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weqwewe [10]
2 years ago
8

How can light energy solve our real life problem?

Physics
2 answers:
anastassius [24]2 years ago
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]2 years ago
4 0

Answer:

It gives our light which we need for probably everything.

Explanation:

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Sauron [17]
E = 0.25 = m*g*h 
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2 years ago
Not too long ago houses were protected from excessive currents by fuses rather than circuit breakers. sometimes a fuse blew out
olganol [36]

Answer: Resistance = 8.21 \times 10^{-8} \Omega

The approximate diameter of a penny is, <em>d</em> = 20 mm

thickness of penny is, <em>L = </em> 1.5×10^{-3} mm

The area of penny along circular face is,A = \frac{\pi d^2}{4} =\frac{\pi (20 mm\frac{1 m}{1000 m})^2}{4}

= 3.14×10^{-4} m²

The resistivity of copper is <em>ρ</em> = 1.72 x 10-8 Ωm.

Resistance,

R = \rho \frac{L}{A}  = (1.72 \times 10^{-8} \Omega m)\frac{1.5 \times 10^{-3} m}{3.14 \times 10^{-4} m^2} = 8.21 \times 10^{-8} \Omega

5 0
2 years ago
The three forces acting on a hot-air balloon that is moving vertically are its weight, the force due to air resistance and the u
Aleksandr [31]
Let
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T+F-W=0......................(1)

Let w=weight of material dumped so that balloon now travels upwards at constant speed.
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2 years ago
a 100 kg gymnast comes to a stop after tumbling. her feet do -5000J of net work to stop her. Use the work-kinetic energy theorem
VikaD [51]
W=ΔKE , W=-5000j
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6 0
2 years ago
Read 2 more answers
A 7.5 nC point charge and a - 2.9 nC point charge are 3.2 cm apart. What is the electric field strength at the midpoint between
Oduvanchick [21]

Answer:

Net electric field, E_{net}=91406.24\ N/C

Explanation:

Given that,

Charge 1, q_1=7.5\ nC=7.5\times 10^{-9}\ C

Charge 2, q_2=-2.9\ nC=-2.9\times 10^{-9}\ C

distance, d = 3.2 cm = 0.032 m

Electric field due to charge 1 is given by :

E_1=\dfrac{kq_1}{r^2}

E_1=\dfrac{9\times 10^9\times 7.5\times 10^{-9}}{(0.032)^2}

E_1=65917.96\ N/C

Electric field due to charge 2 is given by :

E_2=\dfrac{kq_2}{r^2}

E_2=\dfrac{9\times 10^9\times 2.9\times 10^{-9}}{(0.032)^2}

E_2=25488.28\ N/C

The point charges have opposite charge. So, the net electric field is given by the sum of electric field due to both charges as :

E_{net}=E_1+E_2

E_{net}=65917.96+25488.28

E_{net}=91406.24\ N/C

So, the electric field strength at the midpoint between the two charges is 91406.24 N/C. Hence, this is the required solution.

3 0
2 years ago
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