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3 years ago

A 20kg suitcase on a flat surface is pushed by a force of 100N downwards at an

Physics

1 answer:

uranmaximum [27]3 years ago

8 0

Answer:

282.6 N

Explanation:

Draw a free body diagram of the suitcase. There are three forces:

Weight force mg pulling down,

Normal force N pushing up,

and 100 N down at angle 60° with the horizontal.

Sum of forces in the y direction:

∑F = ma

N − mg − 100 sin 60° = 0

N = mg + 100 sin 60°

N = (20 kg) (9.8 m/s²) + 100 sin 60°

N = 282.6 N

Round as needed.

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Unpolarized light is passed through three successive polaroid filters, each with its transmission axis at 32. 8° to the precedin

Oliga [24]

The percentage of light gets through three successive polarized filters is 24.9 %

From the question,

Given that,

Angle of transmission axis = 32.8°

The intensity of light emerging from the first polarizer is determined by the equation

I₁ = I₀ / 2

where I₀ ⇒ intensity of unpolarized light

The light emerging from the second polarizer can be mathematically represented by,

I₂ = I₁ x cos²θ

Substituting the values,

I₂ =( I₀/2) x cos²θ

= (I₀/2) x cos² (32.8 )

= (I₀/2) x 0.706

= (0.706 / 2 ) x I₀

The light emerging from the third polarizer is represented as,

I₃ = I₂ x cos²θ

Substituting the values in the above equation,

I₃ = ( 0.706 / 2 ) I₀ x cos² (32.8)

= (0.706 / 2 ) I₀ x 0.706

= 0.249 I₀

The percentage of intensity of light that gwts through with respect to the intensity of unpolarized light is given by the equation,

(I₃ / I₀) x 100

Substituting the values

[(0.249 x I₀) / I₀ ] x 100 = 24.9 %

Hence the percentage of light gets through is 24.9%

To learn more about unpolarized light: brainly.com/question/17164167

#SPJ4

6 0

2 years ago

An ice cube can slide around the inside of a vertical circular hoop of radius . It undergoes small-amplitude oscillations if dis

klasskru [66]

Answer:

$T=2\pi \times \sqrt {\frac {g}{\mu R}$

Explanation:

$\mu mw^{2}R=mg$

where m is the mass, g is acceleration due to gravity

The masses m are on both sides hence they cancel

$W=\sqrt {\frac {g}{\mu R}$

We know that T=2\pi W and substituting W with the above equation then

$T=2\pi \times \sqrt {\frac {g}{\mu R}$

4 0

4 years ago

The sound strikes the cantina walls and bounces off at the same angle as it struck in a process known as _____.

PIT_PIT [208]

<span>Reflection

Reflection is the change in direction of a wavefront at an interface between two different media so that the wavefront returns into the medium from which it originated.</span>

6 0

3 years ago

Read 2 more answers

The drawing shows a tire of radius R on a moving car

masha68 [24]

Answer:

H / R = 2/3

Explanation:

Let's work this problem with the concepts of energy conservation. Let's start with point P, which we work as a particle.

Initial. Lowest point

Em₀ = K = 1/2 m v²

Final. In the sought height

$Em_{f}$ = U = mg h

Energy is conserved

Em₀ = $Em_{f}$

½ m v² = m g h

v² = 2 gh

Now let's work with the tire that is a cylinder with the axis of rotation in its center of mass

Initial. Lower

Em₀ = K = ½ I w²

Final. Heights sought

Emf = U = m g R

Em₀ = $Em_{f}$

½ I w² = m g R

The moment of inertial of a cylinder is

I = $I_{cm}$ + ½ m R²

I= ½ $I_{cm}$ + ½ m R²

Linear and rotational speed are related

v = w / R

w = v / R

We replace

½ $I_{cm}$ w² + ½ m R² w² = m g R

moment of inertia of the center of mass

$I_{cm}$ = ½ m R²

½ ½ m R² (v²/R²) + ½ m v² = m gR

m v² ( ¼ + ½ ) = m g R

v² = 4/3 g R

As they indicate that the linear velocity of the two points is equal, we equate the two equations

2 g H = 4/3 g R

H / R = 2/3

7 0

3 years ago

Two charged particles are a distance of 1.82 m from each other. One of the particles has a charge of 7.52 nC, and the other has

elena55 [62]

Answer:

(a) $F=9.10*10^{-8}N$

(b) The force is repulsive

Explanation:

a) According to Coulomb's law, the magnitude of the electrice force that one particle exerts on the other is defined as:

$F=\frac{kq_1q_2}{d^2}$

Here k is the coulomb constant, $q_1$ and $q_2$ are the signed magnitudes of the charges and d is the distance between them.

$F=\frac{8.99*10^{9}\frac{N\cdot m^2}{C^2}(7.52*10^{-9}C)(4.46*10^{-9}C)}{(1.82m)^2}\\F=9.10*10^{-8}N$

b) According to Coulomb's law, if the two charges have the same sign, the electrostatic force between them is repulsive.

5 0

4 years ago