Ask question

Ask question

All categories

3 years ago

12

This chart shows characteristics of three different waves, all with the same wavelength of 10 m but moving at different Which st

atement is best supported by the information in the chart?
A 3-column table with 2 rows titled Waves at Different Frequencies. The first column has entries Wave X and 6 hertz. The second column has entries Wave Y and 0.5 hertz. The third column has entries Wave Z and 2 hertz.
Wave Y is moving the fastest.

Wave X is moving the fastest.

All of the waves are moving at the same speed.

All of the waves have the same value of wavelengths per second.

2 answers:

Ainat [17]3 years ago

7 0

Answer:

Wave X is moving the fastest.

JulijaS [17]3 years ago

3 0

Answer:

Wave X is moving the fastest.

Explanation:

As we know that the speed of the wave is given as the product of frequency and wavelength

so here we will have

$\lambda = 10 m$

now for wave X

$f = 6 Hz$

now the speed of this wave is

$v_x = 10 \times 6 = 60 m/s$

now for wave Y

$f = 0.5 Hz$

now the speed of this wave is

$v_y = 10 \times 0.5 = 5 m/s$

now for wave Z

$f = 2 Hz$

now the speed of this wave is

$v_z = 10 \times 2 = 20 m/s$

So correct answer will be

Wave X is moving the fastest.

You might be interested in

A thin slice of silicon that contains many solid-state components is a(an)

Tasya [4]

It's an intrigated circuit

8 0

3 years ago

Read 2 more answers

Bart stole a watermelon and ran 5,000 feet from the cops and they chase lasted 0.1 hours how fast was Bart running in miles per

liraira [26]

I am not as sure but I think it is 9.469 miles

5 0

3 years ago

A nonconducting sphere of diameter 10.0 cm carries charge distributed uniformly inside with charge density of +5.50 µC/m3 . A pr

VLD [36.1K]

Answer:

t = 2.58*10^-6 s

Explanation:

For a nonconducting sphere you have that the value of the electric field, depends of the region:

$rR:\\\\E=k\frac{Q}{r^2}$

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

R: radius of the sphere = 10.0/2 = 5.0cm=0.005m

In this case you can assume that the proton is in the region for r > R. Furthermore you use the secon Newton law in order to find the acceleration of the proton produced by the force:

$F=m_pa\\\\qE=m_pa\\\\k\frac{qQ}{r^2}=m_pa\\\\a=k\frac{qQ}{m_pr^2}$

Due to the proton is just outside the surface you can use r=R and calculate the acceleration. Also, you take into account the charge density of the sphere in order to compute the total charge:

$Q=\rho V=(5.5*10^{-6}C/m^3)(\frac{4}{3}\pi(0.05m)^3)=2.87*10^{-9}C\\\\a=(8.98*10^9Nm^2/C^2)\frac{(1.6*10^{-19}C)(2.87*10^{-9}C)}{(1.67*10^{-27}kg)(0.05m)^2}=9.87*10^{11}\frac{m}{s^2}$

with this values of a you can use the following formula:

$a=\frac{v-v_o}{t}\\\\t=\frac{v-v_o}{a}=\frac{2550*10^3m/s-0m/s}{9.87*10^{11}m/s^2}=2.58*10^{-6}s$

hence, the time that the proton takes to reach a speed of 2550km is 2.58*10^-6 s

3 0

3 years ago

If your apparatus were to be dropped from a mile above the ground, describe the forces acting upon your apparatus as it fell. Ho

kvv77 [185]

Answer:

An accelerometer is a tool that measures proper acceleration.[1] Proper acceleration is the acceleration (the rate of change of velocity) of a body in its own instantaneous rest frame;[2] this is different from coordinate acceleration, which is acceleration in a fixed coordinate system. For example, an accelerometer at rest on the surface of the Earth will measure an acceleration due to Earth's gravity, straight upwards[3] (by definition) of g ≈ 9.81 m/s2. By contrast, accelerometers in free fall (falling toward the center of the Earth at a rate of about 9.81 m/s2) will measure zero.

Accelerometers have many uses in industry and science. Highly sensitive accelerometers are used in inertial navigation systems for aircraft and missiles. Vibration in rotating machines is monitored by accelerometers. They are used in tablet computers and digital cameras so that images on screens are always displayed upright. In unmanned aerial vehicles, accelerometers help to stabilise flight.

When two or more accelerometers are coordinated with one another, they can measure differences in proper acceleration, particularly gravity, over their separation in space—that is, the gradient of the gravitational field. Gravity gradiometry is useful because absolute gravity is a weak effect and depends on the local density of the Earth, which is quite variable.

Single- and multi-axis accelerometers can detect both the magnitude and the direction of the proper acceleration, as a vector quantity, and can be used to sense orientation (because the direction of weight changes), coordinate acceleration, vibration, shock, and falling in a resistive medium (a case in which the proper acceleration changes, increasing from zero). Micromachined microelectromechanical systems (MEMS) accelerometers are increasingly present in portable electronic devices and video-game controllers, to detect changes in the positions of these devices.

Explanation:

hope this helps !!!!

7 0

2 years ago

A uniform disk with a mass of 5.0 kg and diameter 30 cm rotates on a frictionless fixed axis through its center and perpendicula

Marina86 [1]

Answer: $\alpha =10.66 rad/s^2$

Explanation:

Given

mass of disk $m=5 kg$

diameter of disc $d=30 cm$

Force applied $F=4 N$

Now this force will Produce a torque of magnitude

$T=F\cdot r$

$T=4\dot 0.15$

$T=0.6 N-m$

And Torque is given Product of moment of inertia and angular acceleration $(\alpha )$

$T=I\cdot \alpha$

Moment of inertia for Disc $I= \frac{Mr^2}{2}$

$I=0.05625 kg-m^2$

$0.6=0.05625\cdot \alpha$

$\alpha =10.66 rad/s^2$

3 0

3 years ago