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4 years ago

A frequency generator sends a 550 Hz sound wave through both water and ice.

Physics

2 answers:

Brums [2.3K]4 years ago

3 0

Δλ = 3.103 m.

To solve this problem we have to know the speed of sound in both elements water and ice.

Element Speed of sound

Water (25°C) 1493 m/s

Ice 3200 m/s

The velocity of a wave is given by the equation v = λf, where λ is the wavelength, and f is the frecuency of the wave.

In order to calculate the wavelength we have to clear λ in the equation v = λf, resulting:

λ = v/f

Calculating the wavelength in both elements:

λ(water) = 1493 m/s / 550 Hz = 2.715 m

λ(ice) = 3200 m/s / 550 Hz = 5.818 m

So, the difference in wavelength between the wave produced in ice and the wave produced in water is:

Δλ = λ(water) - λ(ice) = 5.818 m - 2.715 m

Δλ = 3.103 m

Ghella [55]4 years ago

3 0

Answer:

3.1 m

Explanation:

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_____ variables are manipulated by the experiments

nikitadnepr [17]

Answer:

dependent variables

Explanation:

dependent varibeles are the thing you're measuring and independent variables are the thing you change in the exeriment to get a different dependent variable.

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4 years ago

If the depth of water in a well is 10 m, what is the pressure exerted by it on the

Dahasolnce [82]

Answer:

Let d be the density of the water (1000 kg / m^3 eq to 1 gm / cm^3)

P = d g h for the pressure due to a column at the bottom of the column.

P = 1000 kg / m^3 * 10 m/s^2 * 10 m = 10^5 kg / m * s^2 = 10^5 N/m

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3 years ago

Charge q is 1 unit of distance away from the source charge S. Charge p is two times further away. The force exerted between S an

insens350 [35]

Answer : The correct option is, (d) 4 times

Solution :

According to the Coulomb's law, the electrostatic force of attraction or repulsion between two charges is directly proportional to the product of the charges and is inversely proportional to the square of the distance between the the charges.

Formula used :

$F=k_e\frac{q_1q_2}{r^2}$

where,

F = electrostatic force of attraction or repulsion

$k_e$ = Coulomb's constant

$q_1$ and $q_2$ are the charges

r = distance between two charges

First we have to calculate the force exerted between S and q when the distance between the charge is 1 unit and let us assumed that the charge be 'q'

$F_{sq}=k_e\frac{qq}{1^2}=k_e\times q^2$ ..........(1)

Now we have to calculate the force exerted between S and p when the distance between the charge is 2 unit at the same charge.

$F_{sp}=k_e\frac{qq}{2^2}=k_e\frac{q^2}{4}$ ...........(2)

Equation equation 1 and 2, we get

$\frac{F_{sq}}{F_{sp}}=\frac{1}{4}$

$F_{sq}=4\times F_{sp}$

Therefore, the force exerted between S and q is 4 times the force exerted between S and p.

5 0

3 years ago

Read 2 more answers

15) A 328-kg car moving at 19.1 m/s in the +x direction hits from behind a second car moving at 13 m/s in the same direction. If

kolbaska11 [484]

Answer:

Explanation:

Given that,

Mass of first car

M1= 328kg

The car is moving in positive direction of x axis with velocity

U1 = 19.1m/s

Velocity of second car

U2 = 13m/s, in the same direction as the first car..

Mass of second car

M2 = 790kg

Velocity of second car after collision

V2 = 15.1 m/s

Velocity of first car after collision

V1 =?

This is an elastic collision,

And using the conservation of momentum principle

Momentum before collision is equal to momentum after collision

P(before) = P(after)

M1•U1 + M2•U2 = M1•V1 + M2•V2

328 × 19.1 + 790 × 13 = 328 × V1 + 790 × 15.1

16534.8 = 328•V1 + 11929

328•V1 = 16534.8—11929

328•V1 = 4605.8

V1 = 4605.8/328

V1 = 14.04 m/s

The velocity of the first car after collision is 14.04 m/s

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3 years ago

PLEASE HELP WILL GIVE BRAINLIEST!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

iVinArrow [24]

Answer:

Explanation: BRAINLIEST ME

6 0

3 years ago