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4 years ago

MIXED MES AGE

Physics

1 answer:

Mashcka [7]4 years ago

6 0

Answer:

I'm not super sure what your asking for. But these look like elements of communication from a speech or communications class.

--------------message-------->

O O

\ | / ------ channel ------ \ | /

/\ /\

sender (encoder) receiver (decoder)

<----------feedback-------------

The sender (or encoder) encodes the message and sends it.
The receiver (or decoder) decodes the message and receives it, then gives feedback to the sender.
The channel is the medium through which the message is sent: this could be social media, email, text, talking, you name it.
Body language can assist (or not) in sending the message.

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A rock is dropped from a vertical cliff. The rock takes 3.00 s to reach the ground below the cliff. A second rock is thrown vert

Phantasy [73]

Answer:

Velocity v= 12.25 $\frac{m}{s}$

Explanation:

The first rock dropped give the distance Y in meters

$Y_{f}=Y_{o}+v_{o}*t +\frac{1}{2}*a*t^{2}\\ Y_{f}=\frac{1}{2}* 9.8 \frac{m}{s^{2} }* 3^{2} \\Y_{f}=44.1 m$

Now the motion of the second rock the time change so to know the velocity

$Y_{f}= Y_{o} +v_{o}*t +\frac{1}{2}*a *t^{2} \\v_{o}*t= -Y_{f} +\frac{1}{2} *a*t^{2} \\v_{o} =\frac{-44.1 +0.5 * 9.8* s^{2} }{2} \\v_{o}= 12.25 \frac{m}{s}$

7 0

3 years ago

Where is the independent variable usually located in a scientific table? First column Second column Third through fifth columns

Reika [66]

Explanation:

On the horizontal side left side

4 0

3 years ago

A car of 1000 kg with good tires on a dry road can decelerate (slow down) at a steady rate of about 5.0 m/s2 when braking. If a

vredina [299]

(a) 4.0 s

The acceleration of the car is given by

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time interval

For this car, we have

v = 0 (the final speed is zero since the car comes to a stop)

u = 20 m/s is the initial velocity

$a=-5.0 m/s^2$ is the deceleration of the car

Solving the equation for t, we find the time needed to stop the car:

$t=\frac{v-u}{a}=\frac{0-(20 m/s)}{-5.0 m/s^2}=4 s$

(b) 40 m

The stopping distance of the car can be calculated by using the equation

$v^2 - u^2 = 2ad$

where

v = 0 is the final velocity

u = 20 m/s is the initial velocity

a = -5.0 m/s^2 is the acceleration of the car

d is the stopping distance

Solving the equation for d, we find

$d=\frac{v^2-u^2}{2a}=\frac{0^2-(20 m/s)^2}{2(-5.0 m/s^2)}=40 m$

(c) $-5.0 m/s^2$

The deceleration is given by the problem, and its value is $-5.0 m/s^2$ .

(d) 5000 N

The net force applied on the car is given by

$F=ma$

where

m is the mass of the car

a is the magnitude of the acceleration

For this car, we have

m = 1000 kg is the mass

$a=5.0 m/s^2$ is the magnitude of the acceleration

Solving the formula, we find

$F=(1000 kg)(5.0 m/s^2)=5000 N$

(e) $2.0\cdot 10^5 J$

The work done by the force applied by the car is

$W=Fd$

where

F is the force applied

d is the total distance covered

Here we have

F = 5000 N

d = 40 m (stopping distance)

So, the work done is

$W=(5000 N)(40 m)=2.0\cdot 10^5 J$

(f) The kinetic energy is converted into thermal energy

Explanation:

when the breaks are applied, the wheels stop rotating. The car slows down, as a result of the frictional forces between the brakes and the tires and between the tires and the road. Due to the presence of these frictional forces, the kinetic energy is converted into thermal energy/heat, until the kinetic energy of the car becomes zero (this occurs when the car comes to a stop, when v = 0).

8 0

4 years ago

dean is slowing down on his skateboards. he starts at a speed of 5.5 m/s and slows to 1.0 m/s over a time of 3.0 seconds. what i

vekshin1

Use one of the equations of accelerated motion; V2=V1 + at ...see attached