Answer:
k = 5.05 N/m
Explanation:
In order to calculate the spring mass of the system, you use the following formula:
(1)
T: period of oscillation of the system
m: mass of the air-track glider = 200g = 0.200 kg
k: spring constant = ?
You first calculate the period of oscillation:
Next, you solve the equation (1) for k, and then you replace the values of the other parmateres:
The spring constant of the spring is 5.05 N/m
Explanation:
initial velocity U = 20m/s
Final velocity V = 35m/s
time = 15.0 secs
change in velocity = 35 - 15
= 20m/s
acceleration a = change in velocity/time V/t
a = (35-20)/15
a= 15/15
Hence, your acceleration is 1m/s^2
B. opposite charge and smaller mass
Answer:
a) S = 2.35 10³ J/m²2
,
b)and the tape recorder must be in the positive Z-axis direction.
the answer is 5
c) the direction of the positive x axis
Explanation:
a) The Poynting vector or intensity of an electromagnetic wave is
S = 1 /μ₀ E x B
if we use that the fields are in phase
B = E / c
we substitute
S = E² /μ₀ c
let's calculate
s = 941 2 / (4π 10⁻⁷ 3 10⁸)
S = 2.35 10³ J/m²2
b) the two fields are perpendicular to each other and in the direction of propagation of the radiation
In this case, the electro field is in the y direction and the wave propagates in the ax direction, so the magnetic cap must be in the y-axis direction, and the tape recorder must be in the positive Z-axis direction.
the answer is 5
C) The poynting electrode has the direction of the electric field, by which or which should be in the direction of the positive x axis
Answer:
Change in electric potential energy ∆E = 365.72 kJ
Explanation:
Electric potential energy can be defined mathematically as:
E = kq1q2/r ....1
k = coulomb's constant = 9.0×10^9 N m^2/C^2
q1 = charge 1 = -2.1C
q2 = charge 2 = -5.0C
∆r = change in distance between the charges
r1 = 420km = 420000m
r2 = 160km = 160000m
From equation 1
∆E = kq1q2 (1/r2 -1/r1) ......2
Substituting the given values
∆E = 9.0×10^9 × -2.1 ×-5.0(1/160000 - 1/420000)
∆E = 94.5 × 10^9 (3.87 × 10^-6) J
∆E = 365.72 × 10^3 J
∆E = 365.72 kJ
