At the same amplitude or volume, which sound would have the highest frequency?
1 answer:
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We can solve the problem by using Newton's second law of motion:
where
F is the net force applied to the object
m is the object's mass
a is the acceleration of the object
In this problem, the force applied to the car is F=1050 N, while the mass of the car is m=760 kg. Therefore, we can rearrange the equation and put these numbers in, in order to find the acceleration of the car:
The equation also tells us that the acceleration and the force have same directions: therefore, since the force exerted on the car is horizontal, the correct answer is
<span>B) 1.4 m/s2 horizontally.</span>
where
F is the net force applied to the object
m is the object's mass
a is the acceleration of the object
In this problem, the force applied to the car is F=1050 N, while the mass of the car is m=760 kg. Therefore, we can rearrange the equation and put these numbers in, in order to find the acceleration of the car:
The equation also tells us that the acceleration and the force have same directions: therefore, since the force exerted on the car is horizontal, the correct answer is
<span>B) 1.4 m/s2 horizontally.</span>
Answer:
Amplitude = 0.058m
Frequency = 6.25Hz
Explanation:
Given
Amplitude (A) = 8.26 x 10-2 m
Frequency (f) = 4.42Hz
Conversation of energy before split
½mv² = ½KA²
Make A the subject of formula
A =
Conversation of energy after split
½(m/2)V'² = ½(m/2)V² = ½KA'²
½(m/2)V² = ½KA'²
Make A the subject of formula
First divide both sides by ½
(m/2)V² = KA'²
Divide both sides by K
V² = A'²
= A'
Substitute for A in the above equation
A' = A/√2
A' = 8.26 x 10^-2/√2
A' = 0.05840702012600882
Amplitude after split = 0.058 (Approximated)
Frequency (f') = f√2
f' = 4.42√2
f' = 6.25082394568908011
Frequency after split = 6.25Hz (approximated)