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4 years ago

How does remove coloration interface "light-shadow"?

Physics

1 answer:

tekilochka [14]4 years ago

5 0

Normal force, weight, Kinetic friction, and air resistance are a few I think of the top of my head.
I hope this helps

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A simple pendulum consisting of a bob of mass m attached to a string of length L swings with a period T. If the bob's mass is do

Alborosie

1. B. T

The period of a simple pendulum is given by:

$T=2\pi \sqrt{\frac{L}{g}}$ (1)

where

L is the length of the pendulum

g is the gravitational acceleration

From the formula, we notice that the period of the pendulum does not depend on the mass of the bob. Therefore, when the bob's mass is doubled, the period does not change.

2. C: sqrt(6)*T

In this case, the pendulum is brought to the moon, where the gravitational acceleration is

$g'=\frac{g}{6}$

If we substitute this value into the equation for the period (1), we find the new period of the pendulum:

$T'=2\pi \sqrt{\frac{L}{g'}}=2\pi \sqrt{\frac{L}{(g/6)}}=\sqrt{6}(2\pi \sqrt{\frac{L}{g}})=\sqrt{6} T$

3. B: It will no longer oscillate because there is no gravity in space

Explanation:

The motion (oscillation) of the pendulum is caused by the force of gravity, which "pulls" the bob towards the equilibrium position. If there is no gravity, then there is no force acting on the bob, therefore the pendulum can no longer oscillate.

So, the correct answer is

B: It will no longer oscillate because there is no gravity in space

4 0

4 years ago

Read 2 more answers

A projectile is shot from the edge of a cliff above the ground level with initial velocity of at an angle with the horizontal. (

Xelga [282]

Answer:

t = √2y/g

Explanation:

This is a projectile launch exercise

a) The vertical velocity in the initial instants ( $v_{oy}$ = 0) zero, so let's use the equation

y = $v_{oy}$ t -1/2 g t²

y= - ½ g t²

t = √2y/g

b) Let's use this time and the horizontal displacement equation, because the constant horizontal velocity

x = vox t

x = v₀ₓ √2y/g

c) Speeds before touching the ground

vₓ = vox = constant

$v_{y}$ = $v_{oy}$ - gt

$v_{y}$ = 0 - g √2y/g

$v_{y}$ = - √2gy

tan θ = Vy / vx

θ = tan⁻¹ (vy / vx)

θ = tan⁻¹ (√2gy / vox)

d) The projectile is higher than the cliff because it is a horizontal launch

6 0

4 years ago

How do I find the third nodal for this question, and how do I answer the question c and d?

salantis [7]

Answer:

my answer is C.

Explanation:

I hope my answer can help you

3 0

3 years ago

Read 2 more answers

One isotope of nitrogen has 7 protons and 8 neutrons. Which is the correct reference for this isotope? A. nitrogen-8 B. nitrogen

tamaranim1 [39]

When referring to an isotope, you refer to the mass of its protons and neutrons combined
7 + 8 = 15, so you would refer to it as C. Nitrogen-15

3 0

4 years ago

(4%) Problem 9: A mass is connected to a spring and is allowed to move horizontally. The mass is at a position L when the spring

skad [1K]

Answer:

a) Acceleration is zero , c) Speed is cero

Explanation:

a) the equation that governs the simple harmonic motion is

x = A cos (wt +φφ)

Where A is the amplitude of the movement, w is the angular velocity and φ the initial phase determined by the initial condition

Body acceleration is

a = d²x / dt²

Let's look for the derivatives

dx / dt = - A w sin (wt + φ)

a = d²x / dt² = - A w² cos (wt + φ)

In the instant when it is not stretched x = 0

As the spring is released at maximum elongation, φ = 0

0 = A cos wt

Cos wt = 0 wt = π / 2

Acceleration is valid for this angle

a = -A w² cos π/2 = 0

Acceleration is zero

c) When the spring is compressed x = A

Speed is

v = dx / dt

v = - A w sin wt

We look for time

A = A cos wt

cos wt = 1 wt = 0, π

For this time the speedy vouchers

v = -A w sin 0 = 0

Speed is cero

5 0

4 years ago