Answer:
-4×-2y=14 (1)
-10×+7y=-25 (2)
multiplying eq 1 by 7 and eq 2 by 2 and add eq. 1 and 2
-28×-14y=98
-20×+14y=-50
___________
-28×=48
×=48/-28
×=-12/7
now
-4×-2y=14
-4*-12/7-2y=14
48/7-2y=14
-2y=14-48/7
-2y=(98-48)/7
-2y=50/7
y=-50/14
y=-25/7
Given:
The magnitude of each charge is q1 = q2 = 1 C
The distance between them is r = 1 m
To find the force when distance is doubled.
Explanation:
The new distance is

The force can be calculated by the formula

Here, k is the constant whose value is

On substituting the values, the force will be

Answer:
8.57 Hz
Explanation:
From the question given above, the following data were obtained:
Wavelength (λ) = 3.5 m
Velocity (v) = 30 m/s
Frequency (f) =?
The velocity, wavelength and frequency of a wave are related according to the equation:
Velocity = wavelength × frequency
v = λ × f
With the above formula, we can simply obtain the frequency of the wave as follow:
Wavelength (λ) = 3.5 m
Velocity (v) = 30 m/s
Frequency (f) =?
v = λ × f
30 = 3.5 × f
Divide both side by 3.5
f = 30 / 3.5
f = 8.57 Hz
Thus, the frequency of the wave is 8.57 Hz