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Tamiku [17]
2 years ago
5

Maya uses a tuning fork of frequency 250 Hz to produce transverese waves on a stretched string. Her friend justin measure the di

stance between the.two consecutive crest on the string as 5 cm. What was the velocity of the wave?​
Physics
1 answer:
Goshia [24]2 years ago
6 0

Answer:

The velocity of the wave is 12.5 m/s

Explanation:

The given parameters are;

he frequency of the tuning fork, f = 250 Hz

The distance between successive crests of the wave formed, λ = 5 cm = 0.05 m

The velocity of a wave, v = f × λ

Where;

f = The frequency of the wave

λ = The wavelength of the wave - The distance between crests =

Substituting the known values gives;

v =  250 Hz × 0.05 m = 12.5 m/s

The velocity of the wave, v = 12.5 m/s.

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An object of mass m is initially at rest. After a force of magnitude F acts on it for a time T, the object has a speed v. Suppos
adell [148]

Answer:

It takes \frac{1}{2}T to accelerate the object from rest to the speed v.

Explanation:

From Newton's second law:

F=m\cdot a  (1)

and the definition of acceleration,

\displaystyle{a = \frac{\Delta v}{\Delta t}} (2)

we can solve this problem. Putting (2) in (1) we have:

\displaystyle{F = m\cdot \frac{\Delta v}{\Delta t}} and solving for \Delta t and considering the initial time as zero (t_0=0) and the initial velocity also zero (v_0=0) we have:

\displaystyle{T=\frac{mv}{F}}

Now, for a mass m^*= 2m and the F^*=4F we can wrtie the same equation:

\displaystyle{T^*=\frac{m^*v}{F^*}} and substituting m^* and F^*:

\displaystyle{T^*=\frac{2m\cdotv}{4F}=\frac{2}{4}T=\boxed{\frac{1}{2}T}}

So now, it only takes half the time to accelerate the object from rest to the speed v

5 0
3 years ago
A string is stretched and fixed at both ends, 200 cm apart. If the density of the string is 0.015 g/cm, and its tension is 600 N
Gre4nikov [31]

Answer:

<h2>f₀ = 158.12 Hertz</h2>

Explanation:

The fundamental frequency of the string  f₀ is expressed as f₀ = V/4L where V is the speed experienced by the string.

V = \sqrt{\frac{T}{\mu} } where T is the tension in the string and  \mu is the density of the string

Given T = 600N and \mu = 0.015 g/cm  = 0.0015kg/m

V = \sqrt{\frac{600}{0.0015} }\\ \\V =  \sqrt{400,000}\\ \\V = 632.46m/s

The next is to get the length L of the string. Since the string is stretched and fixed at both ends, 200 cm apart, then the length of the string in metres is 2m.

L = 2m

Substituting the derived values into the formula f₀ = V/2L

f₀ = 632.46/2(2)

f₀ = 632.46/4

f₀ = 158.12 Hertz

Hence the fundamental frequency of the string is 158.12 Hertz

5 0
3 years ago
A ball is falling from the second floor balcony to the floor below
aleksandrvk [35]

Answer:

KE increases and PE decreases

Explanation:

7 0
2 years ago
Read 2 more answers
When a wheel is rotated through an angle of 21◦ , a point on the circumference travels through an arc length of 2.1 m. When the
LiRa [457]

Answer:

R = 5.73 m

Explanation:

For an angle of rotation through 21 degree we know that

arc length is given as

angle \times Radius = Arc

now we know that

Arc = 2.1 m

Angle = 21 degree

Angle = 21\times \frac{\pi}{180}

so now we have

21\times \frac{\pi}{180} = \frac{2.1}{R}

R = \frac{2.1 \times 180}{21 \times \pi}

R = 5.73 m

6 0
3 years ago
1.15) What distance will 650 joules of work move a box weighing 50 newtons?​
son4ous [18]

Answer:

The answer is 13 however make sure if they ask for a certain measurement like meter answer it by saying 13 meters.

Explanation:

This basically turns into basic algebra if you know the formula for work. The formula for work is W=F*d  

Here are the variables that you know 650J=50N*d so you need d.

All you do is divide 650J by 50N and you get a total of 13 (meters since I don't know what they want you to put it in).

6 0
2 years ago
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