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Alexus [3.1K]
3 years ago
7

If you increase the frequency of a sound wave four times, what will happen to its speed?

Physics
1 answer:
likoan [24]3 years ago
8 0

Answer:

d

Explanation:

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A particle moves at a constant speed of 34 m/s in a circular path of radius 6.3 m. From this information, what can be calculated
Yakvenalex [24]

Answer:

Centripetal acceleration

Explanation:

  • The centripetal acceleration is the motion inwards towards the center of a circular path.
  • <em><u>Centripetal acceleration is given by; the square of the velocity, divided by the radius of the circular path. </u></em>
  • That is;

         ac = v²/r

         Where; ac = acceleration, centripetal, m/s², v is the velocity, m/s and r is the  radius, m

3 0
3 years ago
. A huge pile of leaves was wrapped in a tarp in the middle of a lawn. The wrapped leaves weigh 580 newtons. The coefficient of
Rina8888 [55]

The force required is 319 N

Explanation:

The force of static friction is a force that acts an object on a surface, when this object is pushed by another force to put it in motion. The direction of the force of friction is opposite to the direction of the force of push, and its value increases as the force of push increases, up to a maximum value given by:

F_f = \mu W

where

\mu is the coefficient of friction

W is the weight of the object

Therefore, in order to put the object in motion, the force applied must be greater than this value.

For the pile of leaves in this problem, we have:

\mu = 0.55 (coefficient of friction)

W=580 N (weight of the leaves)

Substituting, we find:

F=(0.55)(580)=319 N

Learn more about force of friction:

brainly.com/question/6217246

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brainly.com/question/3017271

brainly.com/question/2235246

#LearnwithBrainly

7 0
4 years ago
9. A statue is to be scaled down isomorphically (It will have its size changed without changing its shape). It starts with an in
Fynjy0 [20]

Answer:

   m = 1.45 kg

Explanation:

For this exercise we look for size reduction in height

              Reduction = y / y₀

              Reduction = 2.15 / 6.75

              Reduction = 0.3185

As the statue should not be deformed, all reduction has the same factor.

Let's use the concept of density

       ρ = m / V

Initial statue

         ρ = m₀ / V₀

         

It is reduced

         V = x y z

         V = 0.3185 x₀ 0.3185 y₀ 0.3185 z₀

         V = 0.3185³ V₀

       

Density is

         ρ = m / V

         ρ = m / 0.3185³ V₀

As the density remains constant we can match them

         m₀ / Vo = m / 0.3185³ V₀

         m = 0.3185³ m₀

Let's calculate

        m = 0.3185³ 45

        m = 0.03231   45

        m = 1.45 kg

5 0
3 years ago
A violin with string length 32 cm and string density 1.5 g/cm resonates in its fundamental with the first overtone of a 2.0-m or
love history [14]

Answer:

T=1022.42 N

Explanation:

Given that

l = 32 cm ,μ = 1.5 g/cm

L =2 m  ,V= 344 m/s

The pipe is closed so n= 3 ,for first over tone

f=\dfrac{nV}{4L}

f=\dfrac{3\times 344}{4\times 2}

f= 129 Hz

The tension in the string given as

T = f²(4l²) μ

Now by putting the values

T = f²(4l²) μ

T = 129² x (4 x 0.32²)  x 1.5 x  10⁻³ x 100

T=1022.42 N

6 0
3 years ago
Suppose one night the radius of the earth doubled but its mass stayed the same. What would be an approximate new value for the f
Alex17521 [72]

Answer:

g' = g/4

Explanation:

  • The value of the free-fall acceleration at the surface of the earth, can be obtained applying Newton's 2nd law, assuming that the only force acting on an object at the surface of the earth, is the one produced by the mass of the Earth, i.e. gravity.
  • This force can be expressed according  the Newton's Universal Law of Gravitation , as follows:

       F_{g} = G*\frac{m_{x} *m_{E} }{r_{E}^{2} }  (1)

  • From Newton's 2nd Law, we have:
  • F = m* a (2)
  • Since the left sides in (1) and (2) are equal each other, both right sides must be equal each other also.
  • Simplifying the mass m, we can write the acceleration a in (2) as the acceleration due to gravity, g, as follows:

       g = G*\frac{m_{E} }{r_{E}^{2} }  (3)

  • Since G is an universal constant, and the mass mE remains constant, if we double the radius of  the Earth, the new value for the acceleration due to gravity (let's call it g'), is as follows:

        g' = G*\frac{m_{E} }{(2r_{E})^{2} } =G*\frac{m_{E} }{4*r_{E}^{2} }  = g*\frac{1}{4}  =\frac{g}{4} (4)

4 0
3 years ago
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