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Andreas93 [3]
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.
Physics
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.

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In Example 2.7, we investigated a jet landing on an aircraft carrier. In a later maneuver, the jet comes in for a landing on sol
igor_vitrenko [27]

Answer:

a) 20 seconds

b) No.

Explanation:

t = Time taken for jet to stop

u = Initial velocity = 100 m/s (given in the question)

v = Final velocity = 0 (because the jet will stop at the end)

s = Displacement of the jet (Distance between the moment the jet touches the ground to the point the point it stops)

a = Acceleration = -5.00 m/s² (slowing down, so it is negative)

a) Equation of motion

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-100}{-5}\\\Rightarrow t=20\ s

The time required for the plane to slow down from the moment it touches the ground is 20 seconds.

s=ut+\frac{1}{2}at^2\\\Rightarrow s=100\times 20+\frac{1}{2}\times -5\times 20^2\\\Rightarrow s=1000\ m

The distance it requires for the jet to stop is 1000 m so in a small tropical island airport where the runway is 0.800 km long the plane would not be able to land. The runway needs to be atleast 1000 m long here the runway on the island is 1000-800 = 200 m short.

5 0
2 years ago
You have a summer job at a company that developed systems to safely lower large loads down ramps. Your team is investigating a m
Fofino [41]

Answer:

Note that the emf induced is

emf = B d v cos (A)

---> v = emf / [B d cos (A)]

where

B = magnetic field

d = distance of two rails

v = constant speed

A = angle of rails with respect to the horizontal

Also, note that

I = emf/R

where R = resistance of the bar

Thus,

I = B d v cos (A) / R

Thus, the bar experiences a magnetic force of

F(B) = B I d = B^2 d^2 v cos (A) / R, horizontally, up the incline.

Thus, the component of this parallel to the incline is

F(B //) = F(B) cos(A) = B I d = B^2 d^2 v cos^2 (A) / R

As this is equal to the component of the weight parallel to the incline,

B^2 d^2 v cos^2 (A) / R = m g sin (A)

where m = the mass of the bar.

Solving for v,

v = [R m g sin (A) / B^2 d^2 cos^2 (A)]   [ANSWER, the constant speed, PART A]

******************************

v = [R m g sin (A) / B^2 d^2 cos^2 (A)]

Plugging in the units,

m/s = [ [ohm * kg * m/s^2] / [T^2 m^2] ]

Note that T = kg / (s * C), and ohm = J * s/C^2

Thus,

m/s = [ [J * s/C^2 * kg * m/s^2] / [(kg / (s * C))^2 m^2] ]

= [ [J * s/C^2 * kg * m/s^2] / [(kg^2 m^2) / (s^2 C^2)]

As J = kg*m^2/s^2, cancelling C^2,,

= [ [kg*m^2/s^2 * s * kg * m/s^2] / [(kg^2 m^2) / (s^2)]

Cancelling kg^2,

= [ [m^2/s^2 * s * m/s^2] / [(m^2) / (s^2)]

Cancelling m^2/s^2,

= [s * m/s^2]

Cancelling s,

=m/s   [DONE! WE SHOWED THE UNITS ARE CORRECT! ]

8 0
3 years ago
A force of 20 N produces an acceleration of 10 m/s² in mass m1 and an acceleration of 5 m/s² in
Scrat [10]

Explanation:

F = 20N m= m1 a=10m/s²

m=m2 a=5m/s²

F = ma

<u>for the first one</u><u>:</u><u> </u>

f=m1 × a

20 = m1 ×10

20=10m1

m1=20/10

m1=2

<u>for</u><u> </u><u>the</u><u> </u><u>second</u><u> </u><u>one</u><u> </u><u>:</u>

f=m2×a

20=m2×5

m2= 20/5

m2= 4

since F=ma

F=(m1+m2) ×a

F =(4+2)×a

F =6×a

F=20(from the question above )

20=6×a

a=20/6

a=3.33

8 0
3 years ago
Read 2 more answers
In an RC circuit, what fraction of the final energy is stored in an initially uncharged capacitor after it has been charging for
4vir4ik [10]

Answer:

The  fraction fraction of the final energy is stored in an initially uncharged capacitor after it has been charging for 3.0 time constants is  

      k  = 0.903

Explanation:

From the question we are told that

     The time  constant  \tau  =  3

The potential across the capacitor can be mathematically represented as

     V  =  V_o  (1 -  e^{- \tau})

Where V_o is the voltage of the capacitor when it is fully charged

    So   at  \tau  =  3

     V  =  V_o  (1 -  e^{- 3})

     V  =  0.950213 V_o

   Generally energy stored in a capacitor is mathematically represented as

             E = \frac{1}{2 } * C  * V ^2

In this equation the energy stored is directly proportional to the the square of the potential across the capacitor

Now  since capacitance is  constant  at  \tau  =  3

        The  energy stored can be evaluated at as

         V^2 =  (0.950213 V_o )^2

       V^2 =  0.903  V_o ^2

Hence the fraction of the energy stored in an initially uncharged capacitor is  

      k  = 0.903

4 0
3 years ago
When applying an arm sling, the injured arm's hand should be placed in a ________ position in the sling, and above the level of
Ierofanga [76]

Answer:

The answer is "thumb-up".

Explanation:

The sling often called the arm sling, is a solution for preventing either shoulder or elbow motion during healing. A triangle bandage could be constructed.

A sling is a strong tissue loop that descends from the head to hold the forearm. The armrests in the sling bent also at the elbow.

The injured hand of the forearm should be positioned inside the pin position in the joint above the level of the elbow when putting the forearm sling.

3 0
2 years ago
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