All categories

3 years ago

These waves are traveling at the same speed. Which wave has the highest frequency?

Physics

2 answers:

Leviafan [203]3 years ago

8 0

Answer: A

Explanation: The relation between velocity and frequency of the waves is:

f = v/λ

where f is frequency, v is the velocity and λ is the wavelenght.

So you can see that, if the 4 waves have the same velocity, then the one with the smallest wavelenght (because it is in the denominator) will have the highest frequency.

The one with the smallest frequency is the first one (you can see it) so the correct option will be A

Margaret [11]3 years ago

5 0

Answer:

A The shorter the wavelength the higher the frequency the longer the lower, hope this helps, thanks.

You might be interested in

A slender rod is 80.0 cm long and has mass 0.120 kg. A small 0.0200-kg sphere is welded to one end of the rod, and a small 0.050

BartSMP [9]

Answer:

Explanation:

Given that,

Slender rod

Length of rod=80cm=0.8m

Mass of slender rod=0.12kg

Sphere Bob at one end

Mass M1=0.02kg

Sphere Bod at the other end

Mass M2 =0.05kg

Linear speed of mass 2 at the lowest point

We need to calculate the change in potential of the complete system. m2 and m3 are the masses at the rod ends. note the rod centre of mass neither gains nor loses potential.

So, at the lowest point,

∆U = M2•g•y2 + M1•g•y1

Note, at the lowest point, the mass 1 is 40cm (0.4m) form the midpoint, Also, the mass 2 is -40cm(-04m) from the midpoint

∆U = M2•g•y2 + M1•g•y1

∆U=0.05•9.81•(-0.4) + 0.02•9.82•0.4

∆U=-0.1962+0.07848

∆U=-0.11772 Nm

Now, the moment of inertia of the rod is given as

I=∫r²dm

dm=2pdr

I= 2p∫r²dr

I= 2 × 0.12/0.8 ∫r²dr; from r=0 to 0.4

I=0.3 [r³/3] from r=0 to 0.4

I= 0.3 [ 0.4³/3 -0] ,from r=0 to 0.4

I=0.3 × 0.02133

I=0.0064kg/m².

calculating of inertia of the end masses.

I(1+2)=Σmr² = (m1+m2)r²

I(1+2)=(0.02+0.05)0.4²

I(1+2)=0.07×0.4²

I(1+2)=0.0112 kg/m²

Now, the Energy of the masses due to angular velocity is given as

K.E=½ (I + I(1+2))w²

K.E=½(0.0064+0.0112)w²

K.E= 0.0088w²

Using conservation of energy

The potential energy is equal to the kinetic energy of the system

K.E=P.E

0.0088w²=0.11772

Then, w²=0.11772/0.0088

w²=13.377

w=√13.377

w=3.66rad/s

Then, the relationship between linear velocity and angular velocity is given by

v=wr

v=3.66×0.4

v=1.463m/s

The required linear speed is 1.46m/s approximately

8 0

3 years ago

A straight trail with a uniform inclination of 15 degrees leads from a lodge at an elevation of 600 feet to a mountain lake at a

9966 [12]

Answer:

The length of the trail = 22796 ft

Explanation:

From the ΔABC

AC = length of the trail = x

AB = 6100 - 600 = 5500 ft

Angle of inclination $\theta$ = 15°

$\sin \theta = \frac{AB}{AC}$

$\sin 15 = \frac{5900}{x}$

$x = \frac{5900}{0.2588}$

x = 22796 ft

Since x = AC = Length of the trail.

Therefore the length of the trail = 22796 ft

7 0

3 years ago

A record is dropped vertically onto a freely rotating (undriven) turntable. Frictional forces act to bring the record and turnta

alexgriva [62]

Answer:

The loss of initial Kinetic energy = 37.88 %

Explanation:

Given:

Rotational inertia of the turntable = $I_t$

Rotational inertia ( $I_r$ ) of the record = $0.61\times I_t$

According to the question:

<em>Frictional forces act to bring the record and turntable to a common angular speed.</em>

So,angular momentum will be conserved as it is an inelastic collision.

Considering the initial and final angular velocity of the turn table as $\omega _i\ ,\ \omega_f$ respectively.

Note :

Angular momentum $(L)$ = Product of moment of inertia $(I)$ and angular velocity $(\omega)$ .

Lets say,

⇒ initial angular momentum = final angular momentum

⇒ $L_i=L_f$

⇒ $(I_t)\times \omega_i = (I_t+I_r)\times \omega_f$

⇒ $\omega _f=\frac{I_t}{I_t+I_r} \times (\omega_i)$ ...equation (i)

Now we will find the ratio of the Kinetic energies.

⇒ $K_i=\frac{I_t\times \omega_i^2}{2}$ ⇒ $K_f=\frac{(I_r+I_t)\times \omega_f^2}{2}$

Their ratios:

⇒ $\frac{K_f}{K_i} =\frac{\frac{(I_t+I_r)\times \omega_f^2}{2} }{\frac{I_t\times \omega_i^2}{2} }$

⇒ $\frac{K_f}{K_i} = {\frac{(I_t+I_r)\times \omega_f^2}{2} } \times {\frac{2}{I_t\times \omega_i^2}}$

Plugging the values of $\omega _f^2$ as $\omega _f^2 =(\frac{I_t}{I_t+I_r} \times \omega_i\ )^2$ from equation (i) in the ratios of the Kinetic energies.

⇒ $\frac{K_f}{K_i} =\frac{(I_t+I_r)\times \frac{(I_t)^2}{(I_t+I_r)^2} \times \omega_i^2}{I_t\times \omega_i^2} =\frac{(I_t)^2}{(I_t+I_r)}\times \frac{1}{I_t}=\frac{I_t}{I_t+I_r}$

Now,

The Kinetic energy lost in fraction can be written as:

⇒ $\frac{K_f-K_i}{K_i}$

Now re-arranging the terms.

$\frac{K_f-K_i}{K_i} =(\frac{K_f}{K_i} -1)= \frac{I_t}{I_t+I_r} -1=\frac{I_t-I_t-I_r}{I_t+I_r} =\frac{-I_r}{(I_t+I_r)}$

Plugging the values of $I_r$ and $I_t$ .

⇒ $\frac{K_f}{K_i} = \frac{-0.61I_t}{0.61I_t+I_t} =\frac{-0.61}{1.61} =-0.3788$

To find the percentage we have to multiply it with $100$ and here negative means for loss of Kinetic energy.

⇒ $\frac{K_f}{K_i} = =-0.3788\times 100= 37.88$

So the percentage of the initial Kinetic energy lost is 37.88

4 0

3 years ago

Dr. Glover's office has one vendor for their practice management software and another for their electronic health record, but th

Sliva [168]

Answer:

Interface

Explanation:

This is a classic example of Interface technology.

An interface allows different software packages to communicate without re-entering data.

Here in this case also systems are able to communicate with one another without duplicating data entry. For example, practice management software and another for their electronic health record.

5 0

3 years ago

A 51-g rubber ball is released from rest and falls vertically onto a steel plate. The ball strikes the plate and is in contact w

Simora [160]

Answer:

Last option 3000N

Explanation:

4 0

3 years ago