Answer:
567.321nm
Explanation:
See attached handwritten document for more details
Complete question:
An air-filled parallel-plate capacitor has plates of area 2.90 cm2 separated by 2.50 mm. The capacitor is connected to a(n) 18.0 V battery. Find the value of its capacitance.
Answer:
The value of its capacitance is 1.027 x 10⁻¹² F
Explanation:
Given;
area of the plate, A = 2.9 cm² = 2.9 x 10⁻⁴ m²
separation distance of the plates, d = 2.5 mm = 2.5 x 10⁻³ m
voltage of the battery, V = 18 V
The value of its capacitance is calculated as;

Therefore, the value of its capacitance is 1.027 x 10⁻¹² F
Answer:
0.465 kgm/s
Explanation:
Given that
Mass of the cart A, m1 = 450 g
Speed of the cart A, v1 = 0.85 m/s
Mass of the cart B, m2 = 300 g
Speed of the cart B, v2 = 1.12 m/s
Now, using the law of conservation of momentum.
It is worthy of note that our cart B is moving in opposite directions to A
m1v1 + m2v2 =
(450 * 0.85) - (300 * 1.12) =
382.5 - 336 =
46.5 gm/s
If we convert to kg, we have
46.5 / 100 = 0.465 kgm/s
Thus, the total momentum of the system is 0.465 kgm/s
Answer:
Explanation:
Given:
U1 = 1.6 m/s
U2 = -1.1 m/s
M1 = 1850 kg
M2 = 1400 kg
V1 = 0.27 m/s
Using momentum- collision equation,
M1U1 + M2U2 = M1V1 + M2V2
1850 × 1.6 - 1400 × 1.1 = 1850 × 0.27 + 1400 × V2
1420 = 499.5 + 1400V2
V2 = 0.6575 m/s
B.
KE = 1/2 × MV^2
KEa1 + KEa2 = KEb1 + KEb2
Delta KE = KE2 - KE1
KEa1 = 2368 J
KEb1 = 847 J
KEa2 = 67.433 J
KEb2 = 302.6 J
KE1 = KEa1 + KEb1
= 3215 J
KE2 = 370.033 J
Delta KE = -2845 J.