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1 year ago

If f(x) = -x2+x-1 then value of f(f(2)) is

Physics

1 answer:

qaws [65]1 year ago

6 0

Answer:

-13

Explanation:

Function

f(x) = -x² + x - 1

Solving

Substitute 2 in place of x
f(2) = -(2)² + 2 - 1
f(2) = -4 + 2 - 1
f(2) = -3
f(f(2)) = f(-3)
f(-3) = -(-3)² - 3 - 1
f(-3) = -9 - 3 - 1
f(f(2)) = -13

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Students are asked to create roller coasters for marbles. Their goal is to design a coaster with the tallest possible hill that

ivann1987 [24]

Answer:

Kinetic Energy.

Explanation:

The movement of a roller coaster is accomplished by the conversion of potential energy to kinetic energy. The roller coaster cars gain potential energy as they are pulled to the top of the first hill. As the cars descend the potential energy is converted to kinetic energy.

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2 years ago

If an objects speed is constant and the direction is in a straight line what type of velocity does the object have

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That's unaccelerated motion, and constant velocity.

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3 years ago

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Which statements are true concerning Newton's law of gravitation? The gravitational force is related to the mass of each object.

OLEGan [10]

Answer:

The gravitational force is related to the mass of each object.

The gravitational force is an attractive force.

Explanation:

Gravitational force is a long range force of attraction between any two masses.

Mathematically given as :

$F=G.\frac{m_1.m_2}{r^2}$

where:

$m_1 & m_2$ are the masses

r= distance between the center of mass of the two objects.

G= gravitational constant = $6.67\times 10^{-11} m^3.kg^{-1}.s^{-2}$

From the above relation of eq. (1) it is clear that,

Gravitational force is inversely proportional to the square of the distance and directly proportional to the masses.

The mass of an object is independent of its size due to the fact that density may vary for different objects.

The force of gravity varies with height as:

$\frac{g}{g_x} =(\frac{r_x}{r} )^2$

where:

$g=9.8\,m.s^{-2}$

$g_x=$ gravity at height $r_x$ of the center of mass of the object from the center of mass of the earth.

and we know that force:

$F=m\times g$

where: m= mass of the object.

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3 years ago

A light flashes at position x=0m. One microsecond later, a light flashes at position x=1000m. In a second reference frame, movin

padilas [110]

Answer:

To the right relative to the original frame.

Explanation:

In first reference frame S,

Spatial interval of the event, $\rm \Delta x=1000\ m-0\ m=1000\ m.$

Temporal interval of the event, $\rm \Delta t = 1\ \mu s=10^{-6}\ s.$

In the second reference frame S', the two flashes are simultaneous, which means that the temporal interval of the event in this frame is $\rm \Delta t'=0\ s.$

The speed of the frame S' with respect to frame S = v.

According to the Lorentz transformation,

$\rm \Delta t'=\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}\left( \Delta t-\dfrac{v\Delta x}{c^2}\right ).\\\\Since,\ \Delta t'=0,\\\therefore \dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}\left( \Delta t-\dfrac{v\Delta x}{c^2}\right )=0\\\Rightarrow \Delta t-\dfrac{v\Delta x}{c^2}=0\\\dfrac{v\Delta x}{c^2}=\Delta t\\v=\dfrac{c^2\Delta t}{\Delta x }.\\\\Also, \ \Delta t,\ \Delta x>0\ \Rightarrow v>0.$

And positive v means the velocity of the second frame S' is along the positive x-axis direction, i.e., to the right direction relative to the original frame S.

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2 years ago

Particle A of charge 2.70 10-4 C is at the origin, particle B of charge -6.36 10-4 C is at (4.00 m, 0), and particle C of charge

Serhud [2]

Answer:

FC vector representation

$F_{C} =(19.03i+13.78j)N$

Magnitude of FC

$F_{C}=23.495N$

Vector direction FC

$\beta=35.91$ degrees: angle that forms FC with the horizontal

Explanation:

Conceptual analysis

Because the particle C is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

The directions of the individual forces exerted by qA and qB on qC are shown in the attached figure; The force (FAC) of qA over qC is repulsive because they have equal signs and the force (FBC) of qB over qC is attractive because they have opposite signs.

The FAC force is up in the positive direction and the FBC force forms an α angle with respect to the x axis.

$\alpha = tan^{-1}(\frac{3}{4}) = 36.86$ degrees