Answer:
78.4 m
Explanation:
Using newton's equation of motion,
S = ut + 1/2gt²......................... Equation 1
Where S = Height, t = time, u = initial velocity, g = acceleration due to gravity.
Note: Taking upward to be negative, and down ward positive
Given: u = 49 m/s, t = 2.0 s, g = -9.8 m/s²
Substitute into equation 1
S = 49(2) - 1/2(9.8)(2)²
S = 98 - 19.6
S = 78.4 m
Hence the height of the ball two seconds later = 78.4 m
Wave speed = frequency * wavelength
Input the numbers into this equation :
Wave speed = 200 * 3
Work it out and you will get the answer :
Wave speed = 600 m/s
Answer:
The value of charge q₃ is 40.46 μC.
Explanation:
Given that.
Magnitude of net force 
Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.
We need to calculate the distance
Using Pythagorean theorem
Put the value into the formula
We need to calculate the magnitude of the charge q₃
Using formula of net force
Put the value into the formula
Hence, The value of charge q₃ is 40.46 μC.
Answer:
2.6×10⁻³ N
Explanation:
From coulomb's law,
F = kq'q/r²................ Equation 1
Where F = Repulsive force, q' = charge on the first sugar grain, q = charge on the second sugar grain, r = distance of separation between the sugar grain, k = proportionality constant.
From the question,
since q' = q
Then,
F = kq²/r²..................... Equation 2
Given: q = 1.79×10⁻¹¹ C, r = 3.45×10⁻⁵ m,
Constant: k = 9×10⁹ Nm²/kg².
Substitute into equation 2
F = 9×10⁹(1.79×10⁻¹¹)²/(3.45×10⁻⁵ )²
F = 9×10⁹(3.2041×10⁻²²)/(11.9025×10⁻¹⁰)
F = (28.8369×10⁻¹³)/(11.9025×10⁻¹⁰)
F = 2.6×10⁻³ N.
