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3 years ago

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!

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what would the net force be on the box in the problems shown below.( both force and direction). for all four diagrams. please e

DIA [1.3K]

Answer:
A) object moves 20 N [West] or -20 N [East]
B) object moves 6 N [South] or -6 N [North]
C) object moves 90 N [West] or -90 N [East]
D) object does not move and is at rest*

*Rest means 0

Why:

A)both forces from north and south that are pushing against the object neutralize each other. Assume that north is positive and south is negative: 20 [N] + (-20) [S] = 0
On West and east, you can see that west has a greater force. Assume that west is negative and east is positive: 50 [E] + (-70) [W] = -20 [E]

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3 years ago

Density=2g/mL and volume=20mL what is mass

leva [86]

Answer:

40g

Explanation:

Mass = density x volume

= 2 x 20

= 40g

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3 years ago

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A ladder 7.85 m long leans against the side of a building. If the ladder is inclined at an angle of 79.5° to the horizontal, wha

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4 years ago

I need the answer right now pls help me

yaroslaw [1]

The AREA of the shaded region is the moving object's displacement.

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3 years ago

A loaded ore car has a mass of 950 kg. and rolls on rails ofnegligible friction. It starts from rest ans is pulled up a mineshaf

stiks02 [169]

(a) 10241 W

In this situation, the car is moving at constant speed: this means that its acceleration along the direction parallel to the slope is zero, and so the net force along this direction is also zero.

The equation of the forces along the parallel direction is:

$F - mg sin \theta = 0$

where

F is the force applied to pull the car

m = 950 kg is the mass of the car

$g=9.8 m/s^2$ is the acceleration of gravity

$\theta=30.0^{\circ}$ is the angle of the incline

Solving for F,

$F=mg sin \theta = (950)(9.8)(sin 30.0^{\circ})=4655 N$

Now we know that the car is moving at constant velocity of

v = 2.20 m/s

So we can find the power done by the motor during the constant speed phase as

$P=Fv = (4655)(2.20)=10241 W$

(b) 10624 W

The maximum power is provided during the phase of acceleration, because during this phase the force applied is maximum. The acceleration of the car can be found with the equation

$v=u+at$

where

v = 2.20 m/s is the final velocity

a is the acceleration

u = 0 is the initial velocity

t = 12.0 s is the time

Solving for a,

$a=\frac{v-u}{t}=\frac{2.20-0}{12.0}=0.183 m/s^2$

So now the equation of the forces along the direction parallel to the incline is

$F - mg sin \theta = ma$

And solving for F, we find the maximum force applied by the motor:

$F=ma+mgsin \theta =(950)(0.183)+(950)(9.8)(sin 30^{\circ})=4829 N$

The maximum power will be applied when the velocity is maximum, v = 2.20 m/s, and so it is:

$P=Fv=(4829)(2.20)=10624 W$

(c) $5.82\cdot 10^6 J$

Due to the law of conservation of energy, the total energy transferred out of the motor by work must be equal to the gravitational potential energy gained by the car.

The change in potential energy of the car is:

$\Delta U = mg \Delta h$