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3 years ago

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

Physics

2 answers:

Anika [276]3 years ago

8 0

Answer: The correct answer is "the speed of the wave becomes four times".

Explanation:

The relation between the speed, frequency and the wavelength is as follows:

v=f\lambda

Here, v is the speed of the wave, f is the frequency and \lambda is the wavelength.

The speed of the sound wave is directly proportional to the frequency.

In the given problem, if the speed of the sound wave is increased four times then the speed of the sound becomes four times.

Therefore, the speed of the sound wave becomes four times.

Serggg [28]3 years ago

4 0

A w a v e ' s f r e q u e n c y c a n b e m a t h e m a t i c a l l y e x p r e s s e d a s f = v / λ, w h e r e v i s t h e s p e e d a n d λ i s t h e w a v e l e n g t h . W e c a n s e e t h a t f r e q u e n c y a n d s p e e d a r e d i r e c t l y p r o p o r t i o n a l to e a c h o t h e r , t h a t m e a n s w h e n f r e q u e n c y i s i n c r e a s e d f o u r t i m e s, t h e s p e e d w i l l a l s o i n c r e a s e f o u r t i m e s<span>.</span>

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What is the the steadiest firing position?

Paha777 [63]

The closer you are to the ground the more accurate you'll be. That's why most snipers are in the "prone" position.

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3 years ago

8TH GRADE SCIENCE I NEED NOW DO NOT SKIP

yan [13]

Explanation:

1. Force=mass*acceleration

acceleration=force/mass

=100/50

=2m/s^2

2. Gravitational force for downward acceleration= mg-ma=m(g-a) , since a is less than g,

So it will be= 50(9.8-2)

=50(7.8)= 390N

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3 years ago

Blood in a carotid artery carrying blood to the head is moving at 0.15 m/s when it reaches a section where plaque has narrowed t

sp2606 [1]

Answer:

26.9 Pa

Explanation:

We can answer this question by using the continuity equation, which states that the volume flow rate of a fluid in a pipe must be constant; mathematically:

$A_1 v_1 = A_2 v_2$ (1)

where

$A_1$ is the cross-sectional area of the 1st section of the pipe

$A_2$ is the cross-sectional area of the 2nd section of the pipe

$v_1$ is the velocity of the 1st section of the pipe

$v_2$ is the velocity of the 2nd section of the pipe

In this problem we have:

$v_1=0.15 m/s$ is the velocity of blood in the 1st section

The diameter of the 2nd section is 74% of that of the 1st section, so

$d_2=0.74d_1$

The cross-sectional area is proportional to the square of the diameter, so:

$A_2=(0.74)^2 A_1=0.548 A_1$

And solving eq.(1) for v2, we find the final velocity:

$v_2=\frac{A_1 v_1}{A_2}=\frac{A_1 (0.15)}{0.548 A_1}=0.274 m/s$

Now we can use Bernoulli's equation to find the pressure drop:

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2$

where

$\rho=1025 kg/m^3$ is the blood density

$p_1,p_2$ are the initial and final pressure

So the pressure drop is:

$p_1 - p_2 = \frac{1}{2}\rho (v_2^2-v_1^2)=\frac{1}{2}(1025)(0.274^2-0.15^2)=26.9 Pa$

8 0

3 years ago

A football player, with a mass of 69.0 kg, slides on the ground after being knocked down. At the start of the slide, the player

White raven [17]

Answer:

(a) -472.305 J

(b) 1 m

Explanation:

(a)

Change in mechanical energy equals change in kinetic energy

Kinetic energy is given by $0.5mv^{2}$

Initial kinetic energy is $0.5\times 69\times 3.7^{2}=472.305 J$

Since he finally comes to rest, final kinetic energy is zero because the final velocity is zero

Change in kinetic energy is given by final kinetic energy- initial kinetic energy hence

0-472.305 J=-472.305 J

(b)

From fundamental kinematic equation

$v^{2}=u^{2}+2as$

Where v and u are final and initial velocities respectively, a is acceleration, s is distance

Making s the subject we obtain

$s=\frac {v^{2}-u^{2}}{-2a}$ but a=\mu g hence

$s=\frac {v^{2}-u^{2}}{-2\mu g}=\frac {0^{2}-3.7^{2}}{-2*0.7*9.81}=0.996796272\approx 1 m$

7 0

3 years ago

A 2.8 kg grinding wheel is in the form of a solid cylinder of radius 0.1 m. a) What constant torque will bring it from rest to a

kogti [31]

Answer:

a) τ = 0.672 N m , b) θ = 150 rad , c) W = 100.8 J

Explanation:

a) for this part let's start by finding angular acceleration, when the angular velocity stops it is zero (w = 0)

w = w₀ + α t

α = -w₀ / t

α = 120 / 2.5

α = 48 rad / s²

The moment of inertia of a cylinder is

I = ½ M R²

Let's calculate the torque

τ = I α

τ = ½ M R² α

τ = ½ 2.8 0.1² 48

τ = 0.672 N m

b) we look for the angle by kinematics

θ = w₀ t + ½ α t2

θ = ½ α t²

θ = ½ 48 2.5²

θ = 150 rad

c) work in angular movement

W = τ θ

W = 0.672 150

W = 100.8 J

7 0

3 years ago