Four boys push on the front back and sides of a shopping cart. the boys in the front on the two sides push with a force of 60N.
the boy in back pushes with a force of 50N. which way does the cart move
*ASAP
*ASAP
1 answer:
Answer:
The cart will move backwards
Explanation:
To solve this problem we must make a free body diagram of the shopping cart and the forces acting on them. You have two boys pushing on the sides with a force of 60 [N] and the child on the front also pushes with a force of 60[N], the child behind pushes with a force of 50 [N].
In the attached image we can see the free body diagram.
As shown in the attached image the shopping cart is displayed from above, the forces of children pushing on the sides, are canceled with each other as they are equal in magnitude and opposite in direction. While the force of the child pushing with 60 [N] is greater than the force of the child pushing with 50 [N] therefore the cart will move backwards, that is to the left as seen in the attached scheme.
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Answer:
The induced emf in the short coil during this time is 1.728 x 10⁻⁴ V
Explanation:
The magnetic field at the center of the solenoid is given by;
B = μ(N/L)I
Where;
μ is permeability of free space
N is the number of turn
L is the length of the solenoid
I is the current in the solenoid
The rate of change of the field is given by;
The induced emf in the shorter coil is calculated as;
where;
N is the number of turns in the shorter coil
A is the area of the shorter coil
Area of the shorter coil = πr²
The radius of the coil = 2.5cm / 2 = 1.25 cm = 0.0125 m
Area of the shorter coil = πr² = π(0.0125)² = 0.000491 m²
E = 14 x 0.000491 x 0.02514
E = 1.728 x 10⁻⁴ V
Therefore, the induced emf in the short coil during this time is 1.728 x 10⁻⁴ V