A :-) A.a.) Given - u = 0.00 m/s
v = 10.0 m/s
t = 1.5 sec
m = 75 kg
Solution -
a = v - u by t
a = 10 - 0 by 1.5
a = 10 by 1.5
a = 6.6 m/s^2
A.b.) sorry ! I don’t no how to do this question
Modern space suits augment the basic pressure garment with a complex system of equipment and environmental systems designed to keep the wearer comfortable, and to minimize the effort required to bend the limbs, resisting a soft pressure garment's natural tendency to stiffen against the vacuum. A self-contained oxygen supply and environmental control system is frequently employed to allow complete freedom of movement, independent of the spacecraft.
Three types of spacesuits exist for different purposes: IVA (intravehicular activity), EVA (extravehicular activity), and IEVA (intra/extravehicular activity). IVA suits are meant to be worn inside a pressurized spacecraft, and are therefore lighter and more comfortable. IEVA suits are meant for use inside and outside the spacecraft, such as the Gemini G4C suit. They include more protection from the harsh conditions of space, such as protection from micrometeorites and extreme temperature change. EVA suits, such as the EMU, are used outside spacecraft, for either planetary exploration or spacewalks. They must protect the wearer against all conditions of space, as well as provide mobility and functionality.
Flux=q/e0
e0=8.85*10^-12
!!! Cube has 6 sides, so flux through one side is equal to total flux/6
Ans=9.11*10^(-6)/((8.85*10(-12))*6)=(approximately)1.72*10^5Nm^2/C
The molarity of 10% CaCl2 is 0.9%
concentration of the given salt CaCl₂ = 10%
Density of a solution = 1.0835 g/cm³
Volume = m / d
= 100 / 1.0835
= 92.29 litres
Density = mass / volume
1.0835 × 92.29 = mass
mass = 99.99 gram
Thus the molarity can be calculated by = moles of solute / volume of solution multiplied by 100
= 0.9008/ 92.29 X 100 %
= 0.009 X 100 %
= 0.9 %
The molarity of 10% CaCl2 is 0.9%
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A magnetic domain is a group of atoms aligns with magnetic poles. Domains are usually <span>light and dark stripes visible within each grain.</span>
