All categories

3 years ago

How can light energy solve our real life problem?

2 answers:

5 0

One big real-life problem is the inability to SEE stuff, like when it's dark all around you. When that happens and you try to move around, there's a big chance that you could bump into stuff, like walls, or step on stuff, like the dog. And honestly, driving the car is out, altogether.

Light energy can really help us solve this big problem, by bouncing off of the things around us and into our eyes. When light does this, our brain can figure out pretty quick WHERE all the things are, in what direction and how far from us. In that way, we can navigate ourself around, and we are prevented from bumping into something or stepping on it unless we want to. The light gives us our choice back.

kifflom [539]3 years ago

4 0

Answer:

It gives our light which we need for probably everything.

Explanation:

You might be interested in

How can we describe, model, and explain the material world, and are there limitations?

taurus [48]

Answer:· A model is a description of natural phenomenon that scientists can use to make predictions. A good model is both as accurate as possible and as simple as possible, which makes it not only powerful but also easy to understand. However, no matter how good they are, models will almost always have limitations.

Explanation:

5 0

3 years ago

An airplane accelerates down a runmway at 2.9 m/s/s for 38.5 seconds

Serhud [2]

Given,

the initial velocity = 0 m /s.

acceleration = 3.20 m / s^2

time = 32.8 s

According to laws of motion.

s = ut + 1/2 at ^2

s = 1/2 at²

s=1/2(3.20)(32.8)²

s= 1721.344 m

the distance traveled before takeoff is 1731.3m

8 0

3 years ago

Read 2 more answers

A force Fx=cx-3.00x2 acts on a virus as the virus moves along an x axis, with F measured in Newtons, x in meters, and c a consta

andrey2020 [161]

Answer:

Explanation:

We are given that

$F_x=cx-3x^2$

K.E at x=0 m=20 J

K.E at x=3 m=11 J

We have to find the value of c.

By work energy theorem

Work done=Change in kinetic energy

W= $\int_{0}^{3} Fdx=\int_{0}^{3}(cx-3x^2)dx$

$W=[\frac{cx^2}{2}-x^3]^{3}_{0}$

$W=\frac{9c}{2}-27$

$\Delta K=11-20=-9 J$

$-9=\frac{9c}{2}-27$

$-9+27=\frac{9c}{2}$

$18=\frac{9c}{2}$

$c=\frac{2}{9}\times 18=4$

4 0

4 years ago

An electron is projected with an initial speed of 5 3.2 10 / ⋅ m s directly toward a proton that is fixed in place. If the elect

Dvinal [7]

Answer:

The distance is $d =1.66*10^{-9}m$

Explanation:

From the question we are told that

The initial speed of the electron is $v_i = 3.2 *10^5 m/s$

The mass of electron is $m = 9.11*10^{-31}kg$

Let $d$ be the distance between the electron and the proton when the speed of the electron instantaneously equal to twice the initial value

Let $KE_i$ be the initial kinetic energy of the electron \

Let $KE_d$ be the kinetic energy of the electron at the distance $d$ from the proton

Considering that energy is conserved,

The energy at the initial position of the electron = The energy at the final position of the electron

i.e

$KE_i +U_1 = KE_d + U_2$

$U_1 \ and \ U_2$ are the potential energy at the initial position of the electron and at distance d of the electron to the proton

Here $U_1 = 0$

So the equation becomes

$\frac{1}{2} mv_i^2 = \frac{1}{2} mv_d^2 + \frac{kq_1 q_2}{d}$

Here $q_1 \ and \ q_2$ are the charge on the electron and the proton and their are the same since a charge on an electron is equal to charge on a proton