Answer:
Option A is correct - While a guitar string is vibrating, you gently touch the midpoint of the string to ensure that the string does not vibrate at that point. The lowest-frequency standing wave that could be present on the string vibrates at twice the fundamental frequency.
Explanation:
Before touching the midpoint of the string, the string vibrates with one loop.
Fundamental frequency, f1 = v/(2*L)
Now, when the midpoint of the guitar string was touched, the string vibrates with two loops.
Hence, f2 = 2*v/(2*L)
f2 = 2*f1
Therefore, compared to the fundamental frequency the frequency would be double.
Option A is correct - While a guitar string is vibrating, you gently touch the midpoint of the string to ensure that the string does not vibrate at that point. The lowest-frequency standing wave that could be present on the string vibrates at twice the fundamental frequency.
Answer:
correct option is a. 0.2 mA toward D
Explanation:
given data
B carries = 1.5 mA
C carries current = 1.3 mA
solution
we take positive direction of current going away from the point D
and negative direction of current coming towards point D
so we use here kirchoff's current law
that is
iA + iB + iC = 0 ......................1
iA + 1.5 + (-1.3) = 0
iA = - 0.2 mA
so that current in wire A is 0.2 mA towards point D
correct option is a. 0.2 mA toward D
Answer:
a = - 0.3376 g's
Explanation:
The sports car has a constant speed when travelling. Covered 164 m in 13.77 s. Thus, speed = 164/13.77 m/s
It brakes and now comes to a stop in 3.6 s.
Thus final velocity = 0 m/s
Formula for acceleration is;
a = (v - u)/t
a = (0 - (164/13.77))/3.6
a = -3.308 m/s²
In terms of g's", where 1.00 g = 9.80 m/s², we have;
a = -3.308/9.8 g's
a = - 0.3376 g's
Can you reword the sentence?
