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Stella [2.4K]
3 years ago
14

An arrow is launched with an initial velocity of 7.6 m/s at an angle of 42 degrees above the horizontal. What is the horizontal

component of the velocity at the highest point of the trajectory?
Physics
1 answer:
AleksAgata [21]3 years ago
5 0

Explanation:

At any point on the arrow's trajectory, the horizontal component of the velocity is the same. Therefore, the horizontal component of the velocity at the top of its trajectory is

v_x = v_{0x} = v_0\cos{42°} = (7.6\:\text{m/s})\cos{42°}

\:\:\:\:\:=5.6\:\text{m/s}

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A gecko crawls vertically up and down a wall. Its motion is shown on the following graph of vertical position yyy vs. time ttt.
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3 years ago
For a moon orbiting its planet, rp is the shortest distance between the moon and its planet andra is the longest distance betwee
Natasha2012 [34]

Answer: D. 0.57

Explanation:

The formula to calculate the eccentricity e of an ellipse is (assuming the moon's orbit in the shape of an ellipse):

e=\frac{r_{a}-r_{p}}{r_{a}+r_{p}}

Where:

r_{a} is the apoapsis (the longest distance between the moon and its planet)

r_{p}=0.27 r_{a} is the periapsis (the shortest distance between the moon and its planet)

Then:

e=\frac{r_{a}-0.27 r_{a}}{r_{a}+0.27 r_{a}}

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e=0.57 This is the moon's orbital eccentricity

3 0
4 years ago
Assume that the position vector of A is r=i+j+k . Determine the moment about the origin O if the force F=(1)i+(0)j+(5)k . The mo
ddd [48]

Answer:

M₀ = 5i - 4j - k

Explanation:

Using the cross product method, the moment vector(M₀) of a force (F) is about a given point is equal to cross product of the vector A from the point (r) to anywhere on the line of action of the force itself. i.e

M₀ = r x F

From the question,

r = i + j + k

F = 1i + 0j +  5k

Therefore,

M₀ = (i + j + k) x (1i + 0j +  5k)

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M₀ = 5i - 4j - k

Therefore, the moment about the origin O of the force F is

M₀ = 5i - 4j - k

3 0
3 years ago
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5 0
3 years ago
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umka21 [38]

The answer is to the ground.


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8 0
3 years ago
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