Answer:
JA
Explanation:
s of time, (b) the velocity and acceleration at t = 2.0 s, (c) the time at which the position is a maximum, (d) the time at which the velocity is zero, and (e) the maximum position. Assume all variable and constants are in SI units.
Answer:
<em>Maximum=70 m</em>
<em>Minimum=26 m</em>
Explanation:
<u>Vector Addition
</u>
Since vectors have magnitude and direction, adding them takes into consideration not only the magnitudes but also their respective directions. Two vectors can be totally collaborative, i.e., point to the same direction, or be totally opposite. In the first case, the magnitude of the sum is at maximum. Otherwise, it's at a minimum.
Thus, the maximum magnitude of the sum is 48+22 = 70 m and the minimum magnitude of the sum is 48-22= 26 m
Answer:
The value to be reported is 5.48V
Explanation:
The RMS (root mean square) is defined as the value of voltage that will produce the same heating effect, or power dissipation, in circuit, as this AC voltage.
The RMS voltage is also called effective voltage because it is just as effective as DC voltage in providing power to an element.
It is expressed as 
= 
where Vm is the maximum or peak value of the voltage
In calculating the RMS of the voltage , we simply divide the peak voltage by square root of 2 (√2)
= 
= 
= 5.48 V
(a) The work done by the force applied by the tractor is 79,968.47 J.
(b) The work done by the frictional force on the tractor is 55,977.93 J.
(c) The total work done by all the forces is 23,990.54 J.
<h3>
Work done by the applied force</h3>
The work done by the force applied by the tractor is calculated as follows;
W = Fd cosθ
W = (5000 x 20) x cos(36.9)
W = 79,968.47 J
<h3>Work done by frictional force</h3>
W = Ffd cosθ
W = (3500 x 20) x cos(36.9)
W = 55,977.93 J
<h3>Net work done by all the forces on the tractor</h3>
W(net) = work done by applied force - work done by friction force
W(net) = 79,968.47 J - 55,977.93 J
W(net) = 23,990.54 J
Learn more about work done here: brainly.com/question/25573309
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