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boyakko [2]
3 years ago
12

Is a light sensitive screen​

Physics
1 answer:
Fofino [41]3 years ago
8 0

Answer:

The retina is the light sensitive focusing screen. Inside, there's two types of fluid, one watery behind the lens, and further back, a thicker one that helps the eyeball hold its shape.

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A car covers 120 km in 3 hours calculate its speed in m/s​
8_murik_8 [283]

Answer:

11.1 m/s

Explanation:

120 km = 120 x 1,000 = 120,000 m

3 hours = 3 x 60 x 60 = 3600 x 3 = 10,800 s

speed = 120,000 / 10,800

= 1200/108

= 11.1 m/s

6 0
2 years ago
Read 2 more answers
A spring with k = 500 N/m stores 704 J. How far is it extended from the equilibrium position
kaheart [24]

Given that:

k = 500 n/m,

work (W) = 704 J

spring extension (x) = ?

         we know that,

                      Work = (1/2) k x²

                          704 = (1/2) × 500 × x²

                            x = 1.67 m

A spring stretched for 1.67 m distance.

4 0
3 years ago
Can someone tell me the answer to this?
Art [367]

Answer:

A

Explanation:

5 0
2 years ago
Read 2 more answers
The acceleration due to gravity, g , is constant at sea level on the Earth's surface. However, the acceleration decreases as an
blsea [12.9K]

Answer:

  g    = g₀   [1- 2 h / Re + 3 (h / Re)²]

Explanation:

The law of universal gravitation is

        F = G m Me / Re²

Where g is the universal gravitational constant, m and Me are the mass of the body and the Earth, respectively and R is the distance between them

      F = G Me /Re²  m

We call gravity acceleration a

       g₀ = G Me / Re².

When the body is at a height h above the surface the distance is

            R = Re + h

Therefore  the attractive force is

      F = G Me m / (Re + h)²

Let's take Re's common factor

      F = G Me / Re²  m / (1+ h / Re)²

As Re has a value of 6.37 10⁶ m and the height of the body in general is less than 10⁴ m, the h / Re term is very small, so we can perform a series expansion

         (1+ h / Re)⁻² = 1 -2 h / Re + 6/2 (h / Re) 2 + ...

Let's replace

       F = G Me /Re²   m [1- 2 h / Re + 3 (h / Re)²]

       F = g₀   m  [1- 2 h / Re + 3 (h / Re)²]

If we call the force of attraction at height

     m g =g₀ m  [1- 2 h / Re + 3 (h / Re)²]

       g    = g₀   [1- 2 h / Re + 3 (h / Re)²]

3 0
3 years ago
Physics double pivot question​
andriy [413]

Explanation:

Assuming the wall is frictionless, there are four forces acting on the ladder.

Weight pulling down at the center of the ladder (mg).

Reaction force pushing to the left at the wall (Rw).

Reaction force pushing up at the foot of the ladder (Rf).

Friction force pushing to the right at the foot of the ladder (Ff).

(a) Calculate the reaction force at the wall.

Take the sum of the moments about the foot of the ladder.

∑τ = Iα

Rw (3.0 sin 60°) − mg (1.5 cos 60°) = 0

Rw (3.0 sin 60°) = mg (1.5 cos 60°)

Rw = mg / (2 tan 60°)

Rw = (10 kg) (9.8 m/s²) / (2√3)

Rw = 28 N

(b) State the friction at the foot of the ladder.

Take the sum of the forces in the x direction.

∑F = ma

Ff − Rw = 0

Ff = Rw

Ff = 28 N

(c) State the reaction at the foot of the ladder.

Take the sum of the forces in the y direction.

∑F = ma

Rf − mg = 0

Rf = mg

Rf = 98 N

3 0
3 years ago
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