Answer:
a) a = 3.06 10¹⁵ m / s
, b) F= 1.43 10⁻¹⁰ N, c) F_total = 14.32 10⁻²⁶ N
Explanation:
This exercise will average solve using the moment relationship.
a ) let's use the relationship between momentum and momentum
I = ∫ F dt = Δp
F t = m 
- m v₀
F = m (v_{f} -v₀o) / t
in the exercise indicates that the speed module is the same, but in the opposite direction
F = m (-2v) / t
if we use Newton's second law
F = m a
we substitute
- 2 mv / t = m a
a = - 2 v / t
let's calculate
a = - 2 4.59 10²/3 10⁻¹³
a = 3.06 10¹⁵ m / s
b) F= m a
F= 4.68 10⁻²⁶ 3.06 10¹⁵
F= 1.43 10⁻¹⁰ N
c) if we hit the wall for 1015 each exerts a force F
F_total = n F
F_total = n m a
F_total = 10¹⁵ 4.68 10⁻²⁶ 3.06 10¹⁵
F_total = 14.32 10⁻²⁶ N
Risk of not being able to reduce their weight
<em>A statement that is true for ALL of the examples of electromagnetic waves is that;</em>
A) They all move at the same speed in a vacuum
<u>The reason for qualifying 'in vacuum' is because EM waves of different frequencies often propagate at different speeds through material. Generally speaking, we say that light travels in waves, and all electromagnetic radiation travels at the same speed which is about 3.0 * 108 meters per second through a vacuum.</u>
Answer: magnitude of the magnetic field at a distance of 19.4 cm from the wire=4.29mT
Explanation:
According to Biot-Savart law, A magnetic field generated by a current carrying wire at a distance is represented as
B=μ₀I/ 2πr
B = magnetic field intensity 1000 mT =1T, 6.50mT = 6.50 X 10^-3T
μ₀ =permeability of free space 4π × 10−7 H/m
I = current intensity
r = radius, 100cm = 1m, 12.8 cm= 12.8 x 10^-2m
6.50 X 10^-3 = μ₀ x I/ 2 π X 12.8 X 10^-2
I =6.50 X 10 ^-3 X 2π X X 12.8 X 10^-2/ 4π × 10−7 H/m
I= 4160 A
when the magnetic field is at 19.4 cm from the wire
B=μ₀I/ 2πr
= 4π × 10−7 H/m x4160/ 2π x 19.4 x 10^-2
=0.004288
= 4.29x 10 ^-3T
= 4.29mT
Answer:
C2, C1, C4, C5 and C6 are in parallel. Therefore, we use the formula Cp = C1 + C2 + ....
Cp = C2 + C1 + C4 + C5 + C6 = ( 7 * 10 ^-3) + (18 * 10^-6) + (0.8F) + (200 * 10^-3 F) + (750 * 10^-6) = 1.008F
Now, Cp will become one capacitor and it will be aligned with C3, therefore it will now become a circuit in series.
We use the formula: 1/Cs = 1/C1 + 1/C2 + .... + ....1/Cn
Thus,
1/Cs = 1/C3 + 1/Cp
1/Cs = 1/(14 * 10^-3 F) + 1/(1.008F)
Cs = 1.4 * 10 ^-2 or if we do not round too much it will give exactly 0.0138 F
So the answer should be a)
