The most probably answer among the presented options would be the one that says "road map is to passenger".
Let us establish first that we do not know what a thesis statement means and that we are just taking the statement as it is without any prior knowledge to the hard word present. The key to this would be the statement "it lets her know where the writer will take her." Among the ones presented as options, only the road map fits the description. Thus, the answer being road map is to passenger. The car is subject to the driver's commands thus the driver is the one guiding the car, and the student follows what is written in the study guide.
potential is equal to mgh
we know that mass = 1220 kg
g=9.8 0r 10
400=1220*9.8*x
400=11956x
x=400/11956
Answer:
11.962337 × 10^-4 N
Explanation:
Given the following :
Length L = 11.8
Charge = 29nC = 29 × 10^-9 C
Linear charge density λ = 1.4 × 10^-7 C/m
Radius (r) = 2cm = 2/100 = 0.02 m
Using the relation:
E = 2kλ/r ; F =qE
F = 2kλq/L × ∫dr/r
F = 2*k*q*λ/L × (In(0.02 + L) - In(0.02))
2*k*q*λ/L = [2 × (9 * 10^9) * (29 * 10^9) * (1.4 * 10^-7)]/ 0.118] = 6193.2203 × 10^(9 - 9 - 7) = 6193.2203 × 10^-7 = 6.1932203 × 10^-4
In(0.02 + 0.118) - In(0.02) = In(0.138) - In(0.02) = 1.9315214
Hence,
(6.1932203 × 10^-4) × 1.9315214 = 11.962337 × 10^-4 N
Answer:
t = 1.4[s]
Explanation:
To solve this problem we must use the principle of conservation of linear momentum, which tells us that momentum is conserved before and after applying a force to a body. We must remember that the impulse can be calculated by means of the following equation.
where:
P = impulse or lineal momentum [kg*m/s]
m = mass = 50 [kg]
v = velocity [m/s]
F = force = 200[N]
t = time = [s]
Now we must be clear that the final linear momentum must be equal to the original linear momentum plus the applied momentum. In this way we can deduce the following equation.
where:
m₁ = mass of the object = 50 [kg]
v₁ = velocity of the object before the impulse = 18.2 [m/s]
v₂ = velocity of the object after the impulse = 12.6 [m/s]
![(50*18.2)-200*t=50*12.6\\910-200*t=630\\200*t=910-630\\200*t=280\\t=1.4[s]](https://tex.z-dn.net/?f=%2850%2A18.2%29-200%2At%3D50%2A12.6%5C%5C910-200%2At%3D630%5C%5C200%2At%3D910-630%5C%5C200%2At%3D280%5C%5Ct%3D1.4%5Bs%5D)
I hope this helps! I added an example but there are many more!
