Answer:
The intensity of the electric field is
Explanation:
The electric field equation is given by:
Where:
- k is the Coulomb constant
- q is the charge at 0.4100 m from the balloon
- d is the distance from the charge to the balloon
As we need to find the electric field at the location of the balloon, we just need the charge equal to 1.99*10⁻⁷ C.
Then, let's use the equation written above.
I hope it helps you!
Average speed of the car is 4.57 m/s
Explanation:
- Speed is calculated by the rate of change of displacement.
- It is given by the formula, Speed = Distance/Time
- Here, distance = 112 m and time = 24.5 s
Speed of the car = 112/24.5 = 4.57 m/s
Explanation : Explain each characteristic of sound waves.
Intensity : the intensity of the sound wave is understand as the power carry by sound wave per unit area in the direction perpendicular to that area.
Loudness : loudness is the quality of the loud and soft of the sound wave.
Frequency : Human normal hear sound frequency between 20 Hz to kHz.
Pitch : Pitch is the quality of low and high of sound wave . pitch relates to the frequency of the slowest vibration in the sound wave for simple sound.
When looking for distance you multiply speed by time
So 15 x 2 = 30
30 is the distance between his house and school
Given Information:
Pendulum 1 mass = m₁ = 0.2 kg
Pendulum 2 mass = m₂ = 0.6 kg
Pendulum 1 length = L₁ = 5 m
Pendulum 2 length = L₂ = 1 m
Required Information:
Affect of mass on the frequency of the pendulum = ?
Answer:
The mass of the ball will not affect the frequency of the pendulum.
Explanation:
The relation between period and frequency of pendulum is given by
f = 1/T
The period of pendulum is given by
T = 2π√(L/g)
Where g is the acceleration due to gravity and L is the length of the string
As you can see the period (and frequency too) of pendulum is independent of the mass of the pendulum. Therefore, the mass of the ball will not affect the frequency of the pendulum.
Bonus:
Pendulum 1:
T₁ = 2π√(L₁/g)
T₁ = 2π√(5/9.8)
T₁ = 4.49 s
f₁ = 1/T₁
f₁ = 1/4.49
f₁ = 0.22 Hz
Pendulum 2:
T₂ = 2π√(L₂/g)
T₂ = 2π√(1/9.8)
T₂ = 2.0 s
f₂ = 1/T₂
f₂ = 1/2.0
f₂ = 0.5 Hz
So we can conclude that the higher length of the string increases the period of the pendulum and decreases the frequency of the pendulum.
