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3 years ago

Pls help asap. easy science question.

Physics

2 answers:

Solnce55 [7]3 years ago

6 0

This is not a very good set of choices.\

ANY type of electromagnetic wave can be used to send information from place to place, IF we can set up the following conditions:

-- We can only use the electromagnetic waves that are not dangerous to be around.

-- We also can't use the ones that can be blocked or degraded by natural things, like rain, fog, or smoke.

-- We need an easy way to generate them whenever we want them.

-- We need an easy way to change something about them in a pattern that carries the information.

-- We need an easy way to detect them.

-- We need an easy way to accurately detect the changes, so that we can pull the information off of them.

The electromagnetic waves that DO meet all of those requirements, and we DO use to communicate information are:

-- Radio

-- Microwave

-- Infra-red

-- Visible light

(We use visible light to transmit information when we send it through an optical fiber, and also when we wave at our friend across the street !)

Zanzabum3 years ago

3 0

microwaves and radio waves can transmit info. Radio waves can be used for radio. Microwaves are used in tv remotes and etc.

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A car moving at 95 km/h passes a 1.00-km-long train traveling in the same direction on a track that isparallel to the road. If t

victus00 [196]

Answer:

Time to pass the train=0.05 h

How far the car traveled in this time=4.75 Km

Explanation:

We have that the train and the car are moving in the same direction, the difference between the speed of the vehicles is:

$\Delta V=V_{car}-V_{train}=95km/h-75km/h=20km/h$

We will use this difference in the speed of the car an train to calculate how much time take the car to pass the train. For this we have that the train is 1km long and the car is moving with a speed of 20km/h (we use this value because is the speed that the car have in advantage of the train) then for a movement with a constant speed we have:

$V=\dfrac{x}{t}$

Where x is the distance, t is the time and v is the speed. using the data that we have:

$V=\dfrac{x}{t}=\dfrac{1km}{20km/h}=0.05h$

This is the time that the car take to pass the train. Now to calculate how far the car have traveled in this time we have to considered the speed of 95Km/h of the car, then:

$V=\dfrac{x}{t}\\x=v\cdot t\\x=95km/h\cdot 0.05h\\x=4.75km$

7 0

3 years ago

At the end of the adiabatic expansion, the gas fills a new volume V₁, where V₁ > V₀. Find W, the work done by the gas on the

tino4ka555 [31]

Answer:

$W=\frac{p_0V_0-p_1V_1}{\gamma-1}$

Explanation:

An adiabatic process refers to one where there is no exchange of heat.

The equation of state of an adiabatic process is given by,

$pV^{\gamma}=k$

where,

$p$ = pressure

$V$ = volume

$\gamma=\frac{C_p}{C_V}$

$k$ = constant

Therefore, work done by the gas during expansion is,

$W=\int\limits^{V_1}_{V_0} {p} \, dV$

$=k\int\limits^{V_1}_{V_0} {V^{-\gamma}} \, dV$

$=\frac{k}{\gamma -1} (V_0^{1-\gamma}-V_1^{1-\gamma})\\$

(using $pV^{\gamma}=k$ )

$=\frac{p_0V_0-p_1V_1}{\gamma-1}$

4 0

3 years ago

starting at a position of 0 m, if the man is moving at a constant velocity of 2 m/s, it will take seconds for him to reach a pos

Harman [31]

The calculated time is 6 seconds.

Time is defined by physicists as the flow of events from the past through the present and into the future. In essence, a system is timeless if it is unchanging. When describing events that take place in three-dimensional space, time can be thought of as the fourth dimension of reality. Even if time isn't directly connected to energy, it is undoubtedly connected to many other components of energy. For instance, the movement of energy across the universe can affect the direction of time (from the past to the future).

V= 2 m/s

d=12 m

t=?

we know that velocity=displacement / time

time= displacement / velocity

= 12/2

=6 seconds

the calculated time is 6 seconds.

Learn more about time here-

brainly.com/question/28050940

#SPJ4

4 0

1 year ago

A real (non-Carnot) heat engine, operating between heat reservoirs at temperatures of and performs 4.1 kJ of net work, and rejec

Sati [7]

There are some missing data in the problem. The full text is the following:
"A real (non-Carnot) heat engine, operating between heat reservoirs at temperatures of 710 K and 270 K performs 4.1 kJ of net work, and rejects 9.7 kJ of heat, in a single cycle. The thermal efficiency of a Carnot heat engine, operating between the same heat reservoirs, in percent, is closest to.."

Solution:
The efficiency of a Carnot cycle working between cold temperature $T_C$ and hot temperature $T_H$ is given by
$\eta = 1 - \frac{T_C}{T_H}$
and it represents the maximum efficiency that can be reached by a machine operating between these temperatures. If we use the temperatures of the problem, $T_C=270 K$ and $T_H=710 K$ , the efficiency is
$\eta = 1 - \frac{270 K}{710 K}=0.62 = 62%$

Therefore, the correct answer is D) 62 %.

6 0

3 years ago

A stunt cyclist rides on the interior of a cylinder 13 m in radius. The coefficient of static friction between the tires and the

ZanzabumX [31]

Answer:

$v=13.1m/s$

Explanation:

From the question we are told that:

Radius r=13

Coefficient of static friction \mu=0.67

Given the Centripetal concept

Generally the equation for Velocity v is mathematically given by

$v=\sqrt{\frac{gr}{\mu}}$

$v=\sqrt{\frac{9.81*13}{0.67}}$

$v=13.1m/s$

The value of the minimum speed (in m/s) for the cyclist to perform the stunt is Given as

$v=13.1m/s$

7 0

3 years ago