Scientist would check what other scientists have said before they come up with their own scientific question. That way they can record the results from the others and compare them with their own findings when they conduct their own experiments and record the findings.
Answer:
Part a)
Part b)
Explanation:
As we know that the observer is standing in front of one speaker
So here the path difference of the two sound waves reaching to the observer is given as
now phase difference is related with path difference as
here in order to find the wavelength
now we have
Part b)
Now we know that when phase difference is odd multiple of 
then in that case the the sound must be minimum
So nearest value for minimum intensity would be
so we have
so we have
now we have
Answer: Frequency is 0.143 Hz; Period is 7 seconds
Explanation:
Number of vibrations = 8.6
Time required = 60 seconds
Period (T) = ?
Frequency of the vibrations (F) = ?
A) Recall that frequency is the number of vibrations that the Sears tower completes in one second.
i.e Frequency = (Number of vibrations / time taken)
F = 8.6/60 = 0.143Hz
B) Period, T is inversely proportional to frequency. i.e Period = 1/Frequency
T = 1/0.143Hz
T = 7 seconds
Thus, the frequency and period of the vibrations of the Sears Tower are 0.143 Hz
and 7 seconds respectively.
Answer:
a=1.03 m/s²
Explanation:
given required
dist=22.1 cm a=?
m1=2.1 kg
m2=1.56 kg
m3=1.93 kg
g=9.8 m/s²
solution
there are three forces on this system so we can solve by using newton's second law as follows
t1,t2,and weight of are the only forces acted on the system,
if we take the system on the left mass m3
-m3a=t2-m3g
-1.93 kg×a=t2-1.93 kg×9.8 m/s²
-1.93 kg×a=t2-18.91 N................................ eq 1
take at the right mass m2
m2a=t2-m2×g
1.56 kg×a=t2-1.56 kg×9.8 m/s²
1.56 kg×a=t2-15.288 N...................................2
add the two equations simultaneously and then solve for a,
-1×(-1.93 kg×a=t2-18.91 N) multiply by -1 the first and add the two equations
1.56 kg×a=t2-15.288 N
3.49 kg×a=3.622 N
a=1.03 m/s²
The acceleration of the object if the net force is decreased = 0.13 m/s²
<h3>Further explanation</h3>
Given
A net force of 0.8 N acting on a 1.5-kg mass.
The net force is decreased to 0.2 N
Required
The acceleration of the object if the net force is decreased
Solution
Newton's 2nd law :
The mass used in state 1 and 2 remains the same, at 1.5 kg
ΣF=0.8 N
m=1.5 kg
The acceleration, a:
ΣF=0.2 N
m=1.5 kg
The acceleration, a:
