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mylen [45]
3 years ago
9

"Two uniform identical solid spherical balls each of mass M and radius R" and moment of inertia about its center 2/5 MR2 are rel

eased from rest from the same height H above the horizontal ground. Ball A falls straight down (no air resistance) while B rolls down the inclined plane without slipping. Which ball has the greater TME at the bottom of the incline?
Physics
1 answer:
adelina 88 [10]3 years ago
5 0

Answer:

he sphere that uses less time is sphere A

Explanation:

Let's start with ball A, for this let's use the kinematics relations

        v² = v₀² - 2g (y-y₀)

indicate that the sphere is released therefore its initial velocity is zero and when it reaches the floor its height is zero y = 0

         v² = 0 - 2 g (0- y₀)

         v = \sqrt{2g y_o}

         v = \sqrt{2 \ 9.8\ H}

         v = 4.427 √H

Now let's work the sphere B, in this case it rolls down a ramp, let's use the conservation of energy

starting point. At the highest point, before you start to move

         Em₀ = U = m g y

final point. At the bottom of the ramp

         Em_f = K = ½ m v² + ½ I w²

notice that we include the kinetic energy of translation and rotation

energy is conserved

          Em₀ = Em_f

          mg H = ½ m v² + ½ I w²

angular and linear velocity are related

          v = w r

          w = v / r

the momentorot of inertia indicates that it is worth

          I = \frac{2}{5} m r²

we substitute

           m g H = ½ m v² + ½ (\frac{2}{5}  m r²) (\frac{v}{r} )²

           gH = \frac{1}{2}  v² + \frac{1}{5}  v² = \frac{7}{10}  v²

           v = \sqrt{\frac{10}{7} \ g H}

           v = \sqrt{ \frac{10}{7}  \ 9.8 \ H}

           v=3.742 √H

Taking the final speeds of the sphere, let's analyze the distance traveled, sphere A falls into the air, so the distance traveled is H.  The ball B rolls in a plane, so the distance (L) traveled can be found with trigonometry

           sin θ = H / L

           L = H /sin θ

we can see that L> H

In summary, ball A arrives with more speed and travels a shorter distance, therefore it must use a shorter time

Consequently the sphere that uses less time is sphere A

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Time taken by the package to reach the sea level= 13.7 s

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925=0(t)+1/2 (9.8)t^2

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Answer:

The angular velocity at the beginning of the interval is \pi\sqrt{2}\ rad/s.

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Using formula of angular velocity

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Answer:

Answers in solutions.

Explanation:

<u>Question 6:</u>

The density of gold is 19.3 g/cm³

The density of silver is 10.5 g/cm³

  • The density of the substance in Crown A;

Density = mass ÷ volume = \frac{1930}{100} = 19.3 g/cm³

Since the density of gold, given, is 19.3 g/cm³ and the density of the substance in Crown A has a density of 19.3 g/cm³ , then that substance must be gold.

  • The density of the substance in Crown B;

Density = mass ÷ volume = 1930 ÷ 184 = 10.48913043  g/cm³ ≈ 10.5 g/cm³  (answer rounded up to one decimal place)

Since the density of the substance in Crown B is approximately equal to 10.5 g/cm³ , then that substance is Silver.

  • The density of substance in Crown C;

Density = mass ÷ volume = 1930g ÷ 150cm³ = 12.86666667 ≈ 12.9 cm³ (answer rounded up to one decimal place)

<h3><u>The density of the mixture:</u></h3><h3 />

For 2 cm³ of the mixture, its mass equal 19.3 g + 10.5 g = 29.8 g

∴ for 1 cm³ of the mixture, its mass equal to \frac{29.8}{2} = 14.9 g

Hence the density of the mixture = 14.9 g/cm³ and is not equal to the density of the substance in Crown C.

* Crown C is not made up of a mixture of gold and silver.

<u>Question 7:</u>

<u />

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  • Water is poured into measuring cylinder until the liquid level is at the 100 cm³ mark.
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The mass of water that occupied the 100 cm³ space of the container = total mass - mass of the empty container = 850 g - 500 g = 350 g

Density of the liquid (water) poured into the container = mass ÷  volume = 350 g ÷ 100 cm³ = 3.5g/cm³

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A tank filled with water has a volume of 0.02 m³

(a) 1 liter = 0.001 m³

How many liters? = 0.02 m³ ?

Cross multiplying gives:

\frac{0.02 * 1}{0.001} =  20 liters

(b) 1 m³ = 1,000,000 cm³

0.02 m³ = how many cm³ ?

Cross-multiplying gives;

\frac{0.02 * 1,000,000}{1} = 20,000 cm³

(c) 1 cm³ = 1 ml

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The mass of the liquid which overflow = mass of the stone = 20 g

1 gram of the liquid occupies 1 cm³ of space.

20 g of the liquid will occupy; \frac{20 * 1}{1} = 20 cm³

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∴ The volume of the stone = 20 cm³

(b) Mass = density ×  volume

Density of the stone = 5.0 g/cm³

Volume of the stone = 20 cm³

Mass of the stone = 5 g/cm³ × 20 cm³ = 100 g

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Answer:

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Explanation:

From the given information:

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The charge particle Q = 6 mC = 6 × 10⁻³ C

The change in the electric potential from point B to A;

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According to the work-energy theorem:

-Q × ΔV = ΔK

-Q \times ( V_B - V_A) = (K_B - K_A)

-(6\times 10^{-3}\ C) \times ( -2.5 \times 10^3) = (K_B - 1.9000000000000001 \ J)

15 = (K_B - 1.9000000000000001 \ J)

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