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Gnesinka [82]
3 years ago
14

Choose the appropriate words to complete the thought. A solar powered car converts _______________ energy into _________________

energy.
A) light, chemical
B) light, mechanical
C) mechanical, light
D) electrical, mechanical
Physics
1 answer:
Natalija [7]3 years ago
7 0

The best and most correct answer among the choices provided by your question is the second choice or letter C. A solar-powered car converts light energy to mechanical energy.

Solar cars use photovoltaic cells to convert sunlight into energy. Photovoltaic cells are the components in solar panels that convert the sun's energy to electricity<span>. They're made up of semiconductors, usually silicon, that absorb the light.

I hope my answer has come to your help. Thank you for posting your question here in Brainly.
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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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