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Free_Kalibri [48]

4 years ago

15

In a carnival game, the player throws a ball at a haystack. For a typical throw, the ball leaves the hay with a speed exactly on

e-half of the entry speed. If the frictional force exerted by the hay is a constant 6.0 N and the haystack is 1.2 m thick, derive an expression for the typical entry speed as a function of the inertia of the ball. Assume horizontal motion only, and ignore any effects due to gravity. What is the typical entry speed if the ball has an inertia of a 0.35 kg?

1 answer:

8_murik_8 [283]4 years ago

8 0

Answer:

Ve(m) = sqrt (19.2/m)

Ve(0.35) = 7.407 m/s

Explanation:

Given:

- The ball has a mass = m

- The entry speed of the ball is Vi = Ve

- The final speed of the ball Vf = 0.5*Ve

- The constant frictional force on ball due to hay is F = 6 N

- The thickness of hay-stack is s = 1.2 m

- Assume the throw is in horizontal direction and neglect gravity forces

Find:

Derive an expression for the typical entry speed as a function of the inertia of the ball

What is the typical entry speed if the ball has an inertia of a 0.35 kg?

Solution:

- To determine the entry speed as a function of inertia we will use third equation of motion as follows:

Vf^2 = Vi^2 + 2*a*s

Where, a is acceleration of the ball through hay stack. We will use Newton's Law of motion to determine this:

F_net = m*a

The only force acting on the ball in its journey through hay-stack is the frictional force F:

- F = m*a

a = -F/m

- Input all the quantities in the third equation of motion:

(0.5Ve)^2 = Ve^2 - 2*F*s / m

0.75Ve^2 = 2*F*s / m

Ve = sqrt (8*F*s/3*m)

Plug in values:

Ve(m) = sqrt (8*6*1.2/3*m)

Ve(m) = sqrt (19.2/m)

- The entry speed for the inertia of the ball m = 0.35 kg is:

Ve(0.35) = sqrt(19.2/0.35)

Ve(0.35) = 7.407 m/s

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3 years ago

How will the electrostatic force between two electric charges change if the first charge is doubled and the second charge is onl

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Answer:

B) $\frac{2}{3}$

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Let the original charge be represented by q, so that;

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So that,

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4 years ago

We can reasonably model a 90-W incandescent lightbulb as a sphere 7.0cm in diameter. Typically, only about 5% of the energy goes

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c = Speed of light = $3\times 10^8\ m/s$

The intensity is given by

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The visible light intensity at the surface of the bulb is 292.3254055 W/m²

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$u=\frac{1}{2}\epsilon_0E^2$

Energy density is also given by

$\frac{I}{c}=\frac{1}{2}\epsilon_0E^2\\\Rightarrow E=\sqrt{\frac{2I}{c\epsilon_0}}\\\Rightarrow E=\sqrt{\frac{2\times 292.3254055}{3\times 10^8\times 8.85\times 10^{-12}}}\\\Rightarrow E=469.26267\ V/m$

The amplitude of the electric field at this surface is 469.26267 V/m

Amplitude of a magnetic field is given by

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<h3>Answer: The acceleration doubles</h3>

===========================================================

Explanation:

Consider a mass of 10 kg, so m = 10

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F = ma

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20/10 = a

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The acceleration is 2 m/s^2. Every second, the velocity increases by 10 m/s.

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Now let's double the net force on the object

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In summary, if you double the net force applied to the object, then the acceleration doubles as well.

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