Answer:
<h3>
<em>2</em><em>4</em><em>7</em><em>9</em><em> </em><em>Newton</em></h3>
<em>Sol</em><em>ution</em><em>,</em>
<em>Mass</em><em>=</em><em>1</em><em>0</em><em>0</em><em> </em><em>kg</em>
<em>Accele</em><em>ration</em><em> </em><em>due</em><em> </em><em>to</em><em> </em><em>gravity</em><em>(</em><em>g</em><em>)</em><em>=</em><em>2</em><em>4</em><em>.</em><em>7</em><em>9</em><em> </em><em>m</em><em>/</em><em>s^</em><em>2</em>
<em>Now</em><em>,</em><em>.</em>
<em>
</em>
<em>hope</em><em> </em><em>this</em><em> </em><em>helps</em><em> </em><em>.</em><em>.</em>
<em>Good</em><em> </em><em>luck</em><em> on</em><em> your</em><em> assignment</em><em>.</em><em>.</em>
Use this formula to find your answer...
Determine the frequency of a clock waveform whose period is 2us or (micro) and 0.75ms
frequency (f)=1/( Time period).
Frequency of 2 us clock =1/2*10^-6 =10^6/2 =500000Hz =500 kHz.
Frequency of 0..75ms clock =1/0.75*10^-3 =10^3/0.75 =1333.33Hz =1.33kHz.
<u>Given </u><u>:</u><u>-</u>
- An elevator is moving vertically up with an acceleration a.
<u>To </u><u>Find</u><u> </u><u>:</u><u>-</u>
- The force exerted on the floor by a passenger of mass m .
<u>Solution</u><u> </u><u>:</u><u>-</u>
As the man is in a accelerated frame that is <u>non </u><u>inertial</u><u> frame</u><u> </u>, we would have to think of a pseudo force .
- The direction of this force is opposite to the direction of acceleration the frame and its magnitude is equal to the product of mass of the concerned body with the acceleration of the frame .
For the FBD refer to the attachment . From that ,
<u>Hence</u><u> </u><u>option</u><u> </u><u>d </u><u>is </u><u>correct</u><u> </u><u>choice </u><u>.</u>
<em>I </em><em>hope</em><em> this</em><em> helps</em><em> </em><em>.</em>
If you are given distance and a period of time, you can calculate
the speed. The velocity of an object is the rate of change of its position with
respect to a frame of reference, and is a function of time. Velocity is
equivalent to a specification of its speed and direction of motion (e.g. 60
km/h to the north).
(a) Let's convert the final speed of the car in m/s:

The kinetic energy of the car at t=19 s is

(b) The average power delivered by the engine of the car during the 19 s is equal to the work done by the engine divided by the time interval:

But the work done is equal to the increase in kinetic energy of the car, and since its initial kinetic energy is zero (because the car starts from rest), this translates into

(c) The instantaneous power is given by

where F is the force exerted by the engine, equal to F=ma.
So we need to find the acceleration first:

And the problem says this acceleration is constant during the motion, so now we can calculate the instantaneous power at t=19 s: