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trapecia [35]
3 years ago
8

For a science project, you would like to horizontally suspend an 8.5 by 11 inch sheet of black paper in a vertical beam of light

whose dimensions exactly match the paper.
If the mass of the sheet is 0.80g , what light intensity will you need?
Express your answer to two significant figures and include the appropriate units.
I = ________.
Physics
1 answer:
liubo4ka [24]3 years ago
4 0

Answer:

I = 3.9 x 10⁷ W/m²

Explanation:

given,

Sheet of black paper dimension = 8.5 x 11 inch

Area of sheet = 8.5 x 11 = 93.5 inch^2

1 inch =0.0254 m

Area = 0.06032 m²

mass of sheet = 0.80 g

Force = m g =  0.8 x 9.8 x 10⁻³ N

                    =  7.84 x 10⁻³ N

speed of light = c = 3 x 10⁸ m/s

Using equation

F = \dfrac{IA}{c}

where I is the intensity of light

7.84 \times 10^{-3} = \dfrac{I\times 0.06032}{3 \times 10^8}

2.352 \times 10^{6} = I\times 0.06032

I = \dfrac{2.352 \times 10^{6}}{0.06032}

I = 3.9 x 10⁷ W/m²

Intensity of the light is equal to I = 3.9 x 10⁷ W/m²

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Answer:

The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object

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4 years ago
Brainliest if right.
jeka57 [31]

Answer:

the third law (for every action there is an equal and opposite reaction).

Explanation:

The skateboarder pushes backwards on the road (that is he applies a force on the road in a direction opposite the direction of intended motion). By Newton's third law, this action of the skateboarder causes an equal reaction of the road on the skateboarder in the opposite direction. Newton's third law states that action and reaction are equal but opposite in direction. So, the road in response to this backward force pushes the skateboarder in the forward direction causing the skateboarder and the skateboard to move in the forward direction.

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3 years ago
Electric devices use electric _____ to function
Rina8888 [55]

AC or DC

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3 years ago
A horizontal uniform bar of mass 2.7 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to t
Jlenok [28]

Answer:

14.36 N

Explanation:

T_{1} = Tension in string 1

T_{2} = Tension in string 2

m_b = mass of the bar = 2.7 kg

W_b = weight of the bar

weight of the bar is given as

W_b = m_{b} g = (2.7) (9.8) = 26.46N

m_m = mass of the bar = 1.35 kg

W_m = weight of the monkey

weight of the monkey is given as

W_m = m_{m} g = (1.35) (9.8) = 13.23N

Using equilibrium of torque about left end

W_{m} (AB) + W_{b} (AB) = T_{2} (AC)\\W_{m} (AB) + W_{b} (AB) = T_{2} (AD - CD)\\(13.23) (1.5) + (26.46)(1.5) = T_{2} (3 - 0.65)\\\\T_{2} = 25.33 N

Using equilibrium of force in vertical direction

T_{1} + T_{2} = W_{b} + W_{m}\\T_{1} + 25.33 = 26.46 + 13.23\\T_{1} = 14.36 N

7 0
3 years ago
A capacitor in an LC oscillator experiences a maximum potential difference of 88V and a maximum energy of 2002 uJ. At a certain
VLD [36.1K]

Answer:

Explanation:

maximum energy of capacitor

E = 1/2 C V ²

C is capacitance of capacitor and V is potential difference

given V = 88 V

E = 2002 x 10⁻⁶ J

Putting the values

2002 x 10⁻⁶ = 1/2 x C x 88²

C = .517 x 10⁻⁶ F  .

In the second case

Energy E = 125 x 10⁻⁶ J .

C = .517 x 10⁻⁶

V =  ?

E = 1/2 C V ²

125 x 10⁻⁶  = 1/2 x  .517 x 10⁻⁶ x V²

V² = 483.55

V = 21.98 V .

6 0
4 years ago
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