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3 years ago

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An octave is the Question 3 options: a) absolute frequency difference between two notes in the same interval. b) musical distanc

e between two chords. c) amplitude of a piece of music. d) frequency range of a particular piece of music. e) interval between two sound frequencies having a ratio of 2:1.

1 answer:

Scorpion4ik [409]3 years ago

6 0

Answer:

e) interval between two sound frequencies having a ratio of 2:1.

Explanation:

Have a great day! ^w^

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An object in a fluid experiences a buoyant force from the fluid. If the object is completely immersed, on which does the magnitu

loris [4]

Answer:

Weight of the fluid that the object displaces.

Explanation:

When the fluid is completely immersed in a fluid, it experiences pressure from all the direction. While the object is immersed in the fluid a force acts on it in the opposite direction, i.e., upwards. This force is termed as buoyant force.

Also, as per the Archimedes' Principle, the force experience by the object is the same as the weight of the fluid that gets displaced by the object.

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3 years ago

Read 2 more answers

To test the performance of its tires, a car

velikii [3]

<u>Answer</u>:

The coefficient of static friction between the tires and the road is 1.987

<u>Explanation</u>:

<u>Given</u>:

Radius of the track, r = 516 m

Tangential Acceleration $a_r$ = 3.89 m/s^2

Speed,v = 32.8 m/s

<u>To Find:</u>

The coefficient of static friction between the tires and the road = ?

<u>Solution</u>:

The radial Acceleration is given by,

$a_{R = \frac{v^2}{r}$

$a_{R = \frac{(32.8)^2}{516}$

$a_{R = \frac{(1075.84)}{516}$

$a_{R = 2.085 m/s^2$

Now the total acceleration is

$\text{ total acceleration} = \sqrt{\text{(tangential acceleration)}^2 +{\text{(Radial acceleration)}^2$

=> $= \sqrt{ (a_r)^2+(a_R)^2}$

=> $\sqrt{ (3.89 )^2+( 2.085)^2}$

=> $\sqrt{ (15.1321)+(4.347)^2}$

=> $19.4791 m/s^2$

The frictional force on the car will be f = ma------------(1)

And the force due to gravity is W = mg--------------------(2)

Now the coefficient of static friction is

$\mu =\frac{f}{W}$

From (1) and (2)

$\mu =\frac{ma}{mg}$

$\mu =\frac{a}{g}$

Substituting the values, we get

$\mu =\frac{19.4791}{9.8}$

$\mu =1.987$

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3 years ago

Accurate forecasts require careful study of the location, size, movement, and characteristics of large _____.

tankabanditka [31]

Answer: B air masses

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3 years ago

The four tires of an automobile are inflated to a gauge pressure of 2.2 105 Pa. Each tire has an area of 0.023 m2 in contact wit

Stels [109]

Answer:

Force, F = 20240 N

Explanation:

It is given that,

Pressure exerted by the four tires of an automobile, $P=2.2\times 10^5\ Pa$

Area of each tire, $A=0.023\ m^2$

Area of 4 tires, $A=0.092\ m^2$

We know that the pressure exerted by an object is equal to the force per unit area. Its formula is given by :

$P=\dfrac{F}{A}$

$F=P\times A$

$F=2.2\times 10^5\times 0.092$

F = 20240 N

So, the weight of the automobile is 20240 N. Hence, this is the required solution.

7 0

3 years ago