What is the wavelength of a soundwave moving at 340 m/s with a frequency of 265 Hz?

1 answer:

4 0

Wave speed = frequency * wavelength
Rearrange so it's equal to wavelength. Do this by diving both sides by frequency to leave you with:
Wave speed / frequency = wavelength
340 / 265 = 1.2830 m

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An object traveling at constant speed v in a circle of radius R has an acceleration a = 5 m/s2. If both Rand v are doubled, what

tresset_1 [31]

Explanation:

acceleration = change in velocity over the change in time

change in velocity/change in time

8 0

3 years ago

One end of a horizontal spring with force constant 76.0 N/m is attached to a vertical post. A 5.00-kg can of beans is attached t

Jobisdone [24]

Answer:

a. The speed is 2.39 m/s

b. The acceleration of the block is 10.2 $\frac{m}{s^{2} }$

Explanation:

First, we have to do the energy balance where we consider two states, the first where the spring remains still and the second when it is stretched 0.400m:

$K_{1} +U_{1}+W_{ext}=K_{2}+U_{2}\\K_{1}=0\\U_{1}=\frac{1}{2} kx^{2} _{1} =0\\W_{ext}=F$ Δx= $(0.400m)\\$

W_{ext}=20.4 Nm

$U_{2} =\frac{1}{2} kx^{2} =\frac{1}{2} (76.0N/m)0.400^{2}=6.08Nm\\k_{2} =\frac{1}{2}mv^{2} _{2} \\\frac{1}{2} mv^{2} _{2}=W_{ext}-U_{2}\\v_{2}=\sqrt{\frac{W_{ext}-U_{2}}{m} } \\v_{2}=\sqrt{\frac{20.4Nm-6.08Nm}{2.5kg} } \\v_{2}=2.39 \frac{m}{s}$

To determine, the acceleration we solve the following equation for a:

$F=ma\\a=\frac{F}{m} =\frac{51.0N}{5.00kg}\\a=10.2\frac{m}{s^{2} }$

8 0

4 years ago

What is the momentum of a 75-kilogram ice skater gliding across the ice at a speed of 13 m/s? ____ kg-m/s

Damm [24]

Momentum = Mass*Velocity

Momentum = 75*13
Momentum = 975

The answer is 975 kg-m/s
Hope you get help

7 0

4 years ago

Read 2 more answers

A binary star system consists of two stars of masses m1 and m2. The stars, which gravitationally attract each other, revolve aro

mote1985 [20]

Answer: $a_{2}=\frac{m_{1}}{m_{2}} a_{1}$

Explanation:

The rest of the question is below:

Find a2, the magnitude of the centripetal acceleration of the star with mass m2.

Assuming both stars are describing a uniform circular motion, their acceleration vector is directed towards the center of mass of the system (that's why it's called centripetal acceleration).

Now, according to Newton's 2nd law, the force $F$ is directly proportional and in the same direction as the acceleration.