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pshichka [43]
3 years ago
15

You are 15.0m from the source of a sound. At that distance, you hear it at a sound level of 20.0dB. How close must you move to t

he sound to increase the sound level to 60dB?
I think its 45.0cm?

15.0cm
45.0cm
55.0cm
60.0cm
85.0cm
Physics
2 answers:
Andrej [43]3 years ago
5 0
The answer is 45.0cm
USPshnik [31]3 years ago
4 0

Answer:

Explanation:

its 85 cm because it just is in meters it would be .8m  which would be decent distance to hear it

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First extinguish a match or candle by blasting it violently. Why?​
Andreas93 [3]

Answer:

The wax vapor on burning candles could reignite the flame.

Explanation:

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3 years ago
A cart is loaded with a brick and pulled at constant speed along an inclined plane to the height of a seat-top. If the mass of t
grandymaker [24]

Answer:

13.23J

Explanation:

PE = m*g*h

PE = (3 kg ) * (9.8 m/s/s) * (0.45 m)

3 0
3 years ago
Read 2 more answers
How does a sound wave transfer energy to your ears?
inessss [21]

Answer:

I think it's C!

Explanation:

Sound waves travel at 343 m/s through the air and faster through liquids and solids. The waves transfer energy from the source of the sound, e.g. a drum, to its surroundings. Your ear detects sound waves when vibrating air particles cause your eardrum to vibrate. The bigger the vibrations the louder the sound.

Hope this helps!

4 0
3 years ago
A plane accelerates from rest at a constant rate of 5.00 m/s2 along a runway that is 1800 m long. Assume that the plane reaches
tiny-mole [99]

Answer:

26.8 seconds

Explanation:

To solve this problem we have to use 2 kinematics equations: *I can't use subscripts for some reason on here so I am going to use these variables:

v = final velocity

z = initial velocity

x = distance

t = time

a = acceleration

{v}^{2}  =  {z}^{2}  + 2ax

v = z + at

First let's find the final velocity the plane will have at the end of the runway using the first equation:

{v}^{2}  =  {0}^{2}  + 2(5)(1800)

v = 60 \sqrt{5}

Now we can plug this into the second equation to find t:

60 \sqrt{5}  = 0 + 5t

t = 12 \sqrt{5}

Then using 3 significant figures we round to 26.8 seconds

3 0
2 years ago
In 2 1/2 hours an airplane travels 1150 km against the wind. It takes 50 min to travel 450 km with the wind. Find the speed of t
Tamiku [17]

Answer:

Velocity of airplane is 500 km/h

Velocity of wind is 40 km/h

Explanation:

V_a= Velocity of airplane in still air

V_w= Velocity of wind

Time taken by plane to travel 1150 km against the wind is 2.5 hours

V_a-V_w=\frac {1150}{2.5}\\\Rightarrow V_a-V_w=460\quad (1)

Time taken by plane to travel 450 km against the wind is 50 minutes = 50/60 hours

V_a+V_w=\frac {450}{50}\times 60\\\Rightarrow V_a-V_w=540\quad (2)

Subtracting the two equations we get

V_a-V_w-V_a-V_w=460-540\\\Rightarrow -2V_w=-80\\\Rightarrow V_w=40\ km/h

Applying the value of velocity of wind to the first equation

V_a-40=460\\\Rightarrow V_a =500\ km/h

∴ Velocity of airplane in still air is 500 km/h and Velocity of wind is 40 km/h

5 0
3 years ago
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