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Nata [24]
4 years ago
9

One alternative to burning non-renewable fuels, which harm the environment, could be to burn

Physics
1 answer:
marissa [1.9K]4 years ago
8 0
Global warming, acid rain and general air pollution from car exhaust occurs from burning fossil fuels. The global warming occurs because of a "green house effect" where sunlight is trapped inside our atmosphere due to air pollution. Acid rain occurs when the products of combustion, sulfur dioxide, nitrous dioxide and nitrogen dioxide mix with rain water.

Each of these impacts is very extensive. Each fossil fuel has individual impacts related to their use
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7. At room temperature, a substance with a melting
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The first step in responding to any sports injury is to __________.
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A person throws a stone from the corner edge of a building. The stone's initial velocity is 28.0 m/s directed at 43.0° above the
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The stone's acceleration, velocity, and position vectors at time t are

\mathbf a(t)=-g\,\mathbf j

\mathbf v(t)=v_{i,x}\,\mathbf i+\left(v_{i,y}-gt\right)\,\mathbf j

\mathbf r(t)=v_{i,x}t\,\mathbf i+\left(y_i+v_{i,y}t-\dfrac g2t^2\right)\,\mathbf j

where

g=9.80\dfrac{\rm m}{\mathrm s^2}

v_{i,x}=\left(28.0\dfrac{\rm m}{\rm s}\right)\cos43.0^\circ\approx20.478\dfrac{\rm m}{\rm s}

v_{i,y}=\left(28.0\dfrac{\rm m}{\rm s}\right)\sin43.0^\circ\approx19.096\dfrac{\rm m}{\rm s}

and y_i is the height of the building and initial height of the rock.

(a) After 6.1 s, the stone has a height of 5 m. Set the vertical component (\mathbf j) of the position vector to 5 m and solve for y_i:

5\,\mathrm m=y_i+\left(19.096\dfrac{\rm m}{\rm s}\right)(6.1\,\mathrm s)-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.1\,\mathrm s)^2

\implies\boxed{y_i\approx70.8\,\mathrm m}

(b) Evaluate the horizontal component (\mathbf i) of the position vector when t=6.1\,\mathrm s:

\left(20.478\dfrac{\rm m}{\rm s}\right)(6.1\,\mathrm s)\approx\boxed{124.92\,\mathrm m}

(c) The rock's velocity vector has a constant horizontal component, so that

v_{f,x}=v_{i,x}\approx20.478\dfrac{\rm m}{\rm s}

where v_{f,x}

For the vertical component, recall the formula,

{v_{f,y}}^2-{v_{i,y}}^2=2a\Delta y

where v_{i,y} and v_{f,y} are the initial and final velocities, a is the acceleration, and \Delta y is the change in height.

When the rock hits the ground, it will have height y_f=0. It's thrown from a height of y_i, so \Delta y=-y_i. The rock is effectively in freefall, so a=-g. Solve for v_{f,y}:

{v_{f,y}}^2-\left(19.096\dfrac{\rm m}{\rm s}\right)^2=2(-g)(-124.92\,\mathrm m)

\implies v_{f,y}\approx-53.039\dfrac{\rm m}{\rm s}

(where we took the negative square root because we know that v_{f,y} points in the downward direction)

So at the moment the rock hits the ground, its velocity vector is

\mathbf v_f=\left(20.478\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(-53.039\dfrac{\rm m}{\rm s}\right)\,\mathbf j

which has a magnitude of

\|\mathbf v_f\|=\sqrt{\left(20.478\dfrac{\rm m}{\rm s}\right)^2+\left(-53.039\dfrac{\rm m}{\rm s}\right)^2}\approx\boxed{56.855\dfrac{\rm m}{\rm s}}

(d) The acceleration vector stays constant throughout, so

\mathbf a(t)=\boxed{-g\,\mathbf j}

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