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3 years ago

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At a rock concert, the sound intensity 1.0 m in front of the bank of loudspeakers is 0.10 W/m2. A fan is 30 m from the loudspeak

ers. Her eardrums have a diameter of 8.4 mm. How much energy is transferred to each eardrum in 1.0 second?

1 answer:

aliina [53]3 years ago

5 0

Explanation:

Below is an attachment containing the solution.

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1. The whole world is __________________ with rocks.

ArbitrLikvidat [17]

I got it keep it bucks worth u this it tooooo muchhhhhhh

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2 years ago

Read 2 more answers

A solid sphere of mass 8.6 kg, made of metal whose density is 3,400 kg/, hangs by a cord. When the sphere is immersed in a liqui

creativ13 [48]

Answer:

A. $1900 kg/m^3$

Explanation:

We are given that

$m=8.6 kg$

Density, $\rho_s=3400 kg/m^3$

Tension,T=38 N

We have to find the density of liquid.

$T=mg-\rho_l Vg$

$g=9.8 m/s^2$

Volume,V= $\frac{m}{\rho_s}$

$38=8.6\times 9.8-\rho_l\times \frac{8.6}{3400}\times 9.8$

$\rho_l\times \frac{8.6}{3400}\times 9.8=8.6\times 9.8-38$

$\rho_l=\frac{(8.6\times 9.8-38)\times 3400}{8.6\times 9.8}$

$\rho_l=1867kg/m^3\approx 1900 kg/m^3$

Option A is true.

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3 years ago

What is the change in potential energy with respect to the ground of a 0.55 g son of a leech (a very effective fly for steelhead

IceJOKER [234]

Answer:

We will use Potential Energy Formula

Potential Energy = mass × gravitation × height

PE = 0,55 × 10 × 4

PE = 22 J (A)

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3 years ago

he block is released, and it slides 2.0 m (from the point at which it is released) across a horizontal surface before friction s

alex41 [277]

Answer:

0.245

Explanation:

When the block is released, the initial elastic potential energy stored in the spring is entirely converted into kinetic energy of the block.

Therefore, we can calculate the initial speed of the block:

$\frac{1}{2}kx^2 = \frac{1}{2}mv^2$

where the term on the left is the potential energy and where the term on the right is the kinetic energy, and where

k = 4500 N/m is the spring constant

x = 8.0 cm = 0.08 m is the compression of the spring

m = 3.0 kg is the mass of the block

v is the initial velocity

Solving for v,

$v=\sqrt{\frac{kx^2}{m}}=\sqrt{\frac{(4500)(0.08)^2}{3.0}}=3.1 m/s$

Then, after the block is released, all its kinetic energy is converted into thermal energy as the block slows down, due to friction. Therefore, the work done by friction is equal to the initial kinetic energy of the block.

The force of friction is

$F=\mu mg$

where

$\mu$ is the coefficient of friction

$g=9.8 m/s^2$ is the acceleration of gravity

So the work done by it is (in magnitude)

$W=Fd=\mu mg d$

where

d = 2.0 m is the distance covered

Therefore,

$\frac{1}{2}mv^2 = \mu mg d$

And solving for $\mu$ ,

$\mu = \frac{v^2}{2gd}=\frac{3.1^2}{2(9.8)(2.0)}=0.245$

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3 years ago

An object that starts at a position of 5 m and travels for 3 seconds at a velocity of -9 m/s ends up at what position?

Rina8888 [55]

Answer:

Sorry don't know the answer

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2 years ago