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Firdavs [7]
3 years ago
7

What replaces a cold current that sinks to the ocean floor?

Physics
1 answer:
FrozenT [24]3 years ago
5 0
Well simple the warm water then replaces the cold current that sinks to the ocean floor.
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The low-frequency speaker of a stereo set produces 10.0 W of acoustical power. If the speaker projects sound uniformly in all di
Paha777 [63]

Answer:

the required distance is 89.125 m

Explanation:

Given the data in the question;

we know that, sound intensity B in decibels of sound is;

β(dB) = 10log₁₀( I / I_0 )

where intensity I = power / area carried by wave

I_0 = 10⁻¹² W/m² { minimum threshold intensity }

Now,

intensity I = power / area carried by wave = P/A = P/4πr² { spherical  }

given that; β = 80.0 dB and P = 10 W

so

β(dB) = 10log₁₀( I / I_0 )

we substitute

80 = 10log₁₀( P / 4πr²× I_0)

80 = 10log₁₀( 10 / 4πr²× 10⁻¹² )

8 = log₁₀(10) - log₁₀( 4πr²× 10⁻¹² )  

8 = 1 - log₁₀( 4πr²× 10⁻¹² )

8 - 1 = -log₁₀( 4πr²× 10⁻¹² )

7 = -log₁₀( 1.2566 × 10⁻¹¹ × r² )

7 = -[ log₁₀( 1.25 × 10⁻¹¹) + log₁₀( r² ) ]

7 = -[ -10.9 + log₁₀( r² ) ]

7 = 10.9 - log₁₀( r² )

-log₁₀( r² ) = 7 - 10.9

-log₁₀( r² )  = - 3.9

log₁₀( r² ) = 3.9

2log₁₀r = 3.9

log₁₀r  = 3.9 /2

log₁₀r = 1.95

r = 89.125 m

Therefore, the required distance is 89.125 m

6 0
2 years ago
What force is keeping the Earth in orbit around the Sun? * HELP FAST PLEASE.
sveta [45]

Answer:

magnetic force due to gravity

Explanation:

nowno that is because that's what the geography teacher taught us back in don't remember but it's true

6 0
2 years ago
Jerry the mouse is running along a straight desert road at a constant velocity of 18 m/s. If a certain Tom cat wants to capture
Kruka [31]

Answer:

a) t = 1.75 s

b) x = 31.5 m

Explanation:

a) The time at which Tom should drop the net can be found using the following equation:

y_{f} = y_{0} + v_{oy}t - \frac{1}{2}gt^{2}

Where:

y_{f}: is the final height = 0

y₀: is the initial height = 15 m

g: is the gravity = 9.81 m/s²

v_{0y}: is the initial vertical velocity of the net = 0 (it is dropped from rest)

0 = 15m - \frac{1}{2}9.81 m/s^{2}*t^{2}

t = \sqrt{\frac{2*15 m}{9.81 m/s^{2}}} = 1.75 s

Hence, Tom should drop the net at 1.75 s before Jerry is under the bridge.

b) We can find the distance at which is Jerry when Tom drops the net as follows:

v = \frac{x}{t}

x = v*t = 18 m/s*1.75 m = 31.5 m

Then, Jerry is at 31.5 meters from the bridge when Jerry drops the net.

I hope it helps you!                                                                    

3 0
3 years ago
What is the atmospheric pressure 1.00 km above the surface of Venus? Express your answer in Earth-atmospheres.
EleoNora [17]

Below is an attachment containing the solution.

4 0
3 years ago
You move a 25 N object 5 meters. If it takes 8 s how much power did you do?
klasskru [66]

Answer:

15.625 watts

Explanation:

Recall that power is defined as the worked performed per unit of time:

Power = Work / time

The work done is Force * distance, so in our case the work is:

Work = 25 M * 5 m = 125 J

Then the power will be:

Power = 125 J / 8 sec = 15.625 watts

6 0
2 years ago
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