Answer:
42KVA
Explanation:
Given data
High Voltage (HV)= 480V
Low Voltage (LV)= 277V
Fo find
Size of transformer=?
Solution
To find the size of transformer here we use the co-ratio.The Co-ratio is given as:
Co-Ratio= (HV - LV)/HV
where
HV is High Voltage
LV is Low Voltage
Now put the values we get
Co- Ratio=(480-277)/480=.42
So the size of transformer is 42KVA
It can't be less than 250 N or the cart wouldn't move at all. That means there is only 1 answer. It's between not enough info or 250 N. The answer is 250 N. If it was any more, there would be acceleration.
Answer:
<em>The coefficient of static friction between the crate and the floor is 0.41</em>
Explanation:
<u>Friction Force</u>
When an object is moving and encounters friction in the air or rough surfaces, it loses acceleration and velocity because the friction force opposes motion.
The friction force when an object is moving on a horizontal surface is calculated by:
[1]
Where 
is the coefficient of static or kinetics friction and N is the normal force.
If no forces other then the weight and the normal are acting upon the y-direction, then the weight and the normal are equal in magnitude:
N = W = m.g
The crate of m=20 Kg has a weight of:
W = 20*9.8
W = 196 N
The normal force is also N=196 N
We can find the coefficient of static friction by solving [1] for :
The friction force is equal to the minimum force required to start moving the object on the floor, thus Fr=80 N and:
The coefficient of static friction between the crate and the floor is 0.41
Answer:
I_weight = M L²
this value is much larger and with it it is easier to restore balance.I
Explanation:
When man walks a tightrope, he carries a linear velocity, this velocity is related to the angular velocity by
v = w r
For man to maintain equilibrium needs the total moment to be zero
∑τ = I α
S τ = 0
The forces on the home are the weight of the masses, the weight of the man and the support on the rope, the latter two are zero taque the distance to the center of rotation is zero.
Therefore the moment of the masses and the open is the one that must be zero.
If the man carries only the bar, we could approximate it by two open one on each side of the axis of rotation formed by the free of the rope
I = ⅓ m L² / 4
As the length of half the length of the bar and the mass of the bar is small, this moment is small, therefore at the moment if there is some imbalance it is difficult to recover.
If, in addition to the opening, each of them carries a specific weight, the moment of inertia of this weight is
I_weight = M L²
this value is much larger and with it it is easier to restore balance.
Answer:
OK so ik this but what is you question?
Explanation:
