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.
Answer:
- 1.3 x 10⁻¹⁵ C/m
Explanation:
Q = Total charge on the circular arc = - 353 e = - 353 (1.6 x 10⁻¹⁹) C = - 564.8 x 10⁻¹⁹ C
r = Radius of the arc = 5.30 cm = 0.053 m
θ = Angle subtended by the arc = 48° deg = 48 x 0.0175 rad = 0.84 rad (Since 1 deg = 0.0175 rad)
L = length of the arc
length of the arc is given as
L = r θ
L = (0.053) (0.84)
L = 0.045 m
λ = Linear charge density
Linear charge density is given as
Inserting the values
λ = - 1.3 x 10⁻¹⁵ C/m
The emf induced in the second coil is given by:
V = -M(di/dt)
V = emf, M = mutual indutance, di/dt = change of current in the first coil over time
The current in the first coil is given by:
i = i₀
i₀ = 5.0A, a = 2.0×10³s⁻¹
i = 5.0e^(-2.0×10³t)
Calculate di/dt by differentiating i with respect to t.
di/dt = -1.0×10⁴e^(-2.0×10³t)
Calculate a general formula for V. Givens:
M = 32×10⁻³H, di/dt = -1.0×10⁴e^(-2.0×10³t)
Plug in and solve for V:
V = -32×10⁻³(-1.0×10⁴e^(-2.0×10³t))
V = 320e^(-2.0×10³t)
We want to find the induced emf right after the current starts to decay. Plug in t = 0s:
V = 320e^(-2.0×10³(0))
V = 320e^0
V = 320 volts
We want to find the induced emf at t = 1.0×10⁻³s:
V = 320e^(-2.0×10³(1.0×10⁻³))
V = 43 volts