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3 years ago

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For the following questions, imagine the book drops in free fall and falls a distance d. For the system of the book and earth, t

he change in total energy of the system if the book falls a distance d _______.(A) is positive.(B) is negative.(C) is zero.(D) has a sign which cannot be determined.

1 answer:

alukav5142 [94]3 years ago

8 0

Answer:

C) is zero

Explanation:

According to the law of energy conservation, the total mechanical energy of the object is conserved. A book falling a distance d would have a change in potential energy, resulting in the same change in kinetic energy. But the total mechanical energy must be the same. So there's 0 change in total energy of the system.

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peach pie in a 9.00 in diameter plate is placed upon a rotating tray. Then, the tray is rotated such that the rim of the pie pla

zavuch27 [327]

Explanation:

It is given that,

Diameter of the peach pie, d = 9 inches

Radius of the pie, r = 4.5 inches

The tray is rotated such that the rim of the pie plate moves through a distance of 183 inches, d = 183 inches

Let $\theta$ is the angular distance that the pie plate has moved through.

It is given by :

$\theta=\dfrac{d}{r}$

$\theta=\dfrac{183}{4.5}$

$\theta=40.66\ radian$

Since, 1 radian = 57.29 degrees

$\theta=2329.64\ degrees$

Since, 1 radian = 0.159155 revolution

$\theta=6.47124\ revolution$

Hence, this is the required solution.

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3 years ago

What is horoscope? what is its uses

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3 years ago

Read 2 more answers

A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 14.413

natima [27]

Answer:

The value of charge q₃ is 40.46 μC.

Explanation:

Given that.

Magnitude of net force $F=14.413\ N$

Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.

We need to calculate the distance

Using Pythagorean theorem

$r=\sqrt{x_{2}^2+y_{2}^2}$

Put the value into the formula

$r=\sqrt{(8.2\times10^{-2})^2+(3.1\times10^{-2})^2}$

$r=0.0876\ m$

We need to calculate the magnitude of the charge q₃

Using formula of net force

$F_{12}=kq_{2}(\dfrac{q_{3}}{r_{3}^2}+\dfrac{q_{1}}{r_{1}^2})$

Put the value into the formula

$14.413=9\times10^{9}\times7.7\times10^{-6}(\dfrac{q_{3}}{(0.0438)^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})$

$(\dfrac{q_{3}}{(4.38\times10^{-2})^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})=\dfrac{14.413}{9\times10^{9}\times7.7\times10^{-6}}$

$\dfrac{q_{3}}{(0.0438)^2}=207\times10^{-4}+3.909\times10^{-4}$

$q_{3}=0.0210909\times(0.0438)^2$

$q_{3}=40.46\times10^{-6}\ C$

$q_{3}=40.46\ \mu C$

Hence, The value of charge q₃ is 40.46 μC.

5 0

3 years ago

3241004551 [841]

For this case we have that by definition, the kinetic energy is given by the following formula:

$k= \frac {1} {2} * m * v ^ 2$

Where:

m: It is the mass

v: It is the velocity

According to the data we have to:

$m = 100 \ kg\\v = 9 \frac {m} {s}$

Substituting the values we have:

$k = \frac {1} {2} * (100) * (9) ^ 2\\k = \frac {1} {2} * (100) * 81\\k = 50 * 81\\k = 4050$

finally, the kinetic energy is $4050 \ J$

Answer:

Option A

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