Answe
a) Q = 0.820 10⁻⁹ C
, b) Q = -3.52 10⁻⁹ C
Explanation:
The electric field is given by the formula
E = k q / r²
where E is a vector quantity, so it must be added as a vector
E_total = E₁ + E₂
let's look for the two electric fields
E₁ = k q₁ / r₁²
E₁ = 9 10⁹ 5.4 10⁻⁹ / 1.25²
E₁ = 31.10 N / C
E2 = k Q / r₂²
E2 = 9 10⁹ Q / 0.625²
E2 = 23.04 10⁹ Q N / C (1)
now we can solve the two cases presented
a) The total field is
E_total = 50.0 N / C towards + x
since the test charge is positive the electric field E1 points to the right in the direction of the + x axis, so the equation is
E_total = E1 + E₂
E₂ = E_toal - E₁
E₂ = 50.0 -31.10
E2 = 18.9 N /C
With the value of the electric field we can calculate the charge (Q) using equation 1
E₂ = 23.04 10⁹ Q
Q = E₂ / 23.04 10⁹
Q = 18.9 / 23.04 10⁹
Q = 0.820 10⁻⁹ C
the charge on Q is positive
b) E_total = -50.0 N / C
E_total = E₁ + E₂
E₂ = E_total - E₁
E2 = -50.0 - 31.10
E2 = -81.10 N /C
we calculate the charge
Q = E2 / 23.04 10⁹
Q = -81.1 / 23.04 10⁹
Q = -3.52 10⁻⁹ C
for this case the charge is negative
Answer:
a = 0.1 [m/s²]
Explanation:
To solve this problem we must use the following equation of kinematics.
Vf = final velocity = 0.8 [m/s]
Vo = initial velocity = 0.3 [m/s]
a = acceleration [m/s²]
t = time = 5 [s]
![0.8=0.3+a*t\\0.5 = 5*a\\a = 0.1 [m/s^{2}]](https://tex.z-dn.net/?f=0.8%3D0.3%2Ba%2At%5C%5C0.5%20%3D%205%2Aa%5C%5Ca%20%3D%200.1%20%5Bm%2Fs%5E%7B2%7D%5D)
Answer:
Power, P = 924.15 watts
Explanation:
Given that,
Length of the ramp, l = 12 m
Mass of the person, m = 55.8 kg
Angle between the inclined plane and the horizontal, 
Time, t = 3 s
Let h is the height of the hill from the horizontal,
h = 5.07 m
Let P is the power output necessary for a person to run up long hill side as :
P = 924.15 watts
So, the minimum average power output necessary for a person to run up is 924.15 watts. Hence, this is the required solution.
5.8x10^3
3.02x10^8
4.5x10^5
8.6x10^10
Anwser
The strength of a force is usually expressed by its magnitude. We have also to specify the direction in which a force acts.
