Answer:
(A) 1.43secs
(B) -2.50m/s^2
Explanation:
A commuter backs her car out of her garage with an acceleration of 1.40m/s^2
(A) When the speed is 2.00m/s then, the time can be calculated as follows
t= Vf-Vo/a
The values given are a= 1.40m/s^2 , Vf= 2.00m/s, Vo= 0
= 2.00-0/1.40
= 2.00/1.40
= 1.43secs
(B) The deceleration when the time is 0.800secs can be calculated as follows
a= Vf-Vo/t
= 0-2.00/0.800
= -2.00/0.800
= -2.50m/s^2
The magnitude of the air drag is 784 N
Explanation:
An object falling down reaches the terminal velocity when the magnitude of the air drag acting on it becomes equal to the weight of the object. Mathematically, this condition can be written as:
where
is the magnitude of the air drag
m is the mass of the object
g is the acceleration of gravity
In this problem, we have
m = 80 kg is the mass of the airman
is the acceleration of gravity
Substituting into the formula, we find:
Learn more about forces here:
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Answer:
A. Particles in air move in circles as the wave moves forward.
B. Particles in air move forward but not backward as the wave moves
forward.
C. Particles in air move up and down as the wave moves forward.
✔ D. Particles in air move forward and backward as the wave moves
forward.
Explanation:
The waves transfer energy from the source of the sound, e.g. a drum, to its surroundings. Your ear detects sound waves when vibrating air particles cause your ear drum to vibrate. The bigger the vibrations the louder the sound.
