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

Two wires each carry 10.0 A of current (in opposite directions) and are 2.50 mm apart. What is the magnetic field 37.0 cm away a

lyudmila [28]

Answer:

see answer below

Explanation:

Before we do any kind of calculation, we need to convert the proper units of the exercise. All the units of distance must be in meters, so, let's change distance of the wire, and the magnetic field to meters:

Separation between the wires are 2.5 mm:

2.5 mm * (1 m / 1000 mm) = 0.0025 m

The distance of P from the bottom of the wires is 37 cm:

37 cm * (1 m/100 cm) = 0.37 m

The distance of P from the top of the wires is just the sum of the two distances:

R = 0.37 + 0.0025 = 0.3725 m

Now that we have the distance, we can determine the magnetic field, using the following expression:

B = B(bottom) - B(top) or just B₂ - B₁

And B = μ₀ I / 2πR

Replacing in the above expression we have:

B = μ₀ I / 2π ( 1/R₂ - 1/R₁)

Now we can determine the magnetic field:

B = (4πx10⁻⁷ * 10 / 2π) (1/0.37 - 1/0.3725)

Which means that the magnetic field is out of the page.

Hope this helps

4 0

3 years ago

What is the dimensional formula of torque and force

ch4aika [34]

✿━━━━@♥ℳg━━━━✿

<h2> $\boxed{Explained\:Answer}$ </h2>

______________________________

✿━━━━@♥ℳg━━━━✿

Torque = Force X Displacement

= MLT-2 X L = [M1L2T-2]

The dimensional formula of force is MLT^-2

6 0

2 years ago

Read 2 more answers

A flow is isentropically expanded to supersonic speeds in a convergent-divergent nozzle. The reservoir and exit pressures are 1.

Kamila [148]

Answer:

Ae/A* = 1.115

Explanation:

Let the reservoir pressure be $p_0$

Let the exit pressure be $p_e$

Ratio of reservoir pressure and exit pressure

$\frac{p_o}{p_e} = \frac{1}{0.3143}$

= 3.182

For the above value of pressure ratio

Obtain the area ratio from the isentropic flow table

Ae/A* = 1.115

The value of pressure ratio is Ae/A* = 1.115

6 0

2 years ago

slader How much energy is required to move a 1040 kg object from the Earth's surface to an altitude four times the Earth's radiu

andrew-mc [135]

Answer:

ΔU = 5.21 × 10^(10) J

Explanation:

We are given;

Mass of object; m = 1040 kg

To solve this, we will use the formula for potential energy which is;

U = -GMm/r

But we are told we want to move the object from the Earth's surface to an altitude four times the Earth's radius.

Thus;

ΔU = -GMm((1/r_f) - (1/r_i))

Where;

M is mass of earth = 5.98 × 10^(24) kg

r_f is final radius

r_i is initial radius

G is gravitational constant = 6.67 × 10^(-11) N.m²/kg²

Since, it's moving to altitude four times the Earth's radius, it means that;

r_i = R_e

r_f = R_e + 4R_e = 5R_e

Where R_e is radius of earth = 6371 × 10³ m

Thus;

ΔU = -6.67 × 10^(-11) × 5.98 × 10^(24)

× 1040((1/(5 × 6371 × 10³)) - (1/(6371 × 10³))

ΔU = 5.21 × 10^(10) J

7 0

3 years ago

What is the term used to label the energy levels of electrons?

mel-nik [20]

Electron volts...........

4 0

3 years ago