass and volume.

where:
Ro = density [g/cm³]
m = mass = 78 [g]
Vd = displacement volume [cm³]
![Ro = 78/10\\Ro = 7.8 [g/cm^{3} ]](https://tex.z-dn.net/?f=Ro%20%3D%2078%2F10%5C%5CRo%20%3D%207.8%20%5Bg%2Fcm%5E%7B3%7D%20%5D)
Answer:
Pressure,P=6×10^3Pa
Explanation:
The gas has an ideal gas behaviour and ideal gas equation
PV=NKT
T= V/N p/K ...eq1
Average transitional kinetic energy Ktr=1.8×10-23J
Ktr=3/2KT
T=2/3Ktr/K....eq2
Equating eq1 and 2
V/N p/K = 2/3Ktr/K
Cancelling K on both sides
P= 2/3N/V( Ktr)
Substituting the value of N/V and dividing by 10^-6 to convert cm^3 to m^3
P = 2/3 (5.0×10^20)/10^-6 × 1.8×10^-23
P= 6 ×10^3Pa