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Anon25 [30]
3 years ago
12

A very powerful vacuum cleaner has a hose 2.36 cm in diameter. With no nozzle on the hose (no change to hose area), what is the

weight of the heaviest brick it can lift? The atmospheric pressure is 1.00 atm=101,300 Pa. Ignore the buoyancy force of the air on the brick.
Physics
1 answer:
FinnZ [79.3K]3 years ago
8 0

Answer:

Explanation:

Pressure=Force/Area

Force=Pressure * Area

Force=1.013\times 10^5 \times 3.14\times (1.03\times 10^{-2})^2\\\\=33.7N

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Controlled Experiment 

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I Need help ASAP.   A block has a volume of 0.09 m3and a density of 4,000 kg/m3. What's the force of gravity acting on the block
yarga [219]

The correct answer is option A, 3.528 N

In this question the water has nothing to do, the gravity always pull the object with the same magnitude irrespective of the medium.

The force of gravity is calculated as

F=0.09*4000*9.8

=3.528 N

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Scientists are expected to share their results by
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Answer:

Publishing results of research projects in peer-reviewed journals enables the scientific and medical community to evaluate the findings themselves. It also provides instructions so that other researchers can repeat the experiment or build on it to verify and confirm the results.

7 0
4 years ago
Find the mass of the solid cylinder Dequals​{(r,theta​,z): 0less than or equalsrless than or equals2​, 0less than or equalszless
anygoal [31]

The mass of the cylinder <em>D</em> is obtained by integrating the density function over <em>D</em>:

\displaystyle\iiint_D\rho(r,\theta\,z)\,\mathrm dV

With \rho(r,\theta,z)=1+\frac z2, and

D=\left\{(r,\theta,z)\mid 0\le r\le2,0\le z\le10\right\}

the mass would be

\displaystyle\int_0^{2\pi}\int_0^2\int_0^{10}1+\frac z2\,\mathrm dz\,\mathrm dr\,\mathrm d\theta=4\pi\int_0^{10}1+\frac z2\,\mathrm dz

4\pi\left(z+\dfrac{z^2}4\right)\bigg|_0^{10}

=4\pi\left(10+\dfrac{100}4\right)=\boxed{140\pi}

6 0
3 years ago
Two coils have the same number of circular turns and carry the same current. Each rotates in a magnetic field acting perpendicul
ANEK [815]

Answer:

4.14 cm.

Explanation:

Given,

For Coil 1

radius of coil, r₁ = 5.6 cm

Magnetic field, B₁ = 0.24 T

For Coil 2

radius of coil, r₂ = ?

Magnetic field, B₂ = 0.44 T

Using formula of maximum torque

\tau_{max}= NIAB

Since both the coil experience same maximum torques

now,

NIA_1B_1 = NIA_2B_2

A_1B_1 = A_2B_2

r_1^2 B_1 = r_2^2 B_2

5.6^2\times 0.24= r_2^2\times 0.44

r_2 = 4.14\ cm

Radius of the coil 2 is equal to 4.14 cm.

4 0
4 years ago
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