Explanation:
It is given that,
A nerve signal travels 150 meters per second. It is the speed of the nerve signal. We need to convert the number of kilometers that the nerve signal will travel in the same time.
We know that,
1 kilometer = 1000 meter
1 hour = 3600 seconds
So, the nerve signal will travel at the rate of 540 km/h. Hence, this is the required solution.
The Atwood's machine is in motion starting from rest, then Vf = Vo + a(t).
<span>Final Velocity is given as 6.7 m/s and the time is 1.9 s thus 6.7= 0+ a(1.9) </span>
<span>then a = 6.7/1.9 = 3.526 m/s². </span>
<span>The Atwood's Machine also has the formula d= distance = 1/2a(t²) </span>
<span>distance given is 6.365 m , then 6.365 = 1/2 a (1.9)², </span>
<span>a = 3.526 m/s² the same acceleration. </span>
<span>a= g(m1-m2) / m1+m2) </span>
<span>m1a + m2a = m1g - m2g </span>
<span>m1a - m1g = -m2g - m2a </span>
<span>3.526 m1 - 9.81 m1 = -9.81m2 - 3.526 m2 </span>
<span>-6.28 m1 = -13.34 m2 </span>
<span>0.47 m1= m2 </span>
<span>if 24J = 1/2mv² </span>
<span>then 24J = 1/2 m1 ( 6.7)² </span>
<span>48/ 44.89 = m1 </span>
<span>1.069 kg = m1 , then </span>
<span>0.47(1.069) = m2 </span>
<span>0.503 kg = m2</span>
If the lightbulb A in the circuit shown in the image burned out, the path for the current to flow is disrupted because one of its terminals is connected direct to the source. So, there will be no current through the lightbulbs B, C, and D, and they will turn off. Similarly it will happen, if the lightbulb D burned out.
If the lightbulb B burned out the current will continue circulating through the lightbulbs A, C, and D, because lightbulb B is connected in parallel. Similarly it will happen, if the lightbulb C burned out.
There are several information's already given in the question. The answer can be easily deduced using those information's.
Time = 3.0 * 10-3 seconds
Impulse = 0.30 newton
Then
Force = Impulse/Time
= 0.30/3.0 * 10-3
= 1 * 10^3 newtons.
I hope the above procedure is clear for you to understand and it has actually come to your great help.
