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3 years ago

11

Anne is trying to push a tall, heavy cabinet across a carpeted floor. woman pushing a cabinet across the floor Anne is having a

hard time because there is a lot of friction between the cabinet and the floor. How can Anne reduce the amount of friction?

1 answer:

Ne4ueva [31]3 years ago

3 0

Answer:

The carpet is creating the friction.

Explanation:

You can remove the carpet or pick the cabinet up off the carpet.

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A flock of ducks is trying to migrate south for the winter, but they keep being blown off course by a wind blowing from the west

Minchanka [31]

The ducks' flight path as observed by someone standing on the ground is the sum of the wind velocity and the ducks' velocity relative to the wind:

ducks (relative to wind) + wind (relative to Earth) = ducks (relative to Earth)

or equivalently,

$\vec v_{D/W}+\vec v_{W/E}=\vec v_{D/E}$

(see the attached graphic)

We have

ducks (relative to wind) = 7.0 m/s in some direction <em>θ</em> relative to the positive horizontal direction, or

$\vec v_{D/W}=\left(7.0\dfrac{\rm m}{\rm s}\right)(\cos\theta\,\vec\imath+\sin\theta\,\vec\jmath)$

wind (relative to Earth) = 5.0 m/s due East, or

$\vec v_{W/E}=\left(5.0\dfrac{\rm m}{\rm s}\right)(\cos0^\circ\,\vec\imath+\sin0^\circ\,\vec\jmath)$

ducks (relative to earth) = some speed <em>v</em> due South, or

$\vec v_{D/E}=v(\cos270^\circ\,\vec\imath+\sin270^\circ\,\vec\jmath)$

Then by setting components equal, we have

$\left(7.0\dfrac{\rm m}{\rm s}\right)\cos\theta+5.0\dfrac{\rm m}{\rm s}=0$

$\left(7.0\dfrac{\rm m}{\rm s}\right)\sin\theta=-v$

We only care about the direction for this question, which we get from the first equation:

$\left(7.0\dfrac{\rm m}{\rm s}\right)\cos\theta=-5.0\dfrac{\rm m}{\rm s}$

$\cos\theta=-\dfrac57$

$\theta=\cos^{-1}\left(-\dfrac57\right)\text{ OR }\theta=360^\circ-\cos^{-1}\left(-\dfrac57\right)$

or approximately 136º or 224º.

Only one of these directions must be correct. Choosing between them is a matter of picking the one that satisfies <em>both</em> equations. We want

$\left(7.0\dfrac{\rm m}{\rm s}\right)\sin\theta=-v$

which means <em>θ</em> must be between 180º and 360º (since angles in this range have negative sine).

So the ducks must fly (relative to the air) in a direction 224º relative to the positive horizontal direction, or about 44º South of West.

8 0

3 years ago

A jet aircraft is traveling at 260 m/s in horizontal flight. The engine takes in air at a rate of 53.3 kg/s and burns fuel at a

IrinaVladis [17]

Answer:

The thrust of the jet engine is 4188.81 N.

Explanation:

Given that,

Speed = 260 m/s

Rate in air= 53.3 kg/s

Rate of fuel = 3.63 kg/s

Relative speed = 317 m/s

We need to calculate the rate of mass change in the rocket

Using formula of rate of mass

$\dfrac{dM}{dt}=\dfrac{dM_{a}}{dt}+\dfrac{dM_{f}}{dt}$

Put the value into the formula

$\dfrac{dM}{dt}=53.3+3.63$

$\dfrac{dM}{dt}=56.93\ kg/s$

We need to calculate the thrust of the jet engine

Using formula of thrust

$T=\dfrac{dM}{dt}u-\dfrac{dM_{a}}{dt}v$

Put the value into the formula

$T=56.93\times317-53.3\times260$

$T=4188.81\ N$

Hence, The thrust of the jet engine is 4188.81 N.

7 0

3 years ago

8.How long is a day? A year?

prisoha [69]

Answer:

There are a total of 24 hours in a day

3 0

3 years ago

Read 2 more answers

Determine the magnitude of the resultant force FR=F1+F2FR=F1+F2. Assume that F1F1F_1 = 235 lblb and F2F2F_2 = 350 l

DENIUS [597]

Answer:

585lb

Explanation:

Given the following

F1 = 235lb

F2 = 350lb

The resultant is expressed as;

FR = F1+F2

Substitute the given values

FR = 235+350

FR = 585lb

Hence the magnitude of the resultant is 585lb

7 0

3 years ago

A 45.0 - kg girl is standing on a 150. - kg plank. The plank, originally at rest, is free to slide on a frozen lake, which is a

Mrac [35]

Answer:

The velocity relative to the surface of the ice is 6.5 m/s.

Explanation:

Given that,

Mass of girl m= 45.0 kg

Mass of plank M= 150 kg

Velocity = 1.50 m/s

We need to calculate the velocity relative to the surface of the ice

Using conservation of momentum

$mv_{1}+Mv_{2}=(m+M)v$

Here, $v_{2}=0$ because plank at rest

$v_{1}=\dfrac{m+M}{m}\times v$

Put the value into the formula

$v_{1}=\dfrac{45+150}{45}\times1.50$

$v_{1}=6.5\ m/s$

Hence, The velocity relative to the surface of the ice is 6.5 m/s.

6 0

3 years ago