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Dmitriy789 [7]
3 years ago
13

Which type of energy can be sensed by the eyes?

Physics
2 answers:
gayaneshka [121]3 years ago
7 0

Answer:

light energy..

........

Juli2301 [7.4K]3 years ago
3 0

Answer:

light energy

Explanation: that is the only type of energy you can actually see.

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PLEEEASE HELP ME!!! YOU GET 15 PTS
Damm [24]

Answer:

b

Explanation:

4 0
3 years ago
The sphere that refers to Earths water is called the
soldier1979 [14.2K]
Earth Spheres. Earth's Spheres. Everything in Earth's system can be placed into one of four major subsystems: land, water, living things, or air. These four subsystems are called “spheres.” Specifically, they are the lithosphere (land), hydrosphere (water), biosphere (living things), and atmosphere (air).
6 0
3 years ago
Read 2 more answers
A generator connected to an RLC circuit has an rms voltage of 140 V and an rms current of 31 mA. Part A
vredina [299]

Answer:

53.06 ohm

Explanation:

We have given the rms voltage V =140 V

And the rms current i=31 mA

So the impedance Z=\frac{V}{i}=\frac{140}{3\times 10^{-3}}=46.666kohm

Resistance is given as R = 3 kohm

And capacitive reactance X_C=6.5kohm

We know that Z=\sqrt{R^2+(X_L-X_C)^2}

46.666=\sqrt{3^2+(X_L-6.5)^2}

Squaring both side 2177.1556=9+(X_L-6.5)^2

(X_L-6.5)^2=2168.155

(X_L-6.5)=46.5634

X_L=53.063ohm

8 0
3 years ago
A cave rescue team lifts an injured spelunker directly upward and out of a sinkhole by means of a motor-driven cable. The lift i
lys-0071 [83]

Answer:

   W₃ = 3310.49 J ,  W3 = 3310.49 J

Explanation:

We can solve this exercise in parts, the first with acceleration, the second with constant speed and the third with deceleration. Therefore it is work we calculate it in these three sections

We start with the part with acceleration, the distance traveled is y = 5.90 m and the final speed is v = 2.30 m / s. Let's calculate the acceleration with kinematics

       v2 = v₀² + 2 a₁ y

as they rest part of the rest the ricial speed is zero

        v² = 2 a₁ y

        a₁ = v² / 2y

        a₁ = 2.3² / (2 5.90)

        a₁ = 0.448 m / s²

with this acceleration we can calculate the applied force, using Newton's second law

         F -W = m a₁

         F = m a₁ + m g

         F = m (a₁ + g)

         F = 69 (0.448 + 9.8)

         F = 707.1 N

Work is defined by

         W₁ = F.y = F and cos tea

As the force lifts the man, this and the displacement are parallel, therefore the angle is zero

          W₁ = 707.1 5.9

           W₁ = 4171.89 J  W3 = 3310.49 J

Let's calculate for the second part

the speed is constant, therefore they relate it to zero

           F - W = 0

           F = W

           F = m g

           F = 60 9.8

           F = 588 A

the job is

           W² = 588 5.9

            W2 = 3469.2 J

finally the third part

in this case the initial speed is 2.3 m / s and the final speed is zero

           v² = v₀² + 2 a₂ y

            0 = vo2₀² + 2 a₂ y

            a₂ = -v₀² / 2 y

            a₂ = - 2.3²/2 5.9

            a2 = - 0.448 m / s²

we calculate the force

            F - W = m a₂

            F = m (g + a₂)

            F = 60 (9.8 - 0.448)

            F = 561.1 N

we calculate the work

            W3 = F and

            W3 = 561.1 5.9

          W3 = 3310.49 J

total work

          W_total = W1 + W2 + W3

          W_total = 4171.89 +3469.2 + 3310.49

           w_total = 10951.58 J

4 0
3 years ago
Please help with this question! Thanks in advance...
-Dominant- [34]

A.0.6 hope you do good on your test!!

5 0
3 years ago
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