A jogger travels a route that has two parts. The first is a displacement of 3 km due south, and the second involves a displaceme
1 answer:
Answer:
a) 2.41 km
b) 38.8°
Questions c and d are illegible.
Explanation:
We can express the displacements as vectors with origin on the point he started (0, 0).
When he traveled south he moved to (-3, 0).
When he moved east he moved to (-3, x)
The magnitude of the total displacement is found with Pythagoras theorem:
d^2 = dx^2 + dy^2
Rearranging:
dy^2 = d^2 - dx^2
The angle of the displacement vector is:
cos(a) = dx/d
a = arccos(dx/d)
a = arccos(3/3.85) = 38.8°
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Answer:
A) and B) are correct.
Explanation:
Let's take a look at the attached picture. Now
The total voltage across both capacitors is the same as the sum of the voltage from each device, that statement is true for any electrical device connected in series. So a) is TRUE
The equivalent capacitance is going to be:
And that value can be mathematically proven that is always less than any of the values of each capacitor. So b is TRUE
And through both capacitors flow the same current, but the amount of charge depends on the value of the capacitors, so only could be the same if the capacitors are the same value. Otherwise, don't. C) not always, so FALSE
Answer:
The rock must leave the cliff at a velocity of 28.2 m/s
Explanation:
The position vector of the rock at a time t can be calculated using the following equation:
r = (x0 + v0x · t, y0 + 1/2 · g · t²)
Where:
r = position vector at time t.
x0 = initial horizontal position.
v0x = initial horizontal velocity.
t = time.
g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).
Please, see the attached figure for a graphical description of the problem. Notice that the origin of the frame of reference is located at the edge of the cliff so that x0 and y0 = 0.
When the rock reaches the ground, the position vector will be (see r1 in the figure):
r1 = (90 m, -50 m)
Then, using the equation of the vector position written above:
90 m = x0 + v0x · t
-50 m = y0 + 1/2 · g · t²
Since x0 and y0 = 0:
90 m = v0x · t
-50 m = 1/2 · g · t²
Let´s use the equation of the y-component of the vector r1 to find the time it takes the rock to reach the ground and with that time we can calculate v0x:
-50 m = 1/2 · g · t²
-50 m = -1/2 · 9.81 m/s² · t²
-50 m / -1/2 · 9.81 m/s² = t²
t = 3.19 s
Now, using the equation of the x-component of r1:
90 m = v0x · t
90 m = v0x · 3.19 s
v0x = 90 m / 3.19 s
v0x = 28.2 m/s
