When a monochromatic light falls on a metallic surface, a number of electrons is

At t=0, a block A of mass 8 kg and block B of mass 16 kg are both at position x=0 . Block A is at rest, and block B is moving at

love history [14]

The center of mass of the two objects is the average position of the parts of the two object system

The center of mass of block A, and block B after displacement of block B is at 20 m from block A

Reason:

The given parameters are;

The position of block A and block B at t = 0 is x = 0

The mass of block A, m₁ = 8 kg

Mass of block B, m₂ = 16 kg

Speed of block A = 0 m/s

Speed of block B, v₂ = 10 m/s

Location of the center of mass of the two object at t = 3 s; Required

Solution;

The location of block A, after 3 s is x₁ = 0 (block A is at rest)

The location of block B, = v₂ × t

The location of block B, after 3 s is x₂ = 10 m/s × 3 s = 30 m

The center of mass of two masses are given as follows;

$x_{cm} = \dfrac{m_1 \cdot x_1 +m_2\cdot x_2}{m_1 + m_2}$

$x_{cm} = \dfrac{8 \times0 + 16 \times 30}{8 + 16} = 20$

The center of mass of the two objects is at at the position x = 20 m (from block A)

Learn more about the center of mass here:

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4 0

3 years ago

Sound waves produced by a source pass a point five times every second. Which of the following choices correctly describes the pe

Ugo [173]

The period is 1/5 second, and the frequency is 5 Hz.

Explanation:

Let's start by reviewing some definitions about waves:

- The period of a wave is the time taken for the wave to make one complete cycle. It is indicated with T and it is measured in seconds (s)

- The frequency of a wave is the number of complete cycles made by the wave in one second. It is indicated with f and it is measured in Hertz (Hz). It is equal to the reciprocal of the period:

$f=\frac{1}{T}$

where f is the frequency and T is the period.

In this problem, we have a wave that passes a given point five time per second. This means that the number of oscillations per second is five, and so its frequency is:

$f=\frac{5 cycles}{1 sec}=5 Hz$

It follows than the period of the wave is:

$T=\frac{1}{f}=\frac{1}{5}s$

Therefore, the correct statement is

The period is 1/5 second, and the frequency is 5 Hz.

Learn more about waves, frequency and period:

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3 0

3 years ago

Which statements describe the motion of the object represented by the graph? Check all that apply.

bonufazy [111]

Answer:

The object is moving away from the origin

The object started at 2 meters

The object is traveling at a constant velocity

Explanation:

5 0

3 years ago

If Bert the Bat travels East at 200 meters in 10 seconds, what is his velocity?

dexar [7]

Answer:

20m/s due east

Explanation:

Given parameters:

Displacement eastward = 200m

Time = 10s

Unknown:

Velocity = ?

Solution:

Velocity is the displacement divided by time;

Velocity = $\frac{displacement}{time}$

Velocity = $\frac{200}{10}$ = 20m/s due east

4 0

3 years ago

You are designing a new piece of fishing equipment that will allow you to catch a fish from the surface. You know that water ref

OleMash [197]

Answer:

Law of refraction

Explanation:

An experiment to analyze the refraction of light in water can easily be performed with a laser pointer and protractor.

We throw the fishing rod line into the water, place the protractor at the point where the line touches the water and use the direction of the line for the direction of the laser pointer (on), the laser is visible by the reflection on the particles in the air.

The vertical line is called Normal and all angles must be measured with respect to this reference in optics.

Having these angles and the refractive index of water we can use the law of refraction

n₁ sin θ₁ = n₂ sin θ₂

θ₂ = $sin^{-1} ( \frac{n_1}{n_2} \ sin \ \theta_1 )$

we can repeat several times to analyze several different input points (different angles) and to decrease the errors in the measurements.

the refractive index of air is n1 = 1 and n2= 1.33 (water)

4 0

2 years ago