1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
mr Goodwill [35]
3 years ago
13

Do solar panels create enough energy to power the car throughout the day?

Physics
1 answer:
JulijaS [17]3 years ago
5 0

Answer:

Yes, but only if it's sunny.

Explanation:

As you know, solar panels generate energy through the sun's rays of light (better known as sunlight). Therefore, as long as the sun is shining high in the sky, the car will generate electricity and be able to function. If this vehicle was only powered by solar panels, it would not function during the night, in cloudy areas, and/or in dark places (such as parking garages or home garages).

Hope this helps!

You might be interested in
What are the three places where ribisomes occur in a cell​
irina [24]

Answer: Rough endoplasmic reticulum, cytoplasm, inside mitochondria

Explanation:

4 0
3 years ago
Please help, I do not understand
Anettt [7]
I think the key here is to be exquisitely careful at all times, and
any time we make any move, keep our units with it.

We're given two angular speeds, and we need to solve for a time.

Outer (slower) planet:
Angular speed =  ω  rad/sec
Time per unit angle =  (1/ω)  sec/rad
Angle per revolution = 2π rad
Time per revolution = (1/ω sec/rad) · (2π rad) = 2π/ω seconds .

Inner (faster) planet:
Angular speed =  2ω  rad/sec
Time per unit angle =  (1/2ω)  sec/rad
Angle per revolution = 2π rad
Time per revolution = (1/2ω sec/rad) · (2π rad) = 2π/2ω sec = π/ω seconds.

So far so good.  We have the outer planet taking 2π/ω seconds for one
complete revolution, and the inner planet doing it in only π/ω seconds ...
half the time for double the angular speed.  Perfect !

At this point, I know what I'm thinking, but it's hard to explain.
I'm pretty sure that the planets are in line on the same side whenever the
total elapsed time is something like a common multiple of their periods.
What I mean is:

They're in line, SOMEwhere on the circles, when

     (a fraction of one orbit) = (the same fraction of the other orbit)    
AND
     the total elapsed time is a common multiple of their periods.

Wait !  Ignore all of that.  I'm doing a good job of confusing myself, and
probably you too.  It may be simpler than that.  (I hope so.)  Throw away
those last few paragraphs.

The planets are in line again as soon as the faster one has 'lapped'
the slower one ... gone around one more time.  
So, however many of the longer period have passed, ONE MORE
of the shorter period have passed.  We're just looking for the Least
Common Multiple of the two periods.

      K (2π/ω seconds)  =  (K+1) (π/ω seconds)

                     2Kπ/ω   =    Kπ/ω + π/ω

Subtract  Kπ/ω :    Kπ/ω = π/ω

Multiply by  ω/π :      K  =  1

(Now I have a feeling that I have just finished re-inventing the wheel.)

And there we have it:

     In the time it takes the slower planet to revolve once,
     the faster planet revolves twice, and catches up with it.
    
     It will be  2π/ω  seconds before the planets line up again.
    
     When they do, they are again in the same position as shown
     in the drawing.

To describe it another way . . . 

     When Kanye has completed its first revolution ...

     Bieber has made it halfway around.

     Bieber is crawling the rest of the way to the starting point while ...

     Kanye is doing another complete revolution.

     Kanye laps Bieber just as they both reach the starting point ...

     Bieber for the first time, Kanye for the second time.


You're welcome.  The generous bounty of 5 points is very gracious,
and is appreciated.  The warm cloudy water and green breadcrust
are also delicious.
5 0
3 years ago
A 2 kg box of taffy candy has 40j of potential energy relative to the ground. Its height above the ground is​
lord [1]

Answer:

2.04m

Explanation:

PE=mgh

h= PE/mg

h= 40/2(9.82)

h=40/19.64

h=2.04

8 0
3 years ago
Read 2 more answers
Vector a has a magnitude of 12.3 units and points due west. vector b points due north. what is the magnitude of b if a - b has a
algol [13]

The magnitude of vector b is 8.58 Unit.

Since both the vectors a and b are perpendicular to each other, so we can apply the Pythagoras theorem to calculate the magnitude of the vector b.

Applying the Pythagoras theorem

(a-b)^2=a^2+b^2

15^2=12.3^2-b^2

b=8.58 unit

Therefor the magnitude of the vector b is 8.58 unit.

8 0
3 years ago
Please help in desperate
Rom4ik [11]

Answer:

D. 53°

Explanation:

3 0
3 years ago
Other questions:
  • HELP FAST!!!!!
    13·2 answers
  • A mass m is at rest on top of the frictionless incline of mass M. It is initially at a height h above the surface. After m slide
    9·1 answer
  • A student measures the speed of a rolling ball three times. She adds the measurements and divides by 3. What quantity did the st
    9·2 answers
  • An electron has a velocity of 3.2 x 10^6 m/s. What is its’ momentum? (b) What is its’ wavelength? (c) What other objects/materia
    14·1 answer
  • During wich two time intervals does the particle undergo equal displacement
    14·1 answer
  • The attractive force that acts between all objects is the (5 points)
    5·1 answer
  • Which change will always result in an increase in the gravitational force between two objects?
    6·1 answer
  • If you measure the amount of work accomplished in a particular time interval, u have measured-
    11·1 answer
  • Find the force on an object which has a mass of 20 kg and an acceleration of 10 m/s2.
    6·1 answer
  • The tropical easterlies blow away from the<br> equator<br> true or false?
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!