Describe how the properties of the compound carbon monoxide (CO) differ from the properties of the elements in the compound.
2 answers:
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Answer:
a) 0 b) 29.9 N c) 21.7 N d) -17.4 N e) -13.0 N f) -17.4 N g) 16.9 N
h) 24.3 N θ = 44.2º
Explanation:
a) As the electric force is exerted along the line that joins the charges, due to any of the charges A or C has non-zero x-coordinates, the force has no x components either.
So, Fcax = 0
b) Similarly, as Fx = 0, the entire force is directed along the y-axis, and is going upward, due both charges repel each other.
Fyca = k*qa*qc / rac² = (9.10⁹ N*m²/C²*(2.79)*(1.07)*10⁻⁸ C²) / 9.00 m²
Fyca = 29. 9 N
c) In order to get the magnitude of the force exerted by B on C, we need to know first the distance between both charges:
rbc² = (3.00 m)² + (4.00m)² = 25.0 m²
⇒ Fbc = k*qb*qc / rbc² = (9.10⁹ N*m²/C²*(5.64)*(1.07)*10⁻⁸ C²) / 25.0 m²
⇒ Fbc = 21.7 N
d) In order to get the x component of Fbc, we need to get the projection of Fcb over the x axis, taking into account that the force on particle C is attractive, as follows:
Fbcₓ = Fbc * cos (-θ) where θ, is the angle that makes the line of action of the force, with the x-axis, so we can write:
cos θ = x/r = 4.00 / 5.00 m =
Fcbx = 21.7*(-0.8) = -17.4 N
e) The y component can be calculated in the same way, projecting the force over the y-axis, as follows:
Fcby = Fcb* sin (-θ) = 21.7* (-3.00/5.00) = -13.0 N
f) The sum of both x components gives :
Fcx = 0 + (-17.4 N) = -17.4 N
g) The sum of both y components gives :
Fcy = 29.9 N + (-13.0 N) = 16.9 N
h) The magnitude of the resultant electric force acting on C, can be found just applying Pythagorean Theorem, as follows:
Fc = √(Fcx)²+(Fcy)² = (17.4)² + (16.9)² = 24.3 N
The angle from the horizontal can be found as follows:
Ф = arc tg (16.9 / 17.4) = 44.2º
Answer:
The distance from the base of the building to the landing site is 154 m.
The total flight time is 3.5 s.
At the moment of impact, the velocity vector of the cat is v = (44 m/s, -34.3 m/s) and its magnitude is 55.8 m/s.
Explanation:
The equations for the position and velocity vectors of the cat are as follows:
r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)
v = (v0x, v0y + g · t)
Where:
r = position vector of the cat at time t.
x0 = initial horizontal position.
v0x = initial horizontal velocity.
t = time.
y0 = initial vertical position.
v0y = initial vertical velocity
g = acceleration due to gravity (-9.8 m/s² considering the upward position as positive).
v = velocity vector of the cat at time t.
Please, see the attached figure for a better understanding of the problem. Notice that the origin of the frame of reference is located at the launching point so that x0 and y0 = 0. In a horizontal launch, initially there is no vertical velocity, then, v0y = 0.
When the cat reaches the ground, the position vector of the cat will be r1 (see figure). The vertical component of r1 is -60 m and the horizontal component will be the horizontal distance traveled by the cat (r1x). Then, using the equation of the y-component of the position vector, we can obtain the time of flight and with that time we can obtain the horizontal distance traveled by the cat:
r1y = y0 + v0y · t + 1/2 · g · t²
-60 m = 0 m + 0 m/s · t - 1/2 · 9.8 m/s² · t²
- 60 m = -4.9 m/s² · t²
-60 m / - 4.9 m/s² = t²
t = 3.5 s
The cat reaches the ground in 3.5 s
Now, we can calculate the horizontal component of r1:
r1x = x0 + v0 · t
r1x = 0 m + 44 m/s · 3.5 s
r1x = 154 m
The distance from the base of the building to the landing site is 154 m.
The total flight time was already calculated and is 3.5 s.
The velocity vector of the cat when it reaches the ground will be:
v = (v0x, v0y + g · t)
v = (44 m/s, 0 m/s - 9.8 m/s² · 3.5 s)
v = (44 m/s, -34.3 m/s)
The magintude of the vector "v" is calculated as follows:
At the moment of impact, the velocity vector of the cat is v = (44 m/s, -34.3 m/s) and its magnitude is 55.8 m/s.
Answer:
Explanation:
given,
initial velocity of the ball = 20 m/s
angle of ramp = 22°
ball travel at a distance = 5 m
a) for friction less
v = 17.58 m/s
b) considering the friction
v = 17.16 m/s