Answer:
I hear points of low volume sound and points of high volume of sound.
Explanation:
This is because, since the two sources of sound have the same frequency and are separated by a distance, d = 10 mm, there would be successive points of constructive and destructive interference.
Since their frequencies are similar, we should have beats of high and low frequency.
So, at points of low frequency, the amplitude of the wave is smallest and there is destructive interference. The frequency at this point is the difference between the frequencies from both speakers. Since the frequency from both speakers is 400 Hz, we have, f - f' = 400 Hz - 400 Hz = 0 Hz. So, the volume of the sound is low(zero) at these points.
Also, at points of high frequency, the amplitude of the wave is highest and there is constructive interference. The frequency at this point is the sum between the frequencies from both speakers. Since the frequency from both speakers is 400 Hz, we have, (f + f') = 400 Hz + 400 Hz = 800 Hz. So, the volume of the sound is high at these points.
So, as you wander around the room, I should hear points of high and low sound across the room.
The mechanical efficiency = actual work / ideal work
So ζ = 1540 / 1600 * 100% = 96.25%
Answer:
3.25 × 10^7 m/s
Explanation:
Assuming the electrons start from rest, their final kinetic energy is equal to the electric potential energy lost while moving through the potential difference (ΔV)
Ek = 1/2 mv2 = qΔV .................. 1
Given that V is the electron speed in m/s
Charge of electron = 1.60217662 × 10-19 coulombs
Mass of electron = 9.109×10−31 kilograms
ΔV = 3.0kV = 3000V
Make V the subject of the formula in eqaution 1
V = sqr root 2qΔV/m
V = 2 × 1.60217662 × 10-19 × 3000 / 9.109×10−31
V = 3.25 × 10^7 m/s
Answer:
Explanation:
Given
Force 
one at an angle of 
East of North and another at West of North
Net Force is in North Direction
Forces in horizontal direction will cancel out each other
thus Work done will be by north direction forces
here 
Answer:
6318 N
Explanation:
From the question given above, the following data were obtained:
Acceleration due to gravity of the moon (gₘ) = 1.62 m/s²
Mass (m) of container = 650 kg
Weight (W) of container on the earth =.?
Next, we shall determine the acceleration due to gravity of the earth. This can be obtained as follow:
Acceleration due to gravity of the moon (gₘ) = 1.62 m/s²
Acceleration due to gravity of the earth (gₑ) =.?
gₘ = 1/6 × gₑ
1.62 = 1/6 × gₑ
1.62 = gₑ /6
Cross multiply
gₑ = 1.62 × 6
gₑ = 9.72 m/s²
Finally, we shall determine the weight of the container on the earth as follow:
Mass (m) of container = 650 kg
Acceleration due to gravity of the earth (gₑ) = 9.72 m/s²
Weight (W) of container on the earth =.?
W = m × gₑ
W = 650 × 9.72
W = 6318 N
Therefore, the weight of the container on earth is 6318 N
