Answer:
Explanation:
F = ma
<u>Assuming</u> the 20° is angle θ measured to the horizontal
mgsinθ - μmgcosθ = ma
g(sinθ - μcosθ) = a
at constant velocity, a = 0
g(sinθ - μcosθ) = 0
sinθ - μcosθ = 0
sinθ = μcosθ
μ = sinθ/cosθ
μ = tanθ
μ = tan20
μ = 0.3639702342...
μ = 0.36
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.
Answer:
current in water = 0.924 A
Explanation:
Let the current in each row be i.
Thus, current in water is contributed by each row and total current in water becomes 140i.
We are given;
emf of each electroplaque = 0.15 V
Number of electroplaques = 5000
internal resistance = 0.25 Ω
resistance = 800Ω
Applying Kirchoff's Voltage Law to row and water, we have;
5000E − (5000r)i − 800(140i) = 0
Rearranging;
5000E = (5000r)i + 800(140i)
Plugging in the relevant values;
5000 x 0.15 = i((5000 x 0.25) + 112,000)
750 = 113,250i
i = 750/113,250
i = 0.0066 A
Recall earlier, the current in water is 140i.
Thus, current in water = 140 x 0.0066
= 0.924
Answer:
Now e is due to the ring at a
So
We say
1/4πEo(ea/ a²+a²)^3/2
= 1/4πEo ea/2√2a³
So here E is faced towards the ring
Next is E due to a point at the centre
So
E² = 1/4πEo ( e/a²)
Finally we get the total
Et= E²-E
= e/4πEo(2√2-1/2√2)
So the direction here is away from the ring
Explanation : High air pressure is generally associated with nice weather.
Low air pressure is generally associated with cloudy, rainy, or snowy weather.
Therefore, Cloudy and wet weather .