Answer:
Acceleration a ≤ 3 m/s^2
the greatest acceleration that the truck can have without losing its load is 3 m/s^2
Explanation:
For the truck to accelerate without losing its load.
Acceleration force of truck must be less than or equal to the maximum friction force between the truck bed and the load.
Fa ≤ F(friction)
But;
Fa = mass × acceleration
Fa = ma
ma ≤ F(friction)
a ≤ (F(friction))/m ......1
Given;
Fa = mass × acceleration
Fa = ma
mass m = 800 kg
F(friction) = 2400 N
Substituting the given values into equation 1;
a ≤ F(friction)/m
a ≤ 2400N/800kg
a ≤ 3 m/s^2
the greatest acceleration that the truck can have without losing its load is 3 m/s^2
Stop using it all the time for some useless things.
Answer: 0.817A
Explanation:
Assuming , that one coulomb per second of negative charge alone flow through a conductor and no positive charges flow. I.e Q=It
It means a current of one A flow in the opposite direction.
This is similar to one coulomb per second of positive charge flowing through and there is no negative charge,
In addition, the one coulomb per second of positive charge flows. This is flowing in the current direction of the previous one. Then, the total current is 2 A. Since 2 coulomb of positive charges flow through one due to real positive charge and another due to the negative charge flowing in opposite direction.The charges cannot cancel each other, because even before the current flow the conductor was neutral.
According to this, the current in the given problem is
[2.7 + 2.4] x 10 ^ 18 * 1.602 x 10^ [-19] C/s
= 0.817 A
Answer:
D. Nothing will happen; the seesaw will still be balanced.
Explanation:
D. Nothing will happen; the seesaw will still be balanced. Since both toruqes or momentums respect to the center have changed in the same amount (one-half their original distance) the seesaw will remain balanced, if the children change distance in a different amount then it will be out of balance