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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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A coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged con
Tcecarenko [31]

Complete question:

A 50 m length of coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged conductor (with charge −8.5 µC and radius 9.249 mm).

Required:

What is the magnitude of the electric field halfway between the two cylindrical conductors? The Coulomb constant is 8.98755 × 10^9 N.m^2 . Assume the region between the conductors is air, and neglect end effects. Answer in units of V/m.

Answer:

The magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

Explanation:

Given;

charge of the coaxial capable, Q = 8.5 µC = 8.5  x 10⁻⁶ C

length of the conductor, L = 50 m

inner radius, r₁ = 1.304 mm

outer radius, r₂ = 9.249 mm

The magnitude of the electric field halfway between the two cylindrical conductors is given by;

E = \frac{\lambda}{2\pi \epsilon_o r} = \frac{Q}{2\pi \epsilon_o r L}

Where;

λ is linear charge density or charge per unit length

r is the distance halfway between the two cylindrical conductors

r = r_1 + \frac{1}{2}(r_2-r_1) \\\\r = 1.304 \ mm \ + \  \frac{1}{2}(9.249 \ mm-1.304 \ mm)\\\\r = 1.304 \ mm \ + \ 3.9725 \ mm\\\\r = 5.2765 \ mm

The magnitude of the electric field is now given as;

E = \frac{8.5*10^{-6}}{2\pi(8.85*10^{-12})(5.2765*10^{-3})(50)} \\\\E = 5.793*10^5 \ V/m

Therefore, the magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

5 0
2 years ago
A ball is thrown vertically upward with a speed of 27.9 m/s from a height of 2.0 m. How long does it take to reach its highest p
Bas_tet [7]

Answer:

2.84403 seconds

2.91483 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-27.9}{-9.81}\\\Rightarrow t=2.84403\ s

It takes 2.84403 seconds to reach the highest point

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-27.9^2}{2\times -9.81}\\\Rightarrow s=39.67431 m

The ball will travel 39.67431+2 = 41.67431 m while going down to the ground

s=ut+\frac{1}{2}at^2\\\Rightarrow 41.71479=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{41.67431\times 2}{9.81}}\\\Rightarrow t=2.91483\ s

The ball takes 2.91483 seconds to hit the ground after it reaches its highest point.

6 0
3 years ago
A person in a kayak starts paddling, and it accelerates from 0 to 0.61 m/s in a distance of 0.39 m. If the combined mass of the
Iteru [2.4K]

Answer:

35.3 N

Explanation:

U = 0, V = 0.61 m/s, s = 0.39 m

Let a be the acceleration.

Use third equation of motion

V^2 = u^2 + 2 as

0.61 × 0.61 = 0 + 2 × a × 0.39

a = 0.477 m/s^2

Force = mass × acceleration

F = 74 × 0.477 = 35.3 N

6 0
2 years ago
True or False: We have a limited amount of fossil fuels.<br> True.<br> False.
Nina [5.8K]
True because we have limited amount if it
4 0
3 years ago
A loop of area 0.250 m^2 is in a uniform 0.020 0-T magnetic field. If the flux through the loop is 3.83 × 10-3T· m2, what angle
dangina [55]

Answer:

40.0⁰

Explanation:

The formula for calculating the magnetic flux is expressed as:

\phi = BAcos\theta where:

\phi is the magnetic flux

B is the magnetic field

A is the cross sectional area

\theta is the angle that the normal to the plane of the loop make with the direction of the magnetic field.

Given

A = 0.250m²

B = 0.020T

\phi = 3.83 × 10⁻³T· m²

3.83 × 10⁻³ = 0.020*0.250cosθ

3.83 × 10⁻³ = 0.005cosθ

cosθ = 0.00383/0.005

cosθ = 0.766

θ = cos⁻¹0.766

θ = 40.0⁰

<em>Hence the angle normal to the plane of the loop make with the direction of the magnetic field is 40.0⁰</em>

4 0
2 years ago
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