Answer:
s = 589.3 m
Explanation:
Let the truck and car meet at a distance = s m
The truck is moving at constant velocity = v
so s= v * t ---------- (1)
car:
Vi = 0 m/s
a = 3.9 m/s²
s = Vi* t + 1/2 a t²
s= 0 * t + 1/2 a t²
s = 1/2 a t² ----------- (2)
compare equation (1) and equation (2)
s= v * t = 1/2 a t²
⇒ v * t = 1/2 a t²
⇒ t = 2 * v/ a
⇒ t = (2 * 33.9 )/ 3.9
⇒ t = 17. 38 s
Now
from equation (1)
s= v * t
s= 33.9 * 17.38
⇒ s = 589.3 m
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: 100 suns
Explanation:
We can solve this with the following relation:

Where:
is the diameter of a dime
is the diameter of the Sun
is the distance between the Sun and the pinhole
is the amount of dimes that fit in a distance between the sunball and the pinhole
Finding
:


This is roughly the diameter of the Sun
Now, the distance between the Earth and the Sun is one astronomical unit (1 AU), which is equal to:

So, we have to divide this distance between
in order to find how many suns could it fit in this distance:
