An organism that cannot make its own food called
2 answers:
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Explanation:
It is given that,
Potential difference between the ends of a rod, V = 1.1 V
Length of the rod, l = 10 cm = 0.1 m
Area of cross section of the rod,
The resistivity of graphite,
(a) Let R is the resistance of the rod. It is given by :
So, the resistance of the rod is 0.833 ohms.
(b) Let I is the current flowing in the wire. It can be calculated using the Ohm's law as :
I = 1.32 A
(c) Let E is the electric field inside the rod. The electric field in terms of potential difference is given by :
E = 11 V/m
Hence, this is the required solution.
Answer:
Explanation:
The question is ;
3.80 m-long, 500. kg steel beam extends horizontally from the point
where it has been bolted to the framework of a new building under
construction. A 67.0 kg construction worker stands at the far end of
the beam. What is the magnitude of the torque about the point where the beam is bolted into place?
Solution:
Here two torques are in action
a) torque T1 due to weight of worker at the edge of the beam - at center of the beam
b) torque T2 due to weight of the uniform beam - at the point where the beam is bolted
So we first calculate the torque produced due to weight of the worker;
We can see that;
Distance of worker from the center of the beam = 1.9m
Mass of the worker = 67 kg
Value of g= 9.8 m/sec2
T1=force × distance from point of rotation
Here force is weight of the worker which is = mass × g=67×9.8=656.6 N
So the torque is
T1= 656.6×1.9=12225.892 Nm
or
T1 = 12225.892 Nm
Now torque by the beam itself ;
Length of the beam from its center point to the bolt point = 1.9 m
Weight force of beam acting at center point= mass× g= 500×9.8=4900 N
Torque T2 at bolt point by the beam weight = weight of beam × length of beam from its center to the bolt point = 4900×1.9=9310 Nm
T2=9310Nm
So total torque = T1+T2= 12225.892+9310=21565.892 Nm
