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Crank
2 years ago
8

Suppose students experiment with the tube and a variety of darts. Some darts have higher masses than others but are the same aer

odynamic shape. Assume air resistance is negligible for the darts. Should the students use a dart with large mass or small mass to launch the dart the farthest distance possible? State which dart will travel the farthest distance and explain your reasoning without manipulating equations.
Physics
1 answer:
san4es73 [151]2 years ago
5 0

Answer:

The dart with the small mass will travel the farthest distance.

Explanation:

Acceleration is proportional to force times mass, and inertia is proportional to mass. Inertia is the reluctance of a moving body to stop, and a stationary body to start moving (inertia increses with mass). Assuming they both have the same aerodynamic design, and that they are both launched with the same force applied for the same time duration, the dart with less small mass will accelerate faster than the big mass dart. From this we can see that the small dart will have covered a longer distance before the effect of the force stops, when compared to the more massive dart.

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2 years ago
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A hoop is rolling (without slipping) on a horizontal surface so it has two types of kinetic energy: translational kinetic energy
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Answer:

\dfrac{T}{K}=1

Explanation:

The translational kinetic energy of the hoop is given by :

K=\dfrac{1}{2}Mv^2..................(1)

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The rotational kinetic energy of the hoop is given by :

T=\dfrac{1}{2}I\omega^2

Since, I=MR^2

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T=\dfrac{1}{2}\times MR^2\times (\dfrac{v}{R})^2..............(2)

From equation (1) and (2) :

\dfrac{T}{K}=1

Therefore, the ratio of the translational kinetic energy to the rotational kinetic energy is 1.

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2 years ago
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A small smooth object slides from rest down a smooth inclined plane inclined at 30degrees horizontal.What is the acceleration do
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2 years ago
A siphon pumps water from a large reservoir to a lower tank that is initially empty. The tank also has a rounded orifice 20 ft b
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Answer:

height of the water rise in tank is 10ft

Explanation:

Apply the bernoulli's equation between the reservoir surface (1) and siphon exit (2)

\frac{P_1}{pg} + \frac{V^2_1}{2g} + z_1= \frac{P_2}{pg} + \frac{V_2^2}{2g} +z_2

\frac{P_1}{pg} + \frac{V^2_1}{2g} +( z_1-z_2)= \frac{P_2}{pg} + \frac{V_2^2}{2g}-------(1)

substitute P_a_t_m for P_1, (P_a_t_m +pgh) for P_2

0ft/s for V₁, 20ft for (z₁ - z₂) and 32.2ft/s² for g in eqn (1)

\frac{P_1}{pg} + \frac{V^2_1}{2g} +( z_1-z_2)= \frac{P_2}{pg} + \frac{V_2^2}{2g}

\frac{P_1}{pg} + \frac{0^2_1}{2g} +( 20)= \frac{(P_a_t_m+pgh)}{pg} +\frac{V^2_2}{2\times32.2} \\\\V_2 = \sqrt{64.4(20-h)}

Applying bernoulli's equation between tank surface (3) and orifice exit (4)

\frac{P_3}{pg} + \frac{V^2_3}{2g} + z_3= \frac{P_4}{pg} + \frac{V_4^2}{2g} +z_4

substitute

P_a_t_m for P_3, P_a_t_m for P_4

0ft/s for V₃, h for z₃, 0ft for z₄, 32,2ft/s² for g

\frac{P_a_t_m}{pg} + \frac{0^2}{2g} +h=\frac{P_a_t_m}{pg} + \frac{V_4^2}{2\times32.2} +0\\\\V_4 =\sqrt{64.4h}

At equillibrium Fow rate at point 2 is equal to flow rate at point 4

Q₂ = Q₄

A₂V₂ = A₃V₃

The diameter of the orifice and the siphon are equal , hence there area should be the same

substitute A₂ for A₃

\sqrt{64.4(20-h)} for V₂

\sqrt{64.4h} for V₄

A₂V₂ = A₃V₃

A_2\sqrt{64.4(20-h)} = A_2\sqrt{64.4h}\\\\20-h=h\\\\h= 10ft

Therefore ,height of the water rise in tank is 10ft

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2 years ago
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