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Maru [420]
2 years ago
15

2. What do pitch and loudness have in common?

Physics
1 answer:
DanielleElmas [232]2 years ago
5 0

Answer:

Both are subject to a persons interpretation

Explanation:

We hear people describe this when somebody is making an irresistible sound. usually people say the baby has a pitch scream.

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As a mass on a spring moves farther from the equilibrium position, how do the velocity, acceleration, and force change
Umnica [9.8K]
Refer to the diagram shown below.

m =  the mass of the object
x = the distance of the object from the equilibrium position at time t.
v = the velocity of the object at time t
a = the acceleration of the object at time t
A =  the amplitude ( the maximum distance) of the mass from the equilibrium
        position

The oscillatory motion of the object (without damping) is given by
x(t) = A sin(ωt)
where
ω =  the circular frequency of the motion
T =  the period of the motion so that ω = (2π)/T

The velocity and acceleration are respectively
v(t) = ωA cos(ωt)
a(t) = -ω²A sin(ωt)

In the equilibrium position,
x is zero;
v is maximum;
a is zero.

At the farthest distance (A) from the equilibrium position,
x is maximum;
v is zero;
a is zero.

In the graphs shown, it is assumed (for illustrative purposes) that
A = 1 and T = 1.

6 0
3 years ago
Why are ocean currents important?
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For the considerably longer periods– decades to millennia – which are relevant for climate change, the slightly larger heat capacity of the deep ocean<span> is </span>important. Ocean currents<span> and mixing by winds and waves can transport and redistribute heat to deeper </span>ocean<span> layers.</span>
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2 years ago
A uniform solid cylindrical log begins rolling without slipping down a ramp that rises 28.0° above the horizontal. After it has
Novay_Z [31]

Answer:

3.07 m/s

Explanation:

6 0
2 years ago
A sailboat picks up a gust of wind and accelerates to a speed of 6m/s in 16 seconds.If the initial velocity 1 m/s,what is the ac
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3 years ago
Zero, a hypothetical planet, has a mass of 4.5 x 1023 kg, a radius of 3.2 x 106 m, and no atmosphere. A 10 kg space probe is to
jok3333 [9.3K]

a.) K 2=K 1 +GmM( r 21− r 11)=2.2×10 7J

b.) ​K 2 +GmM( r 11− r 21)=6.9×10 7 J

Applying Law of  Energy conservation :

K 1+U 1

=K 2+U 2

⇒K 1− r 1GmM

=K 2− r 2 GmM

where M=5.0×10 23kg,r1

=> R=3.0×10 6m and m=10kg

(a) If K 1​

=5.0×10 7J and r 2

=4.0×10 6 m, then the above equation leads to

K 2=K 1 +GmM (r 21− r 11)=2.2×10 7J

(b) In this case, we require K 2

=0 and r2

=8.0×10 6m, and solve for K 1:K 1

​=K 2 +GmM (r 11− r 21)=6.9×10 7 J

Learn more about Kinetic energy on:

brainly.com/question/12337396

#SPJ4

7 0
2 years ago
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