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Free_Kalibri [48]

2 years ago

7

What is the effect of cigarette smoke in your family

1 answer:

KIM [24]2 years ago

5 0

The family gets second hand smoke?

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A bat can detect small objects, such as an insect, whose size is approximately equal to one wavelength of the sound the bat make

blsea [12.9K]

Answer:

6.136 mm

Explanation:

given,

frequency emitted by the bat = 5.59 x 10⁴ Hz

speed of sound = 343 m/s

smallest insect bat can hear will be equal to the wavelength of the sound the bat make.

$\lambda = \dfrac{v}{f}$

$\lambda = \dfrac{343}{5.59\times 10^4}$

$\lambda = 6.136\times 10^{-3}}$

λ = 6.136 mm

so, the smallest size of insect that bat can hear is equal to 6.136 mm

8 0

2 years ago

Read 2 more answers

The electric force between two charged balloons is 0.12 newtons. If the distance between the two balloons is halved, what will b

marshall27 [118]

Answer:

The answer is D, I just took the test

Explanation:

5 0

2 years ago

A piston-cylinder device contains Helium gas initially at 150 kPa, 20 o C, and 0.5m 3 . The helium is now compressed in a polytr

Molodets [167]

Answer:

Explanation:

Given

$P_1=150 kPa$

$T_1=20^{\circ}C$

$V_1=0.5 m^3$

$T_2=140^{\circ}C$

$P_2=400 kPa$

R for Helium $R=2.076$

$c_v=3.115 kJ/kg-K$

mass of gas $m=\frac{P_1V_1}{RT_1}$

$m=\frac{150\times 0.5}{2.076\times 293}$

$m=0.123 kg$

Similarly $V_2$ can be found

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

$V_2=0.264 m^3$

Work done $W=\int_{V_1}^{V_2}PdV$

$W=\frac{P_2V_2-P_1V_1}{n-1}$

$W=\frac{mR(T_2_T_1)}{n-1}$

Since it is a polytropic Process

therefore $PV^n=c$

$P_1V_1^n=P_2V_2^n$

$(\frac{V_1}{V_2})^n=\frac{P_2}{P_1}$

$(\frac{0.5}{0.264})^n=\frac{400}{150}$

$n=\frac{\ln 2.66}{\ln 1.893}$

$n=1.533$

$W=\frac{0.123\times 2.076(140-20)}{1.533-1}$

$W=57.48 kJ$

From Energy balance

$E_{in}-E_{out}=\Delta E_{system}$

Neglecting kinetic and Potential Energy change

$Q_{in}+W_{in}=change\ in\ Internal\ Energy$

Change in Internal Energy $\Delta U=u_2-u_1$

$\Delta U=mc_v(T_2-T_1)$

$\Delta U=0.123\times 3.115(140-20)$

$\Delta U=45.977 kJ$

$Q_{in}+57.48=45.977$

$Q_{in}=-11.50 kJ$

i.e. Heat is being removed

3 0

2 years ago

calculate the load placed 10m from the fulcrum that can be balanced by an effort of 5 N applied at a distance of 4 m from the fu

Elena L [17]

Answer:

A = 2 m from fulcrum

Explanation:

Product of anti clockwise = Product of clockwise moment

5 × 4 = 10 × A

20 = 10 x A

A = 20 / 10

A = 2 m from fulcrum

8 0

2 years ago

A bob attached to a string of length L = 1.25 m, initially found at the equilibrium

malfutka [58]

The maximum displacement angle of the bob is 13⁰.

The given parameters;

<em>Length of the pendulum, L = 1.25 m</em>
<em>Initial velocity of the bob, v = 0.8 m/s</em>

The maximum displacement of the bob is calculated by applying the principle of conservation of energy;

$P.E = K.E\\\\mgh = \frac{1}{2} mv^2\\\\h = \frac{v^2}{2g} \\\\h = \frac{0.8^2}{2\times 10} \\\\h = 0.032 \ m$

The maximum displacement angle is calculated as follows;

$cos \theta = \frac{L-h}{L} \\\\cos \theta = \frac{1.25 - 0.032}{1.25} \\\\\cos \theta = \frac{1.218}{1.25} \\\\cos \theta = 0.9744\\\\\theta = cos^{-1}(0.9744)\\\\\theta = 13\ ^0$

Thus, the maximum displacement angle of the bob is 13⁰.

Learn more here:brainly.com/question/13981780

4 0

2 years ago