A diagram of a closed circuit with power source on the left labeled 12 V, a resistor on the top labeled 10 Ohms, a resistor on t
he right labeled 8.0 Ohms and a resistor on the bottom labeled 6.0 Ohms.
Use the circuit diagram to determine the voltage drop across the 8 Ω resistor.
Answer is 4
Use the circuit diagram to determine the voltage drop across the 8 Ω resistor.
Answer is 4
2 answers:
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Answer:
magnitude=34.45 m
direction=
Explanation:
Assuming the initial point P1 of this vector is at the origin:
P1=(X1,Y1)=(0,0)
And knowing the other point is P2=(X2,Y2)=(19.5,28.4)
We can find the magnitude and direction of this vector, taking into account a vector has a initial and a final point, with an x-component and a y-component.
For the magnitude we will use the formula to calculate the distance between two points:
(1)
(2)
(3)
(4) This is the magnitude of the vector
For the direction, which is the measure of the angle the vector makes with a horizontal line, we will use the following formula:
(5)
(6)
(7)
Finding :
(8)
(9) This is the direction of the vector
Answer:
The arrow will hit 112.07 m from the point of release.
Explanation:
The equation for the position of an object in a parabolic movement is as follows:
r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)
Where:
x0 = initial horizontal position
v0 = initial velocity
α = launching angle
y0 = initial vertical position
t = time
g = acceleration due to gravity
We know that at the final time the y-component of the vector "r" (see figure") is -1.88 m. The x-component of that vector will be the horizontal distance traveled by the arrow. Using the equation of the y-component of "r", we can obtain the final time and with that time we can calculate the value of the x-component (horizontal distance).
Then:
y = y0 + v0 · t · sin α + 1/2 · g · t²
Since the origin of the frame of reference is located at the point where the arrow is released, y0 = 0. Notice that the angle α = 26° + 15° = 41° ( see figure)
-1.88 m = 33 m/s · sin 41° · t - 1/2 · 9.8 m/s² · t² (g is downward)
0 = -4.9 m/s² · t² + 33 m/s · sin 41° · t + 1.88 m
Solving the quadratic equation:
t = 4.5 s ( the negative value is discarded)
Now, with this time we can calculate the horizontal distance:
x = x0 + v0 · t · cos α (x0 = 0, the same as y0)
x = 33 m/s · 4.5 s · cos 41° = 112.07 m
