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3 years ago

50 points Two waves combine with constructive interference. What must be true of the

Physics

2 answers:

tresset_1 [31]3 years ago

8 0

Answer:

it has a higher amplitude than that of the original waves

Explanation:

trust me its right

Kamila [148]3 years ago

5 0

Answer:

C. It has a higher frequency than that of the original waves.

Explanation:

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A kettle of water contains 1.2 kg of water. Calculate how much energy is required to hear it from water at 10 degrees to its boi

Darya [45]

Answer:

450 kJ

Explanation:

Q = mCΔT

where Q is heat (energy),

m is mass,

C is specific heat capacity,

and ΔT is the temperature change.

Q = (1.2 kg) (4180 J/kg/°C) (100°C − 10°C)

Q = 451,440 J

Q ≈ 450 kJ

6 0

3 years ago

9. To completely describe the motion of an object, you need

pav-90 [236]

I think it’s D) all of the above

4 0

3 years ago

Read 2 more answers

A horizontal pipe of diameter 1.03m has a smooth constriction to a section of diameter 0.618 m. The density of oil flowing in th

vodka [1.7K]

Velocity of the oil in the pipe: 0.76 m/s, in the constricted section: 5.87 m/s

Explanation:

We can solve this problem by using Bernoulli's equation:

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2$ (1)

where

$p_1 = 7340 N/m^2$ is the pressure in the pipe

$p_2 = 5505 N/m^2$ is the pressure in the constricted section

$\rho = 821 kg/m^3$ is the density of the oil

$v_1$ is the velocity of the oil in the pipe

$v_2$ is the velocity of the oil in the constricted section

Also, according to the continuity equation,

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-sectional area of the pipe, with

$r_1 = \frac{1.03}{2}=0.515 m$ is the radius

$A_2 = \pi r_2^2$ is the cross-sectional area of the constricted section, with

$r_2=\frac{0.618}{2}=0.309 m$ is the radius

So the equation becomes

$r_1^2 v_1 = r_2^2 v_2$

So we can write

$v_2=\frac{r_1^2}{r_2^2}v_1$

Substituting into eq.(1),

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho (\frac{r_1^2}{r_2^2}v_1)^2$

And solving the equation for $v_1$ :

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho \frac{r_1^4}{r_2^4}v_1^2\\v_1=\sqrt{\frac{p_2-p_1}{\frac{1}{2}\rho-\frac{1}{2}\rho \frac{r_1^4}{r_2^4}}}=0.76 m/s$

And the velocity in the constricted section is

$v_2=\frac{r_1^2}{r_2^2}v_1=5.87 m/s$

Learn more about flow rate:

brainly.com/question/9805263

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7 0

3 years ago

Write atleast name of five devices which work under the principles of physics

Colt1911 [192]

You could say almost anything.
For example:
phones,
cars,
computers,
clocks,
hydraulics,
bicycles,
the hadron collider,
Planes,
and so on.

3 0

3 years ago

Read 2 more answers

A 15 N force and a 45 N force act on in object in opposite directions. What is the net force on the object?

vovikov84 [41]

Answer:

30N in the direction the 45N acts.

Explanation:

Fnet = F1 + F2 (the vector sum of the forces)

Assigning a positive direction to the 45N force and a negative direction to the 15N force gives:

Fnet = 45 - 15

Fnet = 30N

Since the answer is positive, it is in the direction the 45N force acts.

7 0

3 years ago