The power delivered is equal to the product between the voltage V and the current I:
This power is delivered for a total time of
, so the total energy delivered to the battery is
This power is delivered for a total time of
The name (blank) means "all land"
1 answer:
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Answer:
12164.4 Nm
Explanation:
CHECK THE ATTACHMENT
Given values are;
m1= 470 kg
x= 4m
m2= 75kg
Cm = center of mass
g= acceleration due to gravity= 9.82 m/s^2
The distance of centre of mass is x/2
Center of mass(1) = x/2
But x= 4 m
Then substitute, we have,
Center of mass(1) = 4/2 = 2m
We can find the total torque, through the summation of moments that comes from both the man and the beam.
τ = τ(1) + τ(2)
But
τ(1)= ( Center of m1 × m1 × g)= (2× 470× 9.81)
= 9221.4Nm
τ(2)= X * m2 * g = ( 4× 75 × 9.81)= 2943Nm
τ = τ(1) + τ(2)
= 9221.4Nm + 2943Nm
= 12164.4 Nm
Hence, the magnitude of the torque about the point where the beam is bolted into place is 12164.4 Nm
Answer:
The moment is -78.4 N-m (clockwise).
Explanation:
Given:
Mass of the object (m) = 20 kg
Distance of the object from the knee joint (d) = 0.4 m
Weight of leg is not considered.
Acceleration due to gravity (g) = 9.8 m/s²
Now, weight of the object is equal to the product of its mass and acceleration due to gravity. So,
Weight = Mass × Acceleration due to gravity
=
We know that, moment of a force about a point is defined as the product of force applied and the perpendicular distance between the point and the line of application of force.
Moment of the given weight about the knee joint is given as:
Moment about knee joint = Weight × Distance from knee joint to weight
Moment about knee joint = 196 × 0.4 = 78.4 Nm
Now, from the diagram below, we can observe that, the weight acts vertically down and thus the sense of rotation about the knee joint at point O is clockwise. So, moment is negative.
Therefore, the moment is -78.4 N-m (clockwise).
