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2 years ago

A rubber ball thrown at a speed of 5 m/s hit a flat wall and

Physics

1 answer:

velikii [3]2 years ago

5 0

Answer:

See the explanion below

Explanation:

Linear momentum is defined as the product of mass by velocity and can be calculated by the following equation.

$P =m*v$

where:

P = momentum [kg*m/s]

m = mass [kg]

v = velocity [m/s]

When the ball hits the wall it receives a force for a very short time, this time and force value is known as momentum and can be calculated by the following expression.

$-(m_{1}*v_{1}) + (F*t)=(m_{1}*v_{2})$

where:

m₁ = mass of the ball [kg]

v₁ = 5 [m/s]

v₂ = 5 [m/s]

v₁ = v₂

$F*t = Imp$

$Imp = (m_{1}*v_{1})+ (m_{1}*v_{2})\\Imp = 2*m*v_{1}$

It can be said that the impulse is equal to 2 times the speed of throwing by the mass.

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(a) If two sound waves, one in a gas medium and one in a liquid medium, are equal in intensity, what is the ratio of the pressur

GarryVolchara [31]

Answer:

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

(b) The ratio of the intensities of the waves is 0.000535

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;

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

3 0

3 years ago

Two parallel-plate capacitors have the same plate area. Capacitor 1 has a plate separation twice that of capacitor 2, and the qu

Luba_88 [7]

Answer:

$V_1=8 V_2$

Explanation:

Given that:

Area of the plate of capacitor 1= Area of the plate of capacitor 2=A

separation distance of capacitor 2, $d_2=d$

separation distance of capacitor 1, $d_1=2d$

quantity of charge on capacitor 2, $Q_2=Q$

quantity of charge on capacitor 1, $Q_1=4Q$

We know that the Capacitance of a parallel plate capacitor is directly proportional to the area and inversely proportional to the distance of separation.

Mathematically given as:

$C=\frac{k.\epsilon_0.A}{d}$ .....................................(1)

where:

k = relative permittivity of the dielectric material between the plates= 1 for air

$\epsilon_0 = 8.85\times 10^{-12}\,F.m^{-1}$

From eq. (1)

For capacitor 2:

$C_2=\frac{k.\epsilon_0.A}{d}$

For capacitor 1:

$C_1=\frac{k.\epsilon_0.A}{2d}$

$C_1=\frac{1}{2} [ \frac{k.\epsilon_0.A}{d}]$

We know, potential differences across a capacitor is given by:

$V=\frac{Q}{C}$ ..........................................(2)

where, Q = charge on the capacitor plates.

for capacitor 2:

$V_2=\frac{Q}{\frac{k.\epsilon_0.A}{d}}$

$V_2=\frac{Q.d}{k.\epsilon_0.A}$

& for capacitor 1:

$V_1=\frac{4Q}{\frac{k.\epsilon_0.A}{2d}}$

$V_1=\frac{4Q\times 2d}{k.\epsilon_0.A}$

$V_1=8\times [\frac{Q.d}{k.\epsilon_0.A}]$

$V_1=8 V_2$

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