Answer:
v_{4}= 80.92[m/s] (Heading south)
Explanation:
In order to calculate this problem, we must use the linear moment conservation principle, which tells us that the linear moment is conserved before and after the collision. In this way, we can propose an equation for the solution of the unknown.
ΣPbefore = ΣPafter
where:
P = linear momentum [kg*m/s]
Let's take the southward movement as negative and the northward movement as positive.

where:
m₁ = mass of car 1 = 14650 [kg]
v₁ = velocity of car 1 = 18 [m/s]
m₂ = mass of car 2 = 3825 [kg]
v₂ = velocity of car 2 = 11 [m/s]
v₃ = velocity of car 1 after the collison = 6 [m/s]
v₄ = velocity of car 2 after the collision [m/s]
![-(14650*18)+(3825*11)=(14650*6)-(3825*v_{4})\\v_{4}=80.92[m/s]](https://tex.z-dn.net/?f=-%2814650%2A18%29%2B%283825%2A11%29%3D%2814650%2A6%29-%283825%2Av_%7B4%7D%29%5C%5Cv_%7B4%7D%3D80.92%5Bm%2Fs%5D)
Answer:
93.125 × 10^(19)
Explanation:
We are told the asteroid has acquired a net negative charge of 149 C.
Thus;
Q = -149 C
charge on electron has a value of:
e = -1.6 × 10^(-19) C
Now, for us to determine the excess electrons on the asteroid, we will just divide the net charge in excess on the asteroid by the charge of a single electron.
Thus;
n = Q/e
n = -149/(-1.6 × 10^(-19))
n = 93.125 × 10^(19)
Thus, it has 93.125 × 10^(19) more electrons than protons
<span>Since d=m/v. 122 g (mass) divided by 10 cm3 (volume)=</span>12.2 g/cm<span>3
Your answer would be 12.2 g/cm3
</span>
Answer:
Up AND Down
Explanation:
a transverse wave goes up and Down