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White raven [17]
4 years ago
15

3.An object that begins at rest has an acceleration of 2 m/s/s What is its instantaneous speed after 3 seconds?​ ​

Physics
1 answer:
neonofarm [45]4 years ago
6 0
The speed will be 6m/s i believe
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An Object, Start from rest w Confront Aiceleration 8m/s2 along a
shutvik [7]

Answer:

A)   v = 40 m / s, B)   v_average = 20 m / s

Explanation:

For this exercise we will use the kinematics relations

         

A) the final velocity for t = 5 s and since the body starts from rest its initial velocity is zero

         v = vo + a t

         v = 0 + 8 5

         v = 40 m / s

B) the average velocity can be found with the relation

         v_average = vf + vo / 2

         v-average = 0+ 40/2

          v_average = 20 m / s

6 0
3 years ago
What is the difference in the pitch when you are in front of the car compared to behind the car?
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Answer:The Doppler effect is a change in the frequency of sound waves that occurs when the source of the sound waves is moving relative to a stationary listener.

As the source of sound waves approaches a listener, the sound waves get closer together, increasing their frequency and the pitch of the sound. The opposite happens when the source of sound waves moves away from the listener.

Explanation:

7 0
3 years ago
I need help the option for D is the golfball isnt accelerating
algol13

Answer:

es b

Explanation:

no se

8 0
3 years ago
If a single-load Series circuit has one 20 Ohm bulb, and a
Alex Ar [27]

At the instant the second bulb is connected ...

... the brightness of the <u>first bulb</u> doesn't change

... the brightness of the <u>second bulb</u> changes from dark (no brightness) to equal to the first bulb

5 0
4 years ago
The efficiency of a squeaky pulley system is 73 percent. The pulleys are usedto raise a mass to a certain height. What force is
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The efficiency of the machine is defined as

\eta = \frac{W_{out}}{W_{in}}

Here

Work out is the work output and Work in is the work input

To find the Work in we have then

W_{in} = \frac{W_{out}}{\eta}

W_{in} = \frac{mgh}{\eta}

Replacing with our values

W_{in} = \frac{(58)(9.8)(3)}{73\%}

W_{in} = 2335.89J

The work done by the applied force is

W = Fd

Here,

F = Force

d = Distnace

Rearranging to find F,

F = \frac{W}{d}

F = \frac{2335.89J }{18}

F = 129.77N

Therefore the force exerted on the machine after rounding off to two significant figures is 130N

8 0
3 years ago
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