Answer:
0.25 m
Explanation:
We can solve the problem by using the lens equation:
where
f is the focal length
p is the distance of the object from the lens
q is the distance of the image from the lens
In this problem, we have
f = +20 cm=+0.20 m (the focal length is positive for a converging lens)
q = +1.0 m (the image distance is positive for a real image)
Solving the equation for p, we find
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.
Answer B. 112 m
Step-by-Step Explanation
initial velocity u = 20 m /s
final velocity v = 36 m /s
time taken t = 4 s
acceleration = (v - U) / t
= (36 - 20) / 4
a=4m/s2
from the formula
7-u2=2as , sis distance covered
putting the values
362-202=2×4×s
1296 - 400 = 8 x S
S= 112 m
Answer:
Explanation:
Energy of signal being radiated per second on all sides = 71 x 10³ J .
At a distance of 220 m it is spread over an area of 4 π x (220)² because it is spreading uniformly on all sides.
So energy crossing per unit area
= 
= 11.67 x 10⁻² Wm⁻²s⁻¹.
This is the intensity of the signal.
At 2200 m this intensity will further reduce by 100 times
So there it becomes equal to
11.67 x 10⁻⁴ Wm⁻² s⁻¹.
Answer: 0.5 m/s
Explanation:
Given
Speed of the sled, v = 0.55 m/s
Total mass, m = 96.5 kg
Mass of the rock, m1 = 0.3 kg
Speed of the rock, v1 = 17.5 m/s
To solve this, we would use the law of conservation of momentum
Momentum before throwing the rock: m*V = 96.5 kg * 0.550 m/s = 53.08 Ns
When the man throws the rock forward
rock:
m1 = 0.300 kg
V1 = 17.5 m/s, in the same direction of the sled with the man
m2 = 96.5 kg - 0.300 kg = 96.2 kg
v2 = ?
Law of conservation of momentum states that the momentum is equal before and after the throw.
momentum before throw = momentum after throw
53.08 = 0.300 * 17.5 + 96.2 * v2
53.08 = 5.25 + 96.2 * v2
v2 = [53.08 - 5.25 ] / 96.2
v2 = 47.83 / 96.2
v2 = 0.497 ~= 0.50 m/s
