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Savatey [412]
2 years ago
10

A water wave has a speed of 23.0 meters/second. If the wave’s frequency is 0.0680 hertz, what is the wavelength?

Physics
2 answers:
stellarik [79]2 years ago
6 0
Wavelength = \frac{speed of wave}{frequency of the wave} =  \frac{23}{0.068} = 338.235 meters. Hope this helps you!
KatRina [158]2 years ago
4 0

Answer:

λ = 338m

Explanation:

We can use the equation that describe the wavelength in function of the velocity and the frequency of the wave

λ = \frac{v}{f}

where v, is the velocity of the wave and f is the frequency of the wave

The problem gives you that v=23.0\frac{m}{s} and f=0.0680Hz, so you should replace this values in the equation for the wavelengt:

λ = \frac{23.0\frac{m}{s}}{0.0680Hz}

λ = 338m

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Dimas [21]

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In Physics, "Work" is the product of Force times Distance.

Since the dresser hasn't budged yet, the Distahce is zero.  So no matter how great the Force may be, it's multiplied by zero, so the <em>Work is zero</em>.

5 0
3 years ago
A 13561 N car traveling at 51.1 km/h rounds
Minchanka [31]

Answer:

a) The centripetal acceleration of the car is 0.68 m/s²

b) The force that maintains circular motion is 940.03 N.

c) The minimum coefficient of static friction between the tires and the road is 0.069.

Explanation:

a) The centripetal acceleration of the car can be found using the following equation:

a_{c} = \frac{v^{2}}{r}

Where:

v: is the velocity of the car = 51.1 km/h

r: is the radius = 2.95x10² m

a_{c} = \frac{(51.1 \frac{km}{h}*\frac{1000 m}{1 km}*\frac{1 h}{3600 s})^{2}}{2.95 \cdot 10^{2} m} = 0.68 m/s^{2}

Hence, the centripetal acceleration of the car is 0.68 m/s².

b) The force that maintains circular motion is the centripetal force:

F_{c} = ma_{c}

Where:

m: is the mass of the car

The mass is given by:

P = m*g

Where P is the weight of the car = 13561 N

m = \frac{P}{g} = \frac{13561 N}{9.81 m/s^{2}} = 1382.4 kg

Now, the centripetal force is:

F_{c} = ma_{c} = 1382.4 kg*0.68 m/s^{2} = 940.03 N

Then, the force that maintains circular motion is 940.03 N.

c) Since the centripetal force is equal to the coefficient of static friction, this can be calculated as follows:

F_{c} = F_{\mu}

F_{c} = \mu N = \mu P

\mu = \frac{F_{c}}{P} = \frac{940.03 N}{13561 N} = 0.069

Therefore, the minimum coefficient of static friction between the tires and the road is 0.069.

I hope it helps you!                

3 0
3 years ago
CAN SOMEONE HELP ME
uranmaximum [27]
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So your answer is "the conversion between the liquid and gaseous states of Oxygen." :)
3 0
2 years ago
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inysia [295]

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CONS OF BIOMASS ENERGY: Emissions, deforestation. cost.

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