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3 years ago

I WILL GIVE BRAINLIEST TO WHOEVER GETS THIS QUESTION CORRECT!!!

Physics

1 answer:

luda_lava [24]3 years ago

4 0

Answer: c) The line shows a change in speed if its angle changes.

I'm not sure

Explanation:

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3. The velocity of sound is 332 m/s. Answer the following questions:

Angelina_Jolie [31]

Answer:

20 Hz, 20000 Hz

0.0166 m, 16.6 m

Explanation:

The minimum frequency that a human ear can hear is 20 Hz

The maximum frequency that a human ear can hear is 20000 Hz.

v = Velocity of sound = 332 m/s

Wavelength is given by

$\lambda=\dfrac{v}{f}\\\Rightarrow \lambda=\dfrac{332}{20}\\\Rightarrow \lambda=16.6\ \text{m}$

The longest wavelength that can be heard by the human ear is 16.6 m

$\lambda=\dfrac{332}{20000}\\\Rightarrow \lambda=0.0166\ \text{m}$

The shortest wavelength that can be heard by the human ear is 0.0166 m.

3 0

3 years ago

A small omnidirectional stereo speaker produces waves in all directions that have an intensity of 8.00 at a distance of 4.00 fro

slavikrds [6]

Answer:

A. We have that radius r = 4.00m intensity I = 8.00 W/m^

total power = power/ Area ( 4πr2)= 8.00 w/m^2( 4π ( 4.00 m)2=1607.68 W

b) I = total power/ 4πr2= 8.00 W/m2 ( 4.00 m/ 9.5 m)2= 1.418 W/m2

c) E = total power x time= 1607 . 68 W x 1s= 1607.68 J

5 0

3 years ago

Suppose that two objects attract each other with a gravitational force of 16 units. If the mass of both objects was doubled, and

Lady_Fox [76]

Answer:

Distance, such that displacement over time is taken into account doesn't change regardless of the length in x meters or mass in kg
So the force will remain constant given 16 units

6 0

3 years ago

A seagull flies at a velocity of 9.00 m/s straight into the wind. (a) if it takes the bird 20.0 min to travel 6.00 km relative t

enot [183]

Here we will the speed of seagull which is v = 9 m/s

this is the speed of seagull when there is no effect of wind on it

now in part a)

if effect of wind is in opposite direction then it travels 6 km in 20 min

so the average speed is given by the ratio of total distance and total time

$v_{avg} = \frac{6000}{20*60}$

$v_{avg} = 5m/s$

now since effect of wind is in opposite direction then we can say

$V_{net} = v_{bird} - v_{wind}$

$5 = 9 - v_{wind}$

$v_{wind}= 4 m/s$

Part b)

now if bird travels in the same direction of wind then we will have

$v_{net}= v_{bird} + v_{wind}$

$v_{net} = 9 + 4 = 13 m/s$

now we can find the time to go back

$time = \frac{distance}{speed}$

$time = \frac{6000}{13}$

$time = 7.7 minutes$

Part c)

Total time of round trip when wind is present

$T = t_1 + t_2$

$T = 20 + 7.7 = 27.7 min$

now when there is no wind total time is given by

$T = \frac{6000}{9} + \frac{6000}{9}$

$T = 22.22 min$

So due to wind time will be more

4 0

4 years ago

Draw the following patterns as they would appear when viewed through a compound microscope.:

goblinko [34]

2 j x b
H c e c hope this helppppsssssssss

3 0

3 years ago

Read 2 more answers