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gulaghasi [49]
3 years ago
12

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.
Physics
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.

Thus on complete immersion of the object in the fluid, it experiences the force same as the weight of the fluid that gets displaced

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Which climate condition is typically found in the tropics due to the interaction of the atmosphere and hydrosphere?
Ilya [14]
The answer is A! Hope I helped! 

7 0
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

8 0
3 years ago
Accurate forecasts require careful study of the location, size, movement, and characteristics of large _____.
tankabanditka [31]

Answer: B air masses

3 0
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
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