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lbvjy [14]
3 years ago
5

Which one of the patterns have the highest pitch?​

Physics
2 answers:
Flura [38]3 years ago
5 0
(D) explained in the picture

Norma-Jean [14]3 years ago
3 0
C that’s the answer
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When a bow is stretched back and an arrow is shot what type of energy conversion has occured
Vesna [10]
Elastic to kinetic energy. Bow elastic transfered to arrow kinetic energy
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4 years ago
What is 34 + (3) × (1.2465) written with the correct number of significant figures?
sergejj [24]
34 + 3.7395 = 37.7395
The answer is 38.
The answer is written to two significant figures because the smallest given  number of significant figure is 2.
6 0
4 years ago
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A large vessel of the diameter of 14 mm divides into two identical smaller vessels but the velocity of blood in
ycow [4]

Answer:

the diameter of the smaller vessels is 9.90 mm

Explanation:

The computation of the diameter of the smaller vessels is given below;

Given that

The larger vessel of the diameter is 14mm

So,

d1 = 7 mm

let us assume the diameter of the smaller artery is d2

Now we used the equation of continuity i.e.

v × pi × d1^2 = 2 × v × pi × d2^2

14^2 = 2 × d2^2

d2 = 9.90 mm

Hence, the diameter of the smaller vessels is 9.90 mm

5 0
3 years ago
What is the main function of a telescope?
OLga [1]
Telescope magnifies light coming from distant objects.
6 0
3 years ago
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A block of mass 0.1 kg is attached to a spring of spring constant 21 N/m on a frictionless track. The block moves in simple harm
bogdanovich [222]

Answer:

A) 2.75 m/s  B) 0.1911 m    C) 0.109 s

Explanation:

mass of block = M =0.1 kg

spring constant = k = 21 N/m

amplitude = A = 0.19 m

mass of bullet = m = 1.45 g = 0.00145 kg

velocity of bullet = vᵇ = 68 m/s

as we know:

Angular frequency of S.H.M = ω₀ = \sqrt\frac{k}{M}

                                                       = \sqrt\frac{21}{0.1}

                                                       = 14.49 rad/sec

<h3>A) Speed of the block immediately before the collision:</h3>

displacement of Simple Harmonic  Motion is given as:

                                x = A sin (\omega t + \phi)\\

Differentiating this to find speed of the block immediately before the collision:

                    v=\frac{dx}{dt}= A\omega_{o} cos (\omega_{o}t =\phi}\\

As bullet strikes at equilibrium position so,

                                  φ = 0

                                   t= 2nπ

                             ⇒ cos (ω₀t + φ) = 1

                             ⇒ v= A\omega_{o}

                                       v=(.19)(14.49)\\v= 2.75 ms^{-1}

<h3>B) If the simple harmonic motion after the collision is described by x = B sin(ωt + φ), new amplitude B:</h3>

S.H.M after collision is given as :

                              x= Bsin(\omega t + \phi)

To find B, consider law of conservation of energy

K.E = P.E\\K.E= \frac{1}{2}(m+M)v^{2}  \\P.E = \frac{1}{2} kB^{2}

\frac{m+M}{k} v^{2} = B^{2} \\B =\sqrt\frac{m+M}{k} v\\B = \sqrt\frac{.00145+0.1}{21} (2.75)\\B = .1911m

<h3>C) Time taken by the block to reach maximum amplitude after the collision:</h3>

Time period S.H.M is given as:

T=2\pi \sqrt\frac{m}{k}\\ for given case\\m= m=M\\then\\T=2\pi \sqrt\frac{m+M}{k}

Collision occurred at equilibrium position so time taken by block to reach maximum amplitude is equal to one fourth of total time period

T=\frac{\pi }{2}\sqrt\frac{m+M}{k} \\T=0.109 sec

5 0
4 years ago
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