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Katen [24]
3 years ago
14

An inclined plane is made out of a short plank of wood. It is used to move a 300N box up onto a tabletop 1m above the floor. Wha

t would happen to the work and mechanical advantage if a longer plank were used to accomplish the same task?
A) The work and mechanical advantage would increase.
B) The work would decrease and mechanical advantage would increase.
C) The work would remain the same and mechanical advantage would increase.
D) The work would remain the same and mechanical advantage would decrease.
Physics
2 answers:
den301095 [7]3 years ago
4 0

C. The work would remain the same and mechanical advantage would increase.

You're still doing the same thing, but having a longer ramp would make the work done easier (mechanical advantage has increased)

Ronch [10]3 years ago
3 0

Answer:

<em>The purpose of an inclinded plane is to make easier to move objects to a certain height.</em>

The technology behind this is about the Work you need to use to move the object upwards. Basically, when we use an inclined plane, we are splitting the net force, making easier to move. All this means, the force needed to move the objecto up will be lower, due to the inclined plane.

So, if the force needed is lower, then the work is also lower, because the work done is defined as the product between the force applied and the distance traveled.

<em>In addition, if we have a longer inclined plane, that means the force needed is even lower,</em> beacuse the distance is increased, but the Work is the same, because it only depends on the initial and final point.

Therefore, in this case, the work would remain the same and the mechanical advantage would increase. As we said before, the work needed will be the same despite the force decreases, because the distance increases, remaining the work as a constant. And the mechanical advantage increases, because it's easier to move if the inclined plane is longer.

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Svetach [21]

I'm assuming the question is what is the robin's speed relative to to the ground...

Create an equation that describes its relative motion.

rVg = rVa + aVg


Substitute values.

rVg = 12 m/s [N] + 6.8 m/s [E]


Use vector addition.

| rVg | = √ | rVa |² + | aVg |²

| rVg | = √ 144 m²/s² + 46.24 m²/s²

| rVg | = √ 19<u>0</u>.24 m²/s²

| rVg | = 1<u>3</u>.78 m/s


Find direction.

tanФ = aVg / rVa

tanФ = 6.8 m/s / 12 m/s

Ф = 29°


Therefore, the velocity of the robin relative to the ground is 14 m/s [N29°E]

4 0
3 years ago
NEED HELP ASAP!!!
sineoko [7]

Answer:

2,500 watts

Explanation:

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5 0
3 years ago
In an oscillating LC circuit, the total stored energy is U and the maximum current in the inductor is I. When the current in the
Deffense [45]

Answer:

The definition of that same given problem is outlined in the following section on the clarification.

Explanation:

The Q seems to be endless (hardly any R on the circuit). So energy equations to describe and forth through the inducer as well as the condenser.  

Presently take a gander at the energy stored in your condensers while charging is Q.

⇒  U =\frac{Qmax^2}{C}

So conclude C doesn't change substantially as well as,

When,

⇒  Q=\frac{Qmax}{2}

⇒  Q^2=\frac{Qmax^2}{4}

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5 0
3 years ago
A circular loop of flexible iron wire has an initial circumference of 167 cm, but its circumference is decreasing at a constant
Kisachek [45]

Answer:

Part a)

EMF = 5.6 \times 10^{-3} V

Part b)

Since the radius is decreasing so induced current will increase the flux through the coil

So it would be clockwise in direction

Explanation:

As we know that magnetic flux linked with the coil is given as

\phi = \pi r^2 B

now the rate of change in flux is given as

\frac{d\phi}{dt} = 2\pi r \frac{dr}{dt} B

now we know that circumference is decreasing at rate of 15 cm/s

so here we know the length of circumference as

C = 2\pi r

So rate of change in circumference is

\frac{dC}{dt} = 2\pi \frac{dr}{dt}

\frac{1}{2\pi}(15 cm) = \frac{dr}{dt}

final length of circumference at t = 8 s

C = 167 - (15)(8) = 47

Part a)

Now the induced EMF is given as

EMF = (2\pi r)(\frac{1}{2\pi})(0.15)(0.5)

EMF = (0.47)(\frac{1}{2\pi})(0.15)(0.5)

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Part b)

Since the radius is decreasing so induced current will increase the flux through the coil

So it would be clockwise in direction

5 0
3 years ago
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