Answer:
A) 35 ft
B) 5 ft
C) Net displacement = distance covered by the dog to retrieve the stick - distance covered before the dog starts chewing the stick
Explanation:
A) Total distance covered by the dog = 20 + 15
= 35 ft
B) Since the other distance covered by the dog before chewing the stick, after the retrieval, was in an opposite direction to the initial direction, then;
total displacement of the dog = 20 - 15
= 5 ft
C) Net displacement = distance covered by the dog to retrieve the stick + distance covered before the dog starts chewing the stick
But, displacement involves a specified direction. The distance covered before the dog starts chewing the stick was in an opposite direction to the initial direction.
Thus,
Net displacement = distance covered by the dog to retrieve the stick - distance covered before the dog starts chewing the stick
Answer:
The force generated by a single muscle fiber can be increased by increasing the frequency of action potentials
Explanation:
The force generated by a muscle fiber is the result of the shortening of the skeletal muscle, and this force is also know as muscle tension. The larger motor units shorten along with the smaller units to produce the muscle force. The time lapsed between the beginning of the action potential in the muscle and the beginning of the contraction is the latent period. Action potential is the result of the difference electrical potential as a result of passage of an impulse along the membrane of a muscle or nerve cell.
<span>carbon, hydrogen, and oxygen onLY</span>
Answer:
(a) f= 622.79 Hz
(b) f= 578.82 Hz
Explanation:
Given Data
Frequency= 600 Hz
Distance=1.0 m
n=120 rpm
Temperature =20 degree
Before solve this problem we need to find The sound generator moves on a circular with tangential velocity
So
Speed of sound is given by
c = √(γ·R·T/M)
............in an ideal gas
where γ heat capacity ratio
R universal gas constant
T absolute temperature
M molar mass
The speed of sound at 20°C is
c = √(1.40 ×8.314472J/molK ×293.15K / 0.0289645kg/mol)
c= 343.24m/s
The sound moves on a circular with tangential velocity
vt = ω·r.................where
ω=2·π·n
vt= 2·π·n·r
vt= 2·π · 120min⁻¹ · 1m
vt= 753.6 m/min
convert m/min to m/sec
vt= 12.56 m/s
Part A
For maximum frequency is observed
v = vt
f = f₀/(1 - vt/c )
f= 600Hz / (1 - (12.56m/s / 343.24m/s) )
f= 622.789 Hz
Part B
For minimum frequency is observed
v = -vt
f = f₀/(1 + vt/c )
f= 600Hz / (1 + (12.56m/s / 343.24m/s) )
f= 578.82 Hz
