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kotegsom [21]
3 years ago
9

A toy car is given a quick push so that it rolls up an inclined ramp. After it is released, it rolls up, reaches its highest poi

nt and rolls back down again. Friction is so small it can be ignored. What net force acts on the car?(A) net force of zero(B) Net constant force up the ramp(C) net increasing force up the ramp(D) net increasing force down the ramp(E) net constant force down the ramp(F) net decreasing force down the ramp(G) net decreasing force up the ramp
Physics
1 answer:
kicyunya [14]3 years ago
5 0

Answer: Option (E)

Explanation:

There are only two forces acting on the car: The force of gravity, which point down, and the normal force generated by the interaction of the car and the ramp, which points perpendicular to the ramp.

N=mgcos(\alpha ),

m:mass of the car ; \alpha:angle of the ramp.

g: gravity's acceleration.

N is constante since all parameters are constants.

So, since the force of gravity is also constant, the net force acting on the car is constant. Since velocity decreases over time and the car stars moving down, the force is pointing down the ramp.

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An original funding source.

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Answer:

(a) The ratio of the pressure amplitude of the waves is 43.21

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Explanation:

Given;

density of gas, \rho _g = 2.27 kg/m³

density of liquid, \rho _l = 972 kg/m³

speed of sound in gas, C_g = 376 m/s

speed of sound in liquid, C_l = 1640 m/s

The of the sound wave is given by;

I = \frac{P_o^2}{2 \rho C} \\\\P_o^2 = 2 \rho C I\\\\p_o = \sqrt{2 \rho CI}

Where;

P_o is the pressure amplitude

P_o_g= \sqrt{2 \rho _g C_gI} -------(1)\\\\P_o_l= \sqrt{2 \rho _l C_lI}---------(2)\\\\\frac{P_o_l}{P_o_g} = \frac{\sqrt{2 \rho _l C_lI}}{\sqrt{2 \rho _g C_gI}} \\\\\frac{P_o_l}{P_o_g} = \sqrt{\frac{2 \rho _l C_lI}{2 \rho _g C_gI} }\\\\ \frac{P_o_l}{P_o_g} = \sqrt{\frac{ \rho _l C_l}{ \rho _g C_g} }\\\\ \frac{P_o_l}{P_o_g} = \sqrt{\frac{ (972)( 1640)}{ (2.27)( 376)} }\\\\\frac{P_o_l}{P_o_g} = 43.21

(b) when the pressure amplitudes are equal, the ratio of the intensities is given as;

I = \frac{P_o^2}{2 \rho C}\\\\I_g = \frac{P_o^2}{2 \rho _g C_g}-------(1)\\\\I_l = \frac{P_o^2}{2 \rho _l C_l}-------(2)\\\\\frac{I_l}{I_g} = (\frac{P_o^2}{2 \rho _l C_l})*(\frac{2\rho_gC_g}{P_o^2} )\\\\\frac{I_l}{I_g} = \frac{\rho _gC_g}{\rho_lC_l} \\\\\frac{I_l}{I_g} = \frac{(2.27)(376)}{(972)(1640)}\\\\ \frac{I_l}{I_g} = 0.000535

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