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Marta_Voda [28]

4 years ago

15

A rope.that is 1.50m long exhibits a standing wave pattern with 6 nodes if the frequency of the waves is 75.0 hz what is the vel

ocity of the waves?

1 answer:

Sonja [21]4 years ago

4 0

I think he answer is 34.32

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The traditional view of mangrove forests as wastelands and unhealthy environments helped promote their degradation because _____

skad [1K]

Raditionally mangrove forests<span> were viewed as </span>wastelands and unhealthy environments<span>. People therefore reasoned that </span>their<span> removal would improve the health ...</span>

5 0

3 years ago

At a particular instant, a proton at the origin has velocity < 5e4, -2e4, 0> m/s. You need to calculate the magnetic field

vesna_86 [32]

Answer:

$9.7\times 10^{-5} T$

Explanation:

Velocity = $5\times 10^4i-2\times 10^4j$

r= $0.03i+0.05j$

r= $\mid r\mid=\sqrt{(0.03)^2+(0.05)^2}=0.058$

v= $\mid V\mid=\sqrt{(5\times 10^4)^2+(-2\times 10^{4})^2}=5.39\times 10^{2}$

We know that

$B=\frac{mv}{qr}$

Where q= $1.6\times 10^{-19} C$

Mass of proton= $1.67\times 10^{-27} kg$

Using the formula

$B=\frac{1.67\times 10^{-27}\times 5.39\times 10^2}{1.6\times 10^{-19}\times 0.058}$

$B=9.7\times 10^{-5} T$

3 0

3 years ago

Two trains on separate tracks move toward each other. Train 1 has a speed of 145 km/h; train 2, a speed of 72.0 km/h. Train 2 bl

tekilochka [14]

Answer:

Therefore,

The frequency heard by the engineer on train 1

$f_{o}=603\ Hz$

Explanation:

Given:

Two trains on separate tracks move toward each other

For Train 1 Velocity of the observer,

$v_{o}=145\ km/h=145\times \dfrac{1000}{3600}=40.28\ m/s$

For Train 2 Velocity of the Source,

$v_{s}=90\ km/h=90\times \dfrac{1000}{3600}=25\ m/s$

Frequency of Source,

$f_{s}=500\ Hz$

To Find:

Frequency of Observer,

$f_{o}=?$ (frequency heard by the engineer on train 1)

Solution:

Here we can use the Doppler effect equation to calculate both the velocity of the source $v_{s}$ and observer $v_{o}$ , the original frequency of the sound waves $f_{s}$ and the observed frequency of the sound waves $f_{o}$ ,

The Equation is

$f_{o}=f_{s}(\dfrac{v+v_{o}}{v -v_{s}})$

Where,

v = velocity of sound in air = 343 m/s

Substituting the values we get

$f_{o}=500(\dfrac{343+40.28}{343 -25})=500\times 1.205=602.64\approx 603\ Hz$

Therefore,

The frequency heard by the engineer on train 1

$f_{o}=603\ Hz$

7 0

3 years ago

A spring with k = 15.3 N/cm is initially stretched 1.81 cm from its equilibrium length. a) How much more energy is needed to fur

leonid [27]

Answer:

2.31J

Explanation:

the energy for a spring system is given by:

$E=\frac{1}{2} kx^2$

where $k$ is the spring constant: $k=15.3N/cm=1530N/m$ and $x$ is the distance stretched from the equilibrium position.

In the first case $x=1.81cm=0.0181m$

so the energy to stretch the spring 1.81cm is:

$K_{1}=\frac{1}{2} (1530N/m)(0.0181m)^2=0.25J$

and for the second case, the energy to stretch the spring 5.79cm:

$x=5.79cm=0.0579m$

$K_{1}=\frac{1}{2} (1530N/m)(0.0579m)^2=2.56J$

so to answer a) we must find the difference between these energies:

$2.56J-0.25J=2.31J$

6 0

4 years ago

A merry-go-round with a a radius of R = 1.63 m and moment of inertia I = 196 kg-m2 is spinning with an initial angular speed of

Ulleksa [173]

Answer:

299.88 kgm²/s

499.758 kgm²/s

Explanation:

R = Radius of merry-go-round = 1.63 m

I = Moment of inertia = 196 kgm²

$\omega_i$ = Initial angular velocity = 1.53 rad/s

m = Mass of person = 73 kg

v = Velocity = 4.2 m/s

Initial angular momentum is given by

$L=I\omega_i\\\Rightarrow L=196\times 1.53\\\Rightarrow L=299.88\ kgm^2/s$

The initial angular momentum of the merry-go-round is 299.88 kgm²/s

Angular momentum is given by

$L=mvR\\\Rightarrow L=73\times 4.2\times 1.63\\\Rightarrow L=499.758\ kgm^2/s$

The angular momentum of the person 2 meters before she jumps on the merry-go-round is 499.758 kgm²/s

5 0

3 years ago