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Scorpion4ik [409]

2 years ago

12

Which amplitude of the following longitudinal waves has the greatest energy?

2 answers:

Rashid [163]2 years ago

8 0

Which amplitude of the following longitudinal waves has the greatest energy?

amplitude = 10 cm; wavelength = 6 cm; period = 4 seconds

sukhopar [10]2 years ago

8 0

Answer: A longitudinal wave has the greatest energy for amplitude 10 cm.

Explanation:

We know that,

The energy is directly proportional to the square of the amplitude.

$E \propto a^2$

(I). Amplitude = 10

then, the energy will be

$E = 100$

(II). Amplitude = 6

then, the energy will be

$E = 36$

(II). Amplitude = 4

then, the energy will be

$E = 16$

Hence, A longitudinal wave has the greatest energy for amplitude 10 cm.

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Which number can each term of the equation be multiplied by to eliminate the fractions before solving? 6 – x + = 6 minus StartFr

zysi [14]

Answer:

We need to multiply 12 to each term to eliminate fractions.

Explanation:

Given expression:

$6-\frac{3}{4}x+\frac{1}{3}=\frac{1}{2}x+5$

To eliminate the fraction we need to multiply each term by least common multiple of the denominators of the fraction.

The denominators in the above expressions are:

4, 3 and 2

The multiples of each can be listed below.

2⇒ 2,4,6,8,10,<u>12</u>,14,16.....

3⇒ 3,6,9,<u>12</u>,15,18

4⇒ 4,8,<u>12</u>.......

From the list of the multiples stated, we can see the least common multiple is 12.

So we will multiply each term by 12.

Multiplying 12 to both sides.

$12(6-\frac{3}{4}x+\frac{1}{3})=12(\frac{1}{2}x+5)$

Using distribution,

$72-9x+4=6x+60$

Thus we successfully eliminated the fractions.

5 0

1 year ago

Read 2 more answers

A 10. g cube of copper at a temperature T1 is placed in an insulated cup containing 10. g of water at a temperature T2. If T1 &g

Anna35 [415]

Answer:

a. The temperature of the copper changed more than the temperature of the water.

Explanation:

Because we're only considering the isolated system cube-water, the heat of the system should be constant, that implies the heat the cube loses is equal the heat the water gains (because by zero law of thermodynamics heat (Q) flows from hot body to cold body until reach thermal equilibrium and T1>T2). So:

$Q_{cube}=Q_{water}$ (1)

But Q is related with mass (m), specific heat (c) and changes in temperature ( $\varDelta T$ )in the next way:

$Q=cm\varDelta T$ (2)

Using (2) on (1):

$c_{cooper}*m_{cooper}*\varDelta T_{cooper}=c_{water}*m_{waterer}*\varDelta T_{water}$

$(10g)(0.385 \frac{J}{g\,C})(\varDelta T_{cooper})=(10g)(4.186 \frac{J}{g\,C})(\varDelta T_{water})$

$(0.385 \frac{J}{g\,C})(\varDelta T_{cooper})=(4.186 \frac{J}{g\,C})(\varDelta T_{water})$

Because we have an equality and 0.385 < 4.186 then $\varDelta T_{cooper}>\varDelta T_{waterer}$ to conserve the equality

4 0

2 years ago

A quarterback throws a football with an initial velocity v at an angle θ above horizontal. Assume the ball leaves the quarterbac

Maru [420]

(a) The y-component or vertical velocity is calculated using:
Vy = Vsin(∅)

(b) The x-component or horizontal velocity is calculated using:
Vx = Vcos(∅)

6 0

1 year ago

Submarine a travels horizontally at 11.0 m/s through ocean water. it emits a sonar signal of frequency f 5 5.27 3 103 hz in the

xeze [42]

Velocity of submarine A is vs = 11.0m/s
frequency emitted by submarine A. F = 55.273 × 10∧3HZ
Velocity of submarine B = vO = 3.00m/s
The given equation is
f' = ((V + vO) ((v - vS)) × f
The observer on submarine detects the frequency f'.
The sign of vO should be positive as the observer of submarine B is moving away from the source of submarine A.
The speed of the sound used in seawater is 1533m/s
The frequency which is detected by submarine B is
fo = fs (V -vO/ v +vs)
= 53.273 × 10∧3hz) ((1533 m/s - 4.5 m/s)/ (1533 m/s +11 m/s)
fo = 5408 HZ

6 0

2 years ago

A nonuniform beam 4.50 m long and weighing 1.40 kN makes an angle of 25.0° below the horizontal. It is held in position by a fri

liubo4ka [24]

Answer:

$T = 7.64 kN$

$F_y = 0.52 kN$ (Downwards)

$F_x = 3.23 kN$ (Towards Left)

Explanation:

As we know that beam is in equilibrium

So here we can use torque balance as well as force balance for the beam

Now by torque balance equation at the pivot we can say

$F(4.50 cos\theta) + mg(2cos\theta) = T \times 3$

As we know that

mg = 1.40 kN

F = 5 kN

so we will have

$5 kN(4.50 cos25) + 1.40 kN(2 cos25) = 3 T$

$T = 7.64 kN$

Now force balance in vertical direction

$F + mg = Tsin65 + F_y$

$5 + 1.40 = 7.64 sin65 + F_y$

$F_y = 0.52 kN$ (Downwards)

Force balance in horizontal direction

$F_x = T cos65$

$F_x = 7.64 cos65$

$F_x = 3.23 kN$ (Towards Left)

7 0

2 years ago