Answer:
(a) 0 (b) 
Explanation:
Given that,
Mass of a supertanker, 
The engine of a generate a forward thrust of, 
(a) As the supertanker is moving with a constant velocity. We need to find the magnitude of the resistive force exerted on the tanker by the water. It is given by :
F = ma, a is the acceleration
For constant velocity, a = 0
So, F = 0
(b) The magnitude of the upward buoyant force exerted on the tanker by the water is equal to the weight of the ship.
F = mg
Hence, this is the required solution.
Answer: to achieve a stable octet of electrons in their outer shell
Explanation:
- Sodium (Na) has an atomic number of 11, and an electronic configuration of 1s2, 2s2 2p6, 3s1.
- Chlorine (CI) Sodium (Na) has an atomic number of 17, and an electronic configuration of 1s2, 2s2 2p6, 3s2 3p5.
Hence, sodium donates its single valence electron to chlorine, thereby achieving a stable octet structure of 1s2, 2s2 2p6 while chlorine accept the single electron also forming a stable octet structure of 1s2, 2s2 2p6, 3s2 3p6. Therefore, the reaction yields NaCl, an ionic compound with ionic (electrovalent) bond.
Na + Cl --> NaCl
Thus, Sodium(Na) and Chlorine (CI) want to form a lonic Bond because both acheive a completely filled outermost shell (octet structure)
Answer:
0.306mm
Explanation:
The radius of the conductor is 3mm, or 0.003m
The area of the conductor is:
A = π*r^2 = π*(.003)^2 = 2.8*10^-5 m^2
The current density is:
J = 130/2.8*10^-5 = 4.64*10^6 A/m
According to the listed reference:
Vd = J/(n*e) = 4.64*10^6 / ( 8.46*10^28 * 1.6*10^-19 ) = 0.34*10^-6 m/s = 0.34mm/s
The distance traveled is:
x = v*t = 0.34 * .90 = 0.306 mm
Answer:
The horizontal displacement of the arrow is not larger than the banana split.
Explanation:
Using y - y₀ = ut - 1/2gt², we find the time it takes the arrow to drop to the ground from the top of mount Everest.
So, y₀ = elevation of Mount Everest = 29029 ft = 29029 × 1ft = 29029 × 0.3048 m = 8848.04 m, y = final position of arrow = 0 m, u = initial vertical speed of arrow = 0 m/s, g = acceleration due to gravity = 9.8 m/s² and t = time taken for arrow to fall to the ground.
y - y₀ = ut - 1/2gt²
0 - y₀ = 0 × t - 1/2gt²
-y₀ = -1/2gt²
t² = 2y₀/g
t = √(2y₀/g)
Substituting the values of the variables, we have
t = √(2y₀/g)
= √(2 × 8848.04 m/9.8 m/s²)
= √(17696.08 m/9.8 m/s²)
= √(1805.72 s²)
= 42.5 s
The horizontal distance the arrow moves is thus d = vt where v = maximum firing speed of arrow = 100 m/s and t = 42.5 s
So, d = vt
= 100 m/s × 42.5 s
= 4250 m
= 4.25 km
Since d = 4.25 km < 7.32 km, the horizontal displacement of the arrow is not larger than the banana split.
