Which of these structures is an architect most likely to design?
2 answers:
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Answer:
The driver hits the stationery dog because the applied force is less than required force
Explanation:
Kinetic energy will be given by
where m is the mass of the vehicle and v is the speed/velocity of the vehicle.
Substituting 800 Kg for m and 20 m/s for v we obtain
Frictional force by vehicle pads is given by
where d is the distance moved
Substituting 160000 for KE and 50 m for d we obtain
Therefore, the vehicle hits the dog since the required force is 3200N but the driver applied only 2000 N
Explanation:
It is known that electric field is responsible for creating electric potential. As a result, it depends only on the electric field and not on the magnitude of charge.
So, when a charge is increased by a factor of 2 then electric potential will remain the same. Since, expression to calculate the electric potential is as follows.
U = qV
Since, the electric potential is directly proportional to the charge. Hence, when 0.2 tends to replaced by 0.4
then charge is increased by a factor of 2. Hence, the electric potential energy is doubled.
Thus, we can conclude that if that charge is replaced by a +0.4 µC charge then electric potential stays the same, but the electric potential energy doubles.
The weight of the meterstick is:
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
from which we find the value of d2:
So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
from which we find the value of d2:
So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.