The first mechanized industry was

What statement about energy transfer in a wave is true?

laila [671]

Answer:

Energy moves between the particle of the medium.

Explanation:

8 0

3 years ago

Listed following are the names and mirror diameters for six of the world’s greatest reflecting telescopes used to gather visible

ziro4ka [17]

Answer:

Large binocular telescope, Keck 1 telescope, Hobby-Ebberly telescope, Subaru telescope, Gemini North telescope, Magellan 2 telescope

Explanation:

How much light a telescope can collect depends on its diameter, since in a bigger area more photons will be collected.

Remember that in a circle the area is defined as:

$A = \pi r^{2}$ (1)

Where A is the area and r is its radius.

However, the radius can be determined by means of its diameter.

$d = 2r$

$r = \frac{d}{2}$ (1)

Where d is its diameter.

An example of this is when a person is collecting raindrops with a bucket and with a cup. Since the bucket has a bigger area than the cup, it will collect more raindrops by unit of time. In this scenario the raindrops represent the photons.

To determine the light collecting area of each telescope, equation 2 will be replaced in equation 1.

$A = \pi (\frac{d}{2})^{2}$ (3)

Case for Large binocular telescope:

$A_{mirror1} = \pi (\frac{8.4m}{2})^{2}$

$A_{mirror1} = 55.41m$

For the second mirror will be the same value

$A = A_{mirror1}+A_{mirror2}$

$A = 55.41m+55.41m$

$A= 110.82m$

Case for Keck 1 telescope:

$A = \pi (\frac{10m}{2})^{2}$

$A = 78.53m$

Case for Hobby-Ebberly telescope:

$A = \pi (\frac{9.2m}{2})^{2}$

$A = 66.47m$

Case for Subaru telescope:

$A = \pi (\frac{8.3m}{2})^{2}$

$A = 54.10m$

Case for Gemini North telescope:

$A = \pi (\frac{8m}{2})^{2}$

$A = 50.26m$

Case for Magellan 2 telescope:

$A = \pi (\frac{6.5m}{2})^{2}$

$A = 33.18m$

Hence, they may be rank in the following way:

Large binocular telescope, Keck 1 telescope, Hobby-Ebberly telescope, Subaru telescope, Gemini North telescope, Magellan 2 telescope.

<em>Photons: particles that constitute light. </em>

3 0

3 years ago

To what extent are disorders the product of a particular environment, and to what extent

mixas84 [53]

Sorry don’t know hope you gave the answer to your questions

8 0

3 years ago

A hockey puck with mass 0.3 kg is shot across an ice-covered pond. Before the hockey puck was hit, the puck was at rest. After t

JulsSmile [24]

Answer:

The net friction force is 8.01 N

Explanation:

Net friction force = mass of hockey puck × acceleration

From the equations of motion

v^2 = u^2 + 2as

v = 40 m/s

u = 0 m/s (puck was initially at rest)

s = 30 m

40^2 = 0^2 + 2×a×30

60a = 1600

a = 1600/60 = 26.7 m/s^2

The acceleration of the puck is 26.7 m/s^2

Net friction force = 0.3 × 26.7 = 8.01 N

3 0

4 years ago

Read 2 more answers

A 10 kg ball strikes a wall with a velocity of 3 m/s to the left. The ball bounces off with a velocity of 3 m/s to the right. If

lord [1]

Answer:

The force is 272.73 newtons

Explanation:

We're going to use impulse-momentum theorem that states impulse is the change on the linear momentum this is:

$\overrightarrow{J}=\overrightarrow{p}_{f}-\overrightarrow{p}_{i}$ (1)

Impulse is also defined as average force times the time the force is applied:

$\overrightarrow{J}=\overrightarrow{F}_{avg}(\varDelta t)$ (2)

By (2) on (1):

$\overrightarrow{F}_{avg}(\varDelta t)= \overrightarrow{p}_{f}-\overrightarrow{p}_{i}$

solving for $\overrightarrow{F}_{avg}$ :

$\overrightarrow{F}_{avg}=\frac{\overrightarrow{p}_{f}-\overrightarrow{p}_{i}}{\varDelta t}$ (3)

We already know Δt is equal to 0.22 s, all we should do now is to find $\overrightarrow{p}_{f}-\overrightarrow{p}_{i}$ and put on (3) ( $\overrightarrow{p_{i}}$ the initial momentum and $\overrightarrow{p_{f}}$ the final momentum). Linear momentum is defined as $\overrightarrow{p}=m\overrightarrow{v}$ , using that on (3):

$\varDelta\overrightarrow{p}=m \overrightarrow{v_{f}}-m \overrightarrow{v_{i}}$ (4)

Velocity (v) are vectors so direction matters, if positive direction is the right direction and negative direction left $\overrightarrow{v_{i}}=+3\, \frac{m}{s}$ and $\overrightarrow{v_{f}}=-3\, \frac{m}{s}$ so (4) becomes:

$\varDelta\overrightarrow{p}=m(-3\frac{m}{s}- (+3\frac{m}{s}))=-(10kg)(6\frac{m}{s})$

$\varDelta\overrightarrow{p}=-60\, \frac{mkg}{s}$ (5)

Using (5) on (3):

$\overrightarrow{F}_{avg}=\frac{-60\, \frac{mkg}{s}}{0.22s}$

$F_{avg}=272.73N$

8 0

4 years ago