Answer:
Explanation:
Given
Force
time interval
Impulse is given by
For two significant Figure
3.20-m-long, 500 kgkg steel uniform beam extends horizontally from the point where it has been bolted to the framework of a new
building under construction. A 70 kgkg construction worker stands at the far end of the beam.What is the magnitude of the torque about the point where the beamis bolted into place?
1 answer:
Answer:
10,035
Explanation:
Length of the beam is 3.2m
Mass of steel beam is 500kg
Weight of steel =mg
Ws=400×9.8=4900N
Also, man of mass =70kg
Weight of man(Wm)=70×9.8=686N
The weight of the man will act as the middle of the beam
Torque is given as Force ×distance
tnet=Wm×L + Ws×L/w
tnet=686×3.2+4900×1.6
tnet=10,035.2Nm
The net torque about where the beam is bolted is 10,035.2Nm
Check attachment for diagram
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Answer:
The final velocity of the second car is 57 m/s south.
Explanation:
This is an elastic collision between two train cars. In this case, the total kinetic energy between the two bodies will remain the same.
The formula to apply is :
where ;
Given in the question that;
Apply the formula as;
{14650*18}+{3825*11} = {14650 *6} + {3825 * v₂f}
263700+42075=87900 + 3825v₂f
305775 =87900 + 3825v₂f
305775-87900 = 3825v₂f
217875=3825v₂f
217875/3825 =v₂f
56.96 = v₂f
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Given data
*The value of battery voltage is V = 10 V
*The current flows through the resistor is I = 5 A
The formula for the resistor is given by the Ohm's law as
Substitute the values in the above expression as