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Crank
3 years ago
5

If a ball is rolled down a frictionless inclined plane from a height of 3 feet, to what height would it roll up to on the other

side?
Physics
1 answer:
mojhsa [17]3 years ago
3 0

Answer:

It will roll up to its initial height from which it rolled from on the first hill

Explanation:

The given information are;

The height from which the ball is rolled = 3 feet[

The equation of the potential energy of the ball as it rolled down the frictionless inclined plane is given by the following equation;

m × g × h₁ = 1/2 × m × v² + 1/2×[2/5× m × r²) × [v/r]

Where;

m = The mass of the ball

h = The from which the ball is rolled down

v = The velocity of the ball

r = The radius of the ball

Therefore, all the potential energy, m × g × h, is converted into kinetic energy when the ball reached the ground

Similarly, as the ball rolls up on the other side, by the principle of energy conservation, all the kinetic energy is transformed into potential energy as follows

1/2 × m × v² + 1/2×[2/5× m × r²) × [v/r] = m × g × h₂, therefore, we have;

1/2 × m × v² + 1/2×[2/5× m × r²) × [v/r] = m × g × h₁ =  m × g × h₂

Therefore since m × g × h₁ =  m × g × h₂, then h₁ = h₂, the ball rises to the same height it started rolling from on the first hill.

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What is the speed vfinal of the electron when it is 10.0 cm from charge 1?
fgiga [73]

Answer:

Two stationary positive point charges, charge 1 of magnitude 3.45 nC and charge 2 of magnitude 1.85 nC, are separated by a distance of 50.0 cm. An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges. What is the speed v(final) of the electron when it is 10.0 cm from

The answer to the question is

The speed v_{final} of the electron when it is 10.0 cm from charge Q₁

= 7.53×10⁶ m/s

Explanation:

To solve the question we have

Q₁ = 3.45 nC = 3.45 × 10⁻⁹C

Q₂ = 1.85 nC = 1.85 × 10⁻⁹ C

2·d = 50.0 cm

a = 10.0 cm

q = -1.6×10⁻¹⁹C

Also initial kinetic energy = 0 and

Initial electric potential energy = k\frac{qQ_1}{d} + k\frac{qQ_2}{d} = kq(\frac{Q_1+Q_2}{d})

Final kinetic energy due to motion = 0.5·m·v²

Final electric potential energy = k\frac{qQ_1}{a} + k\frac{qQ_2}{2d-a} = kq(\frac{Q_1}{a}+\frac{ Q_2}{2d-a})

From the energy conservation principle we have

0+ kq(\frac{Q_1+Q_2}{d})=0.5mv^2+  kq(\frac{Q_1}{a}+\frac{ Q_2}{2d-a})

Solving for v gives

v=\sqrt{\frac{kq(\frac{Q_1+Q_2}{d})-   kq(\frac{Q_1}{a}+\frac{ Q_2}{2d-a})}{0.5m}}

where k = 9.0×10⁹ and m = 9.109×10⁻³¹ kg

gives v =7528188.32769 m/s or 7.53×10⁶ m/s

v_{final} = 7.53×10⁶ m/s

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4 years ago
Newton’s second law of motion states that an object with a heavier mass will have more acceleration than an object with a smalle
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How is the electrical conductivity of a metal explained by metallic bonds.
love history [14]

Answer:

Metallic bonding is the bond that exist between the atoms. The electrons do not only flow at its respective atoms but instead, contribute to a sea of delocalised electrons. As a result, metals can conduct electricity as the delocalised electrons are able to carry charges.

Explanation:

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2 years ago
Why is creativity important in constructing scientific questions?
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A 13.3 kg box sliding across the ground
adelina 88 [10]

Answer:

<em>0.25</em>

Explanation:

According to newtons law of motion

\sum F_x = ma

F_f =  ma

nR = ma

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ng = a

n = a/g

g is the acceleration due to gravity

Given

a = 2.42m/s²

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Substitute into the formula;

n = 2.42/9.8

n = 0.25

<em>Hence the coefficient of kinetic friction is 0.25</em>

<em></em>

5 0
3 years ago
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