O2 is a molecule.
1 answer:
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Answer:
The plane is 2353.7 mi from the starting position.
Explanation:
Please, see the attached figure for a graphic representation of the problem.
We have 2 displacement vectors "a" and "b" and a vector "c" that is the sum of vectors "a" plus "b" (c = a + b). The module of "c" will be the distance of the plane from the starting point.
vector a = (xa, ya)
vector b = (xb, yb)
where “xa” and “xb” are the horizontal components of the vectors “a” and “b” respectively and “ya” and “yb” are the vertical components of each vector.
Then, the vector c = a + b will be:
c = (xa + xb, ya + yb)
The module of a vector is calculated using the following expression for a vector “v”:
module of v =
Then, the module of c will be:
module of c = = distance from starting point
Then, we have to find the components of vectors “a” and “b”
The distance traveled during the first 1.5 hours of the trip is the module of the vector “a”. Then:
module of a = = distance traveled during the first 1.5 hours.
The distance can be calculated using the equation of the position of an object moving in a straight line at constant speed:
x = x0 + v * t
where
x = position at time t
x0 = initial position
v = speed
t = time
Considering x0 as the starting point (x0 = 0)
x = 675 mi/h * 1.5 h = 1012.5 mi
Then:
module of a = = 1012. 5 mi
Since the plane moves only on the horizontal (see figure), the "y" component of the vector, "ya", will be 0.
Then:
(1012.5 mi)² = xa²
xa = 1012. 5mi
a = (1012.5 mi, 0)
In the same way, we have fo find the components of the vector “b”. The module of “b” will be the distance traveled during this part of the flight:
module of b = = x = x0 + v * t
Considering x0 as the point at which the plane turns (x0 = 0)
x = 675 mi / h * 2 h = 1350 mi
Using trigonometry, we can calculate xb and yb (see figure):
sin angle = opposite / hypotenuse
cos angle = adjacent / hypotenuse
In this case:
opposite = yb
adjacent = xb
hypotenuse = module of “b”
Then:
sin 10° = yb / module of “b”
sin 10° * module of “b” = yb
In the same way:
cos 10° * module of “b” = xb
Since module of “b” = 1350 mi
xb = 1329.5 mi
yb = 234.4 mi
b = (1329.5 mi, 234.4 mi)
The vector c = a+b can now be calculated:
c = (xa + xb, ya + yb)
c =(1012.5 mi + 1329.5 mi, 0 mi + 234.4 mi) = (2342 mi, 234.4 mi)
The module of c will be:
module of c = = 2353.7 mi
The plane is 2353.7 mi from the starting position.
Complete question:
if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.
Answer:
The mutual force between the two point charges is 319.64 N
Explanation:
Given;
distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m
value of the charges, q₁ and q₂ = 2 μC and - μ4 C
Apply Coulomb's law;
where;
F is the force of attraction between the two charges
|q₁| and |q₂| are the magnitude of the two charges
r is the distance between the two charges
k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²
Therefore, the mutual force between the two point charges is 319.64 N