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4 years ago

20 point physics question! :D

2 answers:

8 0

B, Light, mechanical. The solar panel takes in light energy (from the sun), and in exchange spits out mechanical energy (turns the wheels/powers the battery).

Colt1911 [192]4 years ago

3 0

Quite simple, actually,

Solar means that it's from the sun. (Sunlight into electricity)

So, answer choices C and D are eliminated. Leaving A and B.

Chemical means it's powered by gas, etc.

So it wouldn't make sense, A solar powered car converts light energy into chemical energy. The car is solar powered.

Leaving, B, as the only logical answer.

A solar powered car converts light energy into mechanical energy.

You might be interested in

Equal amounts of heat are added to equal masses of ice and copper at the same initial temperature. Which substance will have the

pychu [463]

One of the major effects of heat transfer is temperature change: heating increases the temperature while cooling decreases it. We assume that there is no phase change and that no work is done on or by the system. Experiments show that the transferred heat depends on three factors—the change in temperature, the mass of the system, and the substance and phase of the substance.

Figure a shows a copper-colored cylinder of mass m and temperature change delta T. The heat Q, shown as a wavy rightward horizontal arrow, is transferred to the cylinder from the left. To the right of this image is a similar image, except that the heat transferred Q prime is twice the heat Q. The temperature change of this second cylinder, which is also labeled m, is two delta T. This cylinder is surrounded by small black wavy lines radiating outward. Figure b shows the same two cylinders as in Figure a. The left cylinder is labeled m and delta T and has a wavy heat arrow pointing at it from the left that is labeled Q. The right cylinder is labeled two m and delta T and has a wavy heat arrow pointing to it from the left labeled Q prime equals two Q. Figure c shows the same copper cylinder of mass m and with temperature change delta T, with heat Q being transferred to it. To the right of this cylinder, Q prime equals ten point eight times Q is being transferred to another cylinder filled with water whose mass and change in temperature are the same as that of the copper cylinder.

7 0

3 years ago

A uniform solid disk and a uniform hoop are placed side by side at the top of an incline of height h. If they are released from

svlad2 [7]

Answer:

) the uniform disk has a lower moment of inertia and arrives first.

Explanation:

(a) the uniform disk has a lower moment of inertia and arrives first.

(b) Let's say the disk has mass m and radius r, and

the hoop has mass M and radius R.

disk: initial E = PE = mgh

I = ½mr², so KE = ½mv² + ½Iω² = ½mv² + ½(½mr²)(v/r)² = (3/4)mv² = mgh

m cancels, leaving v² = 4gh / 3

hoop: initial E = Mgh

I = MR², so KE = ½MV² + ½(MR²)(V/R)² = MV² = Mgh

M cancels, leaving V² = gh

Vdisk = √(4gh/3) > Vhoop = √(gh)

3 0

4 years ago

When two things are the same temperature:

Rus_ich [418]

Answer:

its obviously D if all the other says "incorrect answer" XD

Explanation:

4 0

3 years ago

A solenoid of length 0.250 m and radius 0.0250 m is comprised of 440 turns of wire. Determine the magnitude of the magnetic fiel

Bad White [126]

Answer: 0.02654

Explanation:

in the attachment

6 0

4 years ago

Read 2 more answers

To practice Problem-Solving Strategy 21.1 Coulomb's Law. Three charged particles are placed at each of three corners of an equil

Salsk061 [2.6K]

Answer:

The force felt by charge 3 is F=(-5.6*10⁻⁶,3.36⁻⁵)N

Explanation:

As the superposition principle applies to static charges, we can find the net electric force as the sum of the two forces felt by q3.

Looking at the drawing and knowing that they form an equilateral triangle of lenght 4 we can conclude that each internal angle is 60°.

So, the positions in our coordinate system are:

$r_1=(0,0)\\r_2=(4\ cos(60\°),4\ sin(60\°))\\r_3=(4,0)\\$

Now using Coulomb's force:

$F_{ij}=\frac{-kq_iq_j}{d^2}(\vv{r}_j-\vv{r}_i)$

Where d=4, q1 = -7.8*10⁻⁹C, q2 = -15.6 *10⁻⁹C, q3 = 8.0 *10⁻⁹C, k=8.98*10⁹, e0=8.8*10¹⁰:

Replacing we get 2 equations:

$F_{13}=\frac{-kq_1q_3}{d^2}(\vv{r}_1-\vv{r}_3)=\frac{-kq_1q_2}{d^2}(-0.04\ cos(60\°),-0.04\ sin(60\°))\\\\F_{23}=\frac{-kq_2q_3}{d^2}(\vv{r}_1-\vv{r}_3)=\frac{-kq_1q_2}{d^2}(0.04-0.04\ cos(60\°),-0.04\ sin(60\°))\\$

To work with the sam

F=∑F_i=3.5*10⁻⁴(0.023,0.032)+7*10⁻⁴(-0.016,0.032)=

=((3.5*10⁻⁴-7*10⁻⁴)*0.016,(3.5*10⁻⁴+7*10⁻⁴)*0.032)=

F=(-5.6*10⁻⁶,3.36⁻⁵)N

7 0

3 years ago