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aksik [14]
3 years ago
7

How are wavelength and frequency related for a wave moving at a constant speed?

Physics
1 answer:
soldier1979 [14.2K]3 years ago
6 0
Mathematically, to compute for the wavelength, λ,  we have

\lambda = \frac{v}{f}

where v is the velocity and f is the frequency.
As we can see here, given that velocity is constant, wavelength increases as frequency decreases. This shows that wavelength is inversely proportional to frequency. 

Answer: Wavelength and frequency are inversely related to each other.
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The pirate ship tie at the amusement park is a giant pendulum that riders sit in. It swings back and forth, with a maximum veloc
kkurt [141]

Here as we know that there is no loss of energy

so we can say that maximum kinetic energy will become gravitational potential energy at its maximum height

So here we have

\frac{1}{2}mv^2 = mgh

here we have

v = 20 m/s

m = 8000 kg

now from above equation we have

\frac{1}{2}(8000)(20^2) = (8000)(9.8)h

h = \frac{200}{9.8)

h = 20.4 m

so maximum height is 20.4 m

4 0
3 years ago
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We wrap a light, nonstretching cable around a 8.00 kg solid cylinder with diameter of 30.0 cm. The cylinder rotates with negligi
antoniya [11.8K]

Answer:

h = 16.67m

Explanation:

If the kinetic energy of the cylinder is 510J:

Kc=510=1/2*Ic*\omega c^2

\omega c=\sqrt{510*2/Ic}

Where the inertia is given by:

Ic=1/2*m_c*R_c^2=1/2*(8)*(0.15)^2=0.0225kg.m^2

Replacing this value:

\omega c=106.46rad/s

Speed of the block will therefore be:

V_b=\omega_c*R_c=106.46*0.15=15.969m/s

By conservation of energy:

Eo = Ef

Eo = 0

Ef = 510+1/2*m_b*V_b^2-m_b*g*h

So,

0 = 510+1/2*m_b*V_b^2-m_b*g*h

Solving for h we get:

h=16.67m

3 0
3 years ago
Why is pluto not a planet?
mihalych1998 [28]
Pluto revolves around a star instead of our SUN. So, it's an "Exo-planet" rather than a Planet. It has declared in 2006 by the scientists.

Hope this helps!
3 0
3 years ago
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Hi pls answer 11 point pls
Lelechka [254]

Answer:

the answer is a. a ball is moving towards the camera faster then slower

6 0
2 years ago
A 3 kg mass object is pushed 0.6 m into a spring with spring constant 210 N/m on a frictionless horizontal surface. Upon release
Svetradugi [14.3K]

Answer with Explanation:

We are given that

Mass of spring,m=3 kg

Distance moved by object,d=0.6 m

Spring constant,k=210N/m

Height,h=1.5 m

a.Work done  to compress the spring initially=\frac{1}{2}kx^2=\frac{1}{2}(210)(0.6)^2=37.8J

b.

By conservation law of energy

Initial energy of spring=Kinetic energy  of object

37.8=\frac{1}{2}(3)v^2

v^2=\frac{37.8\times 2}{3}

v=\sqrt{\frac{37.8\times 2}{3}}

v=5.02 m/s

c.Work done by friction on the incline,w_{friction}=P.E-spring \;energy

W_{friction}=3\times 9.8\times 1.5-37.8=6.3 J

8 0
3 years ago
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