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Daniel [21]
3 years ago
10

Two loudspeakers emit sound waves along the x-axis. The sound has maximum intensity when the speakers are 19 cm apart. The sound

intensity decreases as the distance between the speakers is increased, reaching zero at a separation of 29 cm.What is the wavelength of the sound?Express your answer using two significant figures.
Physics
1 answer:
Vanyuwa [196]3 years ago
7 0

Answer:

Explanation:

The former is the case of constructive interference and the later case relates to destructive interference.

For listener, path difference is the separation of loudspeaker as listener is not standing in between the speaker.

If λ be the wave length

For constructive interference

19 = n λ

For destructive interference

29 = (2n+1) λ / 2

= n λ +  λ / 2

= 19 +  λ / 2

10 =   λ / 2

 λ  = 20 cm

= 0. 20 m

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Explanation:

3 0
3 years ago
a 150 kg roller coaster is released from rest at the top of a 50 m hill. how fast will it be going if the second hill is 10 m hi
pav-90 [236]

Answer:

10 m/s

Explanation:

The potential energy of the roller coaster is converted to kinetic energy. According to law of transformation of energy, potential energy = kinetic energy.

mgh = ½mv².Take g as 10m/s².

150 * 10 * 50 = ½ * 150 * v².

7500 = 75v².

v² = 7500/75.

v = √100.

v= 10 m/s.

8 0
3 years ago
An object travels at constant velocity of 3 m/s for a time period of 7.15 s. What is its displacement over this time?
son4ous [18]

Hi there!

We can use the following equation for constant velocity:

\large\boxed{d = vt}

d = displacement (m)

v = velocity (m/s)

t = time (s)

Plug in the givens:

d = 3 * 7.15 = \boxed{21.45 m}

5 0
3 years ago
(eText prob. 4.25 with some values changed) Air enters a diffuser of a jet engine operating at steady state at 2.65 psia, 389◦R,
krek1111 [17]

Answer:

V_2 = 45.44m/s

Explanation:

We have to many data in different system, so we need transform everything to SI, that is

P_1 = 2.65 Psi = 18.271 kPa\\T_1= 389\°R = 216 K\\V_1 = 869ft/s = 264m/s\\T_2 = 450\°R = 250K

When we have all this values in SI apply a Energy Balance Equation,

\dot{Q}_{cv}-\dot{W}_{cv}+\dot{m}[(h_1-h_2)+(\frac{V_1^2-V_2^2}{2})+g(z_1-z_2)]=0

Solving for V_2

V_2 = \sqrt{V_1^2+2(h_1-h_2)}

From the table of gas properties we calculate for T_1 = 216K and T_2 = 250K

h_1 = 209.97+(219.97-209.97)(\frac{216-210}{220-210})

h_1 = 215.97kJ/kg

For T_2;

h_2 = 250.05kJ/kg

Substituting in equation for V_2

V_2 = \sqrt{V_1^2+2(h_1-h_2)}

V_2 = \sqrt{265^2+2(215.97-250.05)*10^3}\\V_2 = 45.44m/s

4 0
3 years ago
0.3-L glass of water at 20C is to be cooled with ice to 5C. Determine how much ice needs to be added to the water, in grams, i
abruzzese [7]

Answer:

a. m_i_c_e=54.6g\\b. m_i_c_e=48.7g\\m_c_o_l_d_w_a_t_e_r=900g

Explanation:

First we need to state our assumptions:

Thermal properties of ice and water are constant, heat transfer to the glass is negligible, Heat of ice h_i_f=333.7KJkg

Mass of water,m_w=\rho V =1\times0.3=0.3Kg.

Energy balance for the ice-water system is defined as

E_i_n-E_o_u_t=\bigtriangleup E_s_y_s\\0=\bigtriangleup U=\bigtriangleup U_i_c_e+\bigtriangleup U_w

a.The mass of ice at 0\textdegree C is defined as:

[mc(0-T_1)+mh_i_f+mc(T_2-0)]_i_c_e+[mc(T_2-T_1)]_w=0\\\\m_i_c_e[0+333.7+418\times(5-0)]+0.3\times4.18\times(5-20)=0\\m_i_c_e=0.0546Kg=54.6g

b.Mass of ice at 20\textdegree C is defined as:

[mc(0-T_1)+mh_i_f+mc(T_2-0)]_i_c_e+[mc(T_2-T_1)]_w=0\\\\m_i_c_e[2.11\times(0-(20))+333.7+4.18\times(5-0)]+0.3\times4.18\times(5-20)=0\\\\m_i_c_e=0.0487Kg=48.7g

c.Mass of cooled water at T_c_w=0\textdegree C

\bigtriangleup U_c_w+\bigtriangleup U_w=0

[mc(T_2-T_1)]_c_w+[mc(T_2-T_1)]_w=0\\m_c_w\times4.18\times(5-0)+0.3\times4.18\times(5-20)\\m_c_w=0.9kg=900g

8 0
3 years ago
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