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True [87]
2 years ago
15

How much energy was absorbed

Physics
1 answer:
ser-zykov [4K]2 years ago
6 0
. we need like a picture you something what’re you trying to ask
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In a pig caller can produce a sound intensity level of 107 dB. How many pig callers would be needed to generate an intensity lev
myrzilka [38]

Answer:

20 pig callers

Explanation:

Given that:

A pig caller produced intensity level of  a sound = 107 dB

To find how many pig callers required to generate an intensity level of 120 dB;

we have:

120 dB - 107 dB = 13 dB

Taking the logarithm function;

10 \ log \bigg(\dfrac{I}{I_o} \bigg) = 13 \ dB

where;

I_o = initial intensity

log \bigg(\dfrac{I}{I_o} \bigg) = 1.3

\dfrac{I}{I_o}=  10^{1.3 }

I = 19.95I_o

I ≅ 20 pig callers

6 0
3 years ago
When is the best time to take a resting heart break
Olegator [25]

Answer:

I believe D

Explanation:

You need to have a more accurate reading and you want to test it multiple times throughout the week though to get a base resting rate.

I hope this is correct good luck!

5 0
3 years ago
A space station has a large ring-like component that rotates to simulategravity for the crew. This ring has a massM= 2.1×105kg a
ivann1987 [24]

Answer:

Each thruster has to applied a force of 294.5N in tangential direction

Explanation:

Mass of the ring ,M =2.1×105kg

Mass of the ship ,m = 3.5×104kg

Radius of the ring R = 86.0 m

distance of ship from center of the ring, r =31.0 m

Let force applied by each thruster be F

Time taken to reach  gravity ,t = 3hrs = 3600× 3 =10800sec

The movement of ring make the object kept at the edge feel a force of centrifuge in outward direction.

Centrifugal force = weight of the object on earth

Assume the ring is moving with angular speed ω

Centripetalforce of the object kept at ring

m₁R ω²=m₁g  (m₁=mas of object)

⇒Rω² = g

⇒ω = √g/R

The ring start from 0 angular speed with constant angular acceleration

Let the constant angular acceleration be ∝

∝ = ω  / t

(ω = √g/R)

∝ = \frac{1}{t } \sqrt{\frac{g}{R} } }

Consider Torque on the ring and ship system

T = FR + FR = 2FR

Moment of inertia of ring ship system

I = I(ring)+I(ship)+I(ship)

= MR² + mr² + mr² = MR² + 2mr²

angular acceleration of the ring ship system

∝ = \frac{2FR}{MR^2 + 2mr^2}

Now we have ,

∝ = \frac{1}{t} \sqrt{\frac{g}{R} }  , ∝ = \frac{2FR}{MR^2+2mr^2}

equating both values

We have,

F = \frac{MR^2+2mr^2}{2Rt} \sqrt{\frac{g}{R} }

where,

m = 2.1 × 10⁵kg, R = 86m, m = 3.5 × 10₄kg,

r = 31m, g = 9.8m/s² , t = 10800sec

F = \frac{(2.1 \times 10^5 \times86^2)+(2\times3.5v10^4\times31^2) }{2\times86\times10800} \times\sqrt{\frac{9.8}{86} } \\\\F = 294.47N\\\approx294.5N

Each thruster has to applied a force of 294.5N in tangential direction

3 0
3 years ago
Which clue indicates that a speciation has occurred between two populations?
Diano4ka-milaya [45]
I believe the answer is C
4 0
3 years ago
Two tuning forks of frequency 480 hz and 484 hz are struck simultaneously. what is the beat frequency resulting from the two sou
umka21 [38]
Ans: Beat frequency = f_b = 4Hz

Explanation: 
The beat frequency is equal to the absolute value of the difference in frequency of the two waves. In other words, the number of beats per second is equal to the difference in frequency. It is due to the destructive and constructive interference. <span>According to this interference, sound will be soft or loud.

Hence. the formula is:
</span>Beat frequency = f_b = |f_2 - f_1|
<span>
Since,
</span>f_1 = 480Hz
f_2 = 484Hz

Therefore,
Beat frequency = f_b = |484 - 480|

=> Beat frequency = f_b = 4Hz
-i
7 0
3 years ago
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