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lozanna [386]
3 years ago
12

In the final situation below, the 8.0 kg box has been launched with a speed of 10.0 m/s across a frictionless surface. Find the

energy of the spring at the start
Physics
1 answer:
Murljashka [212]3 years ago
8 0

Answer:

the energy of the spring at the start is 400 J.

Explanation:

Given;

mass of the box, m = 8.0 kg

final speed of the box, v = 10 m/s

Apply the principle of conservation of energy to determine the energy of the spring at the start;

Final Kinetic energy of the box = initial elastic potential energy of the spring

K.E = Ux

¹/₂mv² = Ux

¹/₂ x 8 x 10² = Ux

400 J = Ux

Therefore, the energy of the spring at the start is 400 J.

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A compass in a magnetic field will line up __________.
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Option c

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A plastic rod 1.3 m long is rubbed all over with wool, and acquires a charge of -3e-08 coulombs. we choose the center of the rod
Anestetic [448]
(a) The plastic rod has a length of L=1.3m. If we divide by 8, we get the length of each piece:
L/8=1.3m/8=0.1625 m

(b) The center of the rod is located at x=0. This means we have 4 pieces of the rod on the negative side of x-axis, and 4 pieces on the positive side. So, starting from x=0 and going towards positive direction, we have: piece 5, piece 6, piece 7 and piece 8. Each piece is 0.1625 m long. Therefore, the center of piece 5 is at 0.1625m/2=0.0812 m. And the center of piece 6 will be shifted by 0.1625m with respect to this:
c_6 = 0.0812m+0.1625m=0.2437 m

(c) The total charge is Q=-3 \cdot 10^{-8}C. To get the charge on each piece, we should divide this value by 8, the number of pieces:
Q/8=-3\cdot 10^{-8}C/8=-3.75\cdot 10^{-9}C

(d) We have to calculate the electric field at x=0.7 generated by piece 6. The charge on piece 6 is the value calculated at point (c):
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If we approximate piece 6 as a single  charge, the electric field is given by
E=k_e  \frac{q}{d^2}
where k_e=8.99\cdot 10^9Nm^2C^{-2} and d is the distance between the charge (center of piece 6, located at 0.2437m) and point a (located at x=0.7m). Therefore we have
E= 8.99\cdot 10^9 Nm^2C^{-2} \frac{-3.75\cdot 10^9 C}{(0.2437m-0.7m)^2} =-161.9 V/m
poiting towards the center of piece 6, since the charge is negative.

(e) missing details on this question.
5 0
3 years ago
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