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3 years ago

1.- Se desea elevar un cuerpo de 1500kg utilizando una elevadora hidráulica de plato grande

Physics

1 answer:

kvv77 [185]3 years ago

5 0

Answer:

181.48 N

Explanation:

Calculate the area :

Area = pi * r² ;

pi = 3.14 ; r1 = 90cm /100 = 0.9m ; r2 = 10/100 = 0.1m

Area 1, A1 = 3.14 * 0.1² = 0.0314 m²

Area 2, A2 = 3.14 * 0.9² = 2.5434 m²

Force, F = mass * acceleration due to gravity

F2 = 1500 * 9.8 = 14700 N

Force 1 / Area 1 = Force 2 / Area 2

Force 1 = (Force 2 / Area 2), * Area 1

Force 1 = (14700 / 2.5434) * 0.0314

Force = 5779.6650 * 0.0314

= 181.48 N

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A highway patrolman traveling at the speed limit is passed by a car going 20 mph faster than the speed limit. After one minute,

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Answer:

It will take 40 seconds to catch the speeding car

Explanation:

Initial speed of the patrolman = 55 mph

after one minute speed of patrol man = 115 mph

now acceleration of patrolman is given by

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$a = \frac{115 - 55}{1/60} = 3600 m/h^2$

now at this acceleration the distance covered by patrolman in "t" time is given as

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now we know the speed of the speeding car is given as

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now in the same time distance covered by it

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3 years ago

Read 2 more answers

For a polymer that was maintained at constant strain, an initial stress of 2.76 MPa decreases to 1.72 MPa after 6 hr. Estimate t

DedPeter [7]

Answer:

1.48 Mpa

Explanation:

Given that For a polymer that was maintained at constant strain, an initial stress of 2.76 MPa decreases to 1.72 MPa after 6 hr. To estimate the stress remaining after 30 hr, we must first find the decay constant by using the exponential equation.

Therefore, the estimate of the stress remaining after 30 hr, is 1.48Mpa.

Please find the attached file for the solution.

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3 years ago

Approximate the skin friction drag on a 1m long by 60 cm diameter cylinder, located axially in a wind tunnel, when the air speed

TiliK225 [7]

Answer:

The drag force is $1.76\times10^{-2}\ N$

Explanation:

Given that,

Diameter = 60 cm

Length = 1 m

Air speed = 4.5 m/s

Temperature = 50°C

We need to calculate the Reynolds number

Using formula of Reynolds number

$R_{e}=\dfrac{vl}{\mu}$

Put the value into the formula

$R_{e}=\dfrac{4.5\times1}{1.900\times10^{-5}}$

$R_{e}=2.368\times10^{5}$

We need to calculate the drag coefficient

Using formula of drag coefficient

$C_{d}=\dfrac{2\times0.646}{\sqrt{R_{e}}}$

Put the value into the formula

$C_{d}=\dfrac{2\times0.646}{\sqrt{2.368\times10^{5}}}$

$C_{d}=0.002655$

We need to calculate the drag force

Using formula of drag force

$F_{d}=\dfrac{1}{2}\times C_{d}\times\rho\times A\times v^2$

Put the value into the formula

$F_{d}=\dfrac{1}{2}\times0.002655\times1.095\times0.6\times1\times(4.5)^2$

$F_{d}=0.01766\ N$

$F_{d}=1.76\times10^{-2}\ N$

Hence, The drag force is $1.76\times10^{-2}\ N$

3 0

3 years ago