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spin [16.1K]
4 years ago
14

Turning a magnet very quickley would be best used to create

Physics
1 answer:
irakobra [83]4 years ago
7 0
C) an electric current
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In an experiment, a variable, position-dependent force F(x)F(x) is exerted on a block of mass 1.0kg1.0kg that is moving on a hor
leonid [27]

Answer:

The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(C) is correct option.

Explanation:

Given that,

Mass of block = 1.0 kg

Dependent force = F(x)

Frictional force = F(f)

Suppose, the following information would students need to test the hypothesis,

(A) The function F(x) for 0 < x < 5 and the value of F(f).

(B) The function a(t) for the time interval of travel and the value of F(f).

(C) The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(D) The function a(t) for the time interval of travel, the time it takes the block to move 5 m, and the value of F(f).

(E) The block's initial velocity, the time it takes the block to move 5 m, and the value of F(f).

We know that,

The work done by a force is given by,

W=\int_{x_{0}}^{x_{f}}{F(x)\ dx}.....(I)

Where, F(x) = net force

We know, the net force is the sum of forces.

So, \sum{F}=ma

According to question,

We have two forces F(x) and F(f)

So, the sum of these forces are

F(x)+(-F(f))=ma

Here, frictional force is negative because F(f) acts against the F(x)

Now put the value in equation (I)

W=\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}

We need to find the value of \int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}

Using newton's second law

\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{x_{0}}^{x_{f}}{ma\ dx}...(II)

We know that,

Acceleration is rate of change of velocity.

a=\dfrac{dv}{dt}

Put the value of a in equation (II)

\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{x_{0}}^{x_{f}}{m\dfrac{dv}{dt}dx}

\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{v_{0}}^{v_{f}}{mv\ dv}

\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\dfrac{mv_{f}^2}{2}+\dfrac{mv_{0}^2}{2}

Now, the work done by the net force on the block is,

W=\dfrac{mv_{f}^2}{2}+\dfrac{mv_{0}^2}{2}

The work done by the net force on the block is equal to the change in kinetic energy of the block.

Hence, The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(C) is correct option.

7 0
3 years ago
Which of the following lies in the ecliptic plane?
babymother [125]
<h2>Answer: Earth's orbital path around the Sun</h2><h2></h2>

The <u>Ecliptic</u> refers to the orbit of the Earth around the Sun. Therefore, <u>for an observer on Earth it will be the apparent path of the Sun in the sky during the year, with respect to the "immobile background" of the other stars.</u>

<u />

It should be noted that the ecliptic plane (which is the same orbital plane of the Earth in its translation movement) is tilted with respect to the equator of the planet about 23\° approximately. This is due to the inclination of the Earth's axis.

Hence, the correct option is Earth's orbital path around the Sun.

7 0
4 years ago
Heat engines convert which type of energy into mechanical work
KonstantinChe [14]

Answer: Heat engines convert chemical energy to mechanical energy

Explanation:

7 0
3 years ago
What is a lot like gases but the atoms are different because they are made up of free electrons and ions of elements
34kurt

plasmas  are a lot like gases



hope this helps.

7 0
3 years ago
When the north pole of a magnet is moved towards the center of a loop of wire containing a galvanometer, the needle of the galva
andriy [413]

Answer:

Moving the magnet away from the center of the loop with its south pole facing the center of the loop.

Explanation:

Electromagnetic induction is due to a rapidly changing magnetic field, or loop area. The poles of the magnet induce current in the loop but in the opposite direction, depending on the direction of their relative motion. An approaching north pole will induce an anticlockwise current in the loop, while an approaching south pole will do the reverse. To get the galvanometer to flicker in the same direction as of that when the north pole was approaching, we move the magnet away from the center of the loop with its south pole facing the center of the loop.

6 0
3 years ago
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