<span>Since the torque involves the product of force times lever arm, a small force can exert a greater torque than a larger force if the small force has a large enough lever arm.
With a large force exerts a small torque is a gate, hinged in its vertical line (axis). When pushed from a point near to the hinge, a very large amount is needed to open the gate.
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Answer:
Explanation:
An object is at rest along a slope if the net force acting on it is zero. The equation of the forces along the direction parallel to the slope is:
(1)
where
is the component of the weight parallel to the slope, with m being the mass of the object, g the acceleration of gravity, 
the angle of the slope
is the frictional force, with 
being the coefficient of friction and R the normal reaction of the incline
The equation of the forces along the direction perpendicular to the slope is
where
R is the normal reaction
is the component of the weight perpendicular to the slope
Solving for R,
And substituting into (1)
Re-arranging the equation,
This the condition at which the equilibrium holds: when the tangent of the angle becomes larger than the value of , the force of friction is no longer able to balance the component of the weight parallel to the slope, and so the object starts sliding down.
Answer:
a) 200A
b) 10.2V
c) 2.04kW
d)
I=80A
V=4.08V
P=0.326kW
Explanation:
Here we have a circuit of one power source and two resistors in series, the first question is asking for the current, so according to Ohm's Law:
Where R is the equivalent resistance of the resistors in series
![R=0.0510+0.0090=0.0600[ohm]](https://tex.z-dn.net/?f=R%3D0.0510%2B0.0090%3D0.0600%5Bohm%5D)
To calculate the voltage dropped by the motor we have to apply the voltage divider rule:
The power dissipated supplied to the motor is given by:
now solving adding a 0.0900 ohm resistor:
La intensidad de la luz se baja con cada bombillo que agregas
Answer:
R = 715.4 N
L = 166.6 N
Explanation:
ASSUME the painter is standing right of center
Let L be the left rope tension
Let R be the right rope tension
Sum moments about the left end to zero. Assume CCW moment is positive
R[5] - 20(9.8)[5/2] - 70(9.8)[5/2 + 2] = 0
R = 715.4 N
Sum moments about the right end to zero
20(9.8)[5/2] + 70(9.8)[5/2 - 2] - L[5] = 0
L = 166.6 N
We can verify by summing vertical forces
116.6 + 715.4 - (70 + 20)(9.8) ?=? 0
0 = 0 checks
If the assumption about which side of center the paint stood is incorrect, the only difference would be the values of L and R would be swapped.
