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3 years ago

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A robot arm moves so that P travels in a circle about point B, which is not moving. Knowing that P starts from rest, and its spe

ed increases at a constant rate of 10 mm/s2, determine (a) the magnitude of the acceleration when t=4s.

1 answer:

algol133 years ago

4 0

Answer

Acceleration=>10.20mm/s

Explanation

Below in image

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Plzzzzzzzzzz!!!!!!! Hurryyyyy

Scilla [17]

Answer:

student A or B

Explanation:

A common demonstration is to put a ringing alarm clock or bell in the bell jar, and when the vacuum is created, you can no longer hear the sound of the clock/bell.

The bell is connected to a lab pack or batteries and rung to show pupils it can be heard under normal circumstances. The bell jar is then connected to a vacuum pump using a vacuum plate (see Fig 2) and the air is removed from inside creating a near vacuum. The bell is then again rung. This time however, it cannot be heard.

Small low voltage buzzers can be used as a bell replacement for the bell and work in exactly the same way though teachers generally prefer bells as students may be able to see the hammer moving, proving that it is actually ringing even though they cannot hear it.

Some vacuum pumps are better than others at keeping a strong vacuum though if you cannot completely lose the sound, you will at least notice the volume decreasing.

Sound is simply a series of longitudinal waves travelling from the source, through the air to our ears. Without air present, these waves cannot form and therefore sound cannot be conveyed.

In a longitudinal wave the particles oscillate back and forth in the direction of the wave movement unlike transverse waves which like waves on the sea, single particles travel up and down and not in the direction of the wave.

Because you will not be able to create a perfect vacuum, you may still be able to hear the bell ring slightly. Vibrations from the ringing bell can also travel up to the bung in the bell jar which in turn may resonate the jar slightly. This means you may hear the bell ring, however strong the vacuum. To compensate for this, try to insulate the bell as much as possible from the bell jar. Hanging the bell using elastic cord means some of the vibrations will be absorbed by the cord and not be transferred to the bell jar.

3 0

3 years ago

Consider two circular metal wire loops each carrying the same current I as shown below. In what region(s) could the net magnetic

nadezda [96]

Answer:

you counteract the Chromozones with the numerzones which contacts the specimen in the brain cells which then has voltage in the hand that connects with the wire then wallah

Explanation:

7 0

3 years ago

A bullet with mass 1.0kg and velocity 180 m/s is brought to rest in 0.02 s by a sandbag.assuming constant acceleration in the sa

kotykmax [81]

Hello
The bullet is moving by uniformly accelerated motion.
The initial velocity is $v_i=180~m/s$ , the final velocity is $v_i=0~m/s$ , and the total time of the motion is $\Delta t=0.02~s$ , so the acceleration is given by
$a= \frac{v_f-v_i}{\Delta t} = -9000~m/s^2$
where the negative sign means that is a deceleration.
Therefore we can calculate the total distance covered by the bullet in its motion using
$S=v_i t + \frac{1}{2}at^2 = 180~m/s \cdot 0.02~s + \frac{1}{2}(-9000~m/s^2)(0.02~s)^2=1.8~m$
So, the bullet penetrates the sandbag 1.8 meters.

5 0

3 years ago

1. Indicate whether these objects or atoms are positively,<br> negatively or neutrally charged.

vovangra [49]

Answer:

Neutrally charged!!!!!!!!!!!!!!!!!!!!!

Explanation:

6 0

3 years ago

The period of rotation of Mars is 1 day and 37 minutes. Determine its frequency of rotation in Hertz.

Sholpan [36]

The frequency of rotation of Mars is 0.0000113 Hertz.

<u>Given the following data:</u>

Period = 1 day and 37 minutes.

To find the frequency of rotation in Hertz:

First of all, we would convert the the value of period in days and minutes to seconds because the period of oscillation of a physical object is measured in seconds.

<u>Conversion:</u>

1 day = 24 hours

24 hours to minutes = $60$ × $24$ = $1440$ minutes

$1440 + 37 = 1477 \; minutes$

1 minute = 60 seconds

1477 minute = X seconds

Cross-multiplying, we have:

$X = 60$ × $1477$

X = 88620 seconds

Now, we can find the frequency of rotation of Mars by using the formula:

$Frequency = \frac{1}{Period}\\\\Frequency = \frac{1}{88620}$

<em>Frequency </em><em>of rotation</em> = <em>0.0000113 Hertz</em>

Therefore, the frequency of rotation of Mars is 0.0000113 Hertz.

Read more: brainly.com/question/14708169

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3 years ago