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olga nikolaevna [1]
2 years ago
12

In the communications process, the ______ is the originator of the message while the _____ is the medium of transmission that ca

rries the messages to the receiver.
Physics
1 answer:
motikmotik2 years ago
5 0

In the communications process, the <u>sender </u>is the originator of the message while the <u>channel</u> is the medium of transmission that carries the messages to the receiver.

The message sent by the sender is transmitted to the receiver through a medium known as a channel or system of communication. The channel used for communication affects how a receiver will receive the message.

The Communication channel is categorized into three channels; verbal, non-verbal, and written. Every channel has its own pros and cons.

Sometimes disturbance occurs in the communication channels and due to this message is not properly decoded at the receiver end.

If you need to learn about how <u>sender </u>know that the receiver got the message, click here

https://brainly.in/question/49802218?referrer=searchResults

#spj4

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A fire hose has an inside diameter of 6.40 cm. Suppose such a hose carries a flow of 40.0 L/s starting at a gauge pressure of 1.
inessss [21]

Answer:

The Reynolds numbers for flow in the fire hose.

Explanation:

Given that,

Diameter = 6.40 cm

Rate of flow = 40.0 L/s

Pressure P=1.62\times10^{6}\ N/m^2

We need to calculate the Reynolds numbers for flow in the fire hose

Using formula of rate of flow

Q=Av

v=\dfrac{Q}{A}

Where, Q = rate of flow

A = area of cross section

Put the value into the formula

v=\dfrac{40.0\times10^{-3}}{3.14\times(3.2\times10^{-2})^2}

v=12.44\ m/s

We need to calculate the Reynolds number

Using formula of the Reynolds number

n_{R}=\dfrac{2\rho\times v\times r}{\eta}

Where, \eta =viscosity of fluid

\rho =density of fluid

Put the value into the formula

n_{R}=\dfrac{2\times100\times12.44\times3.2\times10^{-2}}{1.002\times10^{-3}}

n_{R}=7.945\times10^{5}

Hence, The Reynolds numbers for flow in the fire hose.

3 0
3 years ago
If a car moves with a uniform speed of 2m/s in a circle of radius 0.4m, what is its angular speed?
REY [17]

Answer:

The car's angular speed is \frac{rad}{s}.

Explanation:

Angular velocity is usually measured with \frac{radians}{sec}, so I'm going to use that to answer your question.

The relationship between tangential velocity and angular velocity (ω)  is given by:

V = \omega R

Using the values from the question, we get:

2 \frac{m}{s} = \omega (0.4m)

\omega = 5 \frac{rad}{s}

Therefore, the car's angular speed is 5 \frac{rad}{s}.

Hope this helped!

3 0
3 years ago
A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.
kow [346]

Before the engines fail, the rocket's altitude at time <em>t</em> is given by

y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2

and its velocity is

v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t

The rocket then reaches an altitude of 1150 m at time <em>t</em> such that

1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2

Solve for <em>t</em> to find this time to be

t=11.2\,\mathrm s

At this time, the rocket attains a velocity of

v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}

When it's in freefall, the rocket's altitude is given by

y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2

where g=9.80\frac{\rm m}{\mathrm s^2} is the acceleration due to gravity, and its velocity is

v_2(t)=124\dfrac{\rm m}{\rm s}-gt

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for y_2(t) to reach 0:

1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

{v_f}^2-{v_i}^2=2a\Delta y

where v_f and v_i denote final and initial velocities, respecitively, a denotes acceleration, and \Delta y the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means y_2 will contain the information we need to find the maximum height.

-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)

Solve for y_{\rm max} and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to y_2(t)) to be about 32.6 s. Plug this into v_2(t) to find the velocity before it crashes:

v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

3 0
4 years ago
A uniform rod is attached to a wall by a hinge at its base. The rod has a mass of 6.5 kg, a length of 1.3 m, is at an angle of 1
liubo4ka [24]

Answer:

tension wire 104 N, horizontal force hinge 104 N, vertical force hinge 63.7 N

Explanation:

The system described is in static equilibrium under the action of several forces, which are shown in the attached diagram, where T is the tension of the wire, W is the weight of the bar, Fx is the horizontal reaction of the wall and Fy It is the vertical reaction of the wall.

The directions of the forces are indicated by the arrows and are marked intuitively, but when solving the problem if one gives a negative value this indicates that the direction is wrong, but does not alter the results of the problem

For the resolution we use Newton's second law, both in translational and rotational equilibrium, if necessary.

We must establish a reference system to assign the positive meaning, we place it with the origin in the hinge, and the positive directions to the right and up. The location of the coordinate system allows us to eliminate the reaction of the hinge by having zero distance to origin. We write the equilibrium equations for each axis

     ∑ Fx = 0

     Fx -T =0

     ∑ Fy =0

     Fy -W =O      W = m g

     ∑τ =0

      -T dy + W dx =0

For rotational equilibrium we take the positive direction as counterclockwise rotation and distance is the perpendicular distance of the force to the axis of the coordinate system

     

       dy = L sin 17

      dx = (L/2) cos 17

Where L is the length of the bar and the weight is applied at the center of it, we write and simplify the equation

    -T L sin 17 + mg (L / 2) cos 17 = 0

   - T sin 17 + mg/2  Cos 17 =0

We write the three equations together

     Fx- T = 0

     Fy -W = 0

     -T sin 17 + (m g / 2) cos 17 = 0

a) With the third equation we can find the wire tension

     T = m g / 2 cos 17 / sin17

     T = 6.5  9.8/2 cotan 17

     T = 104 N

b) We use the first equation to find Fx

       Fx- T =0

       Fx = T = 104 N

c) We use the second equation to find Fy

 

       Fy = W = m g

       Fy = 6.5 9.8

       Fy = 63.7 N

4 0
3 years ago
On April 15, 1999, a South Korean cargo plane crashed due to a confusion over units. The plane was to fly from Shanghai, China,
sattari [20]

Answer:

993 m or 3257 ft

Explanation:

The captain was told to fly at at 1500 ft altitude. At the rate of 3.28 ft to 1 m, this is

\dfrac{1500}{3.28}=457 metres.

Since he was at 1450 m, he thought he was above the correct altitude by

1450 - 457 = 993 m.

In feet, this is

993 * 3.28 = 3257 ft

6 0
4 years ago
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