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2 years ago

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A basketball is tossed upwards with a speed of 5.0\,\dfrac{\text m}{\text s}5.0 s m 5, point, 0, start fraction, start text, m

, end text, divided by, start text, s, end text, end fraction. We can ignore air resistance. What is the maximum height reached by the basketball from its release point?

1 answer:

Hatshy [7]2 years ago

4 0

Answer:

The last one

Explanation:

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The electric potential in a region that is within 2.00 mm of the origin of a rectangular coordinate system is given by V=Axl+Bym

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Answer:

Given the potential, $V = Ax^l+By^m+Cz^n+D$

The components of the electric field are:

$E_x = \frac{-dV}{dx} = -Alx^l^-^1$

$E_y = \frac{-dV}{dy} = - Bmy^m^-^1$

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Let's calculate the potential difference for all given points.

$V(0, 0, 0) = 10V => Ax^l+By^m+Cz^n+D = 10$

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Solving for A, we have:

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$V(0, 1, 0) = 6V => B + 10 = 6$

Solving for B, we have:

$B = 6 - 10$

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$V(0, 0, 1) = 8V => C + 10 = 4$

Solving for C, we have:

$C = 8 - 10$

$C = -2$

For all given points, let's calculate the magnitude of electric field as follow:

$E_x (1, 0, 0) = 16 => - Alx^l^-^1 = 16$

$Al = -16$

Solving for l, we have:

$l = \frac{-16}{A}$

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$l = \frac{-16}{-6}$

$l = \frac{8}{3}$

$E_y (0, 1, 0) = 16=> Bmy^m^-^1 = 16$

$Bm = -16$

Solving for m, we have:

$m = \frac{-16}{A}$

From above, B = -4

$m = \frac{-16}{-4}$

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$E_y (0, 0, 1) = 16=> nCz^n^-^1 = 16$

$nC = - 16$

Solving for n, we have:

$n = \frac{-16}{C}$

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$n = 8$

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Answer:

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Explanation:

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ii)However, when they are touched the charges will be rearranged among the two sphere such that the two sphere have exact same magnitude and sign of charge and now they will repel each other or the magnitude of charges on the two spheres become zero and they neither attract or repel each other.

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Thermal energy in the form of fire is generated by the combustion of fuel. Due to the tendency of hot air to rise upward, the heat generated rises to fill the space of the balloon. One this space is full of trapped hot air, the heat's tendency to rise causes the hot air balloon to be lifted into the air.

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Two students walk in the same direction along a straight path, at a constant speed one at 0.90 m/s and the other at 1.90 m/s. a.

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Answer: a) 456.66 s ; b) 564.3 m

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v= distance/time so t=distance/v

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In order to calculate the distance that faster student should walk

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distance =vslower*time1

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replacing in the above expression we have

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