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jekas [21]
3 years ago
14

In gym class you run 22 m horizontally, then climb a rope vertically for 6.2 m. What is the direction angle of your total displa

cement, as measured from the horizontal?
Express your answer using two significant figures and include the appropriate units.
Physics
2 answers:
Paul [167]3 years ago
8 0

Answer:

\theta = 16 degree

Explanation:

Displacement in horizontal direction(d) = 22 m

displacement in vertical direction(h) = 6.2 m

As we know that total displacement is the straight line which joins the initial position and final position of the object

so here we will have

tan\theta = \frac{h}{d}

now plug in all data in it

tan\theta = \frac{6.2}{22}

\theta = tan^{-1}(0.282)

\theta = 15.74 degree

since we have to give the answer in two significant digit so it is

\theta = 16 degree

miskamm [114]3 years ago
7 0
Im 99% sure
tan^-1 (6.2/22)= 15.7º
with 2 significant figures 16º
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A particle's position is given by z(t) = −(6.50 m/s2)t2k for t ≥ 0. (Express your answer in vector form.) a. Find the particle's
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Answer:

a) z'(t) =v(t) = -13t

Now we can replace the velocity for t=1.75 s

v(1.75s) = -13*1.75 =-22.75 \frac{m}{s}

For t = 3.0 s we have:

v(3.0s) = -13*3.0 =-39 \frac{m}{s}

b) v_{avg}= \frac{z_f - z_i}{t_f -t_i}

And we can find the positions for the two times required like this:

z_f = z(3.0s) = -(6.5 \frac{m}{s^2}) (3.0s)^2=-58.5m

z_i = z(1.75s) = -(6.5 \frac{m}{s^2}) (1.75s)^2=-19.906m

And now we can replace and we got:

V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}

Explanation:

The particle position is given by:

z(t) = -(6.5 \frac{m}{s^2}) t^2, t\geq 0

Part a

In order to find the velocity we need to take the first derivate for the position function like this:

z'(t) =v(t) = -13t

Now we can replace the velocity for t=1.75 s

v(1.75s) = -13*1.75 =-22.75 \frac{m}{s}

For t = 3.0 s we have:

v(3.0s) = -13*3.0 =-39 \frac{m}{s}

Part b

For this case we can find the average velocity with the following formula:

v_{avg}= \frac{z_f - z_i}{t_f -t_i}

And we can find the positions for the two times required like this:

z_f = z(3.0s) = -(6.5 \frac{m}{s^2}) (3.0s)^2=-58.5m

z_i = z(1.75s) = -(6.5 \frac{m}{s^2}) (1.75s)^2=-19.906m

And now we can replace and we got:

V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}

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Answer:

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