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nlexa [21]
3 years ago
7

1. You are driving south on the 405. You are in traffic so your speed is 35 miles per hour,

Physics
1 answer:
Step2247 [10]3 years ago
7 0

Answer:

1. 15.64 m/s

2. 2,354,400 seconds

Explanation:

1. 1mile is equivalent to 1609metres and 1hour is equivalent to 3600sec.

35mph = (35×1609)/3600

= 56315/3600

= 15.64 m/s

2. Note that 1 day ===> 24hrs

60min ==> 1hr

60sec ===> 1min

Thus 27.25 days = 27.25×24×60×60

= 2,354,400 seconds

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What is a non-example of Newton's 3rd law of motion?
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Answer:

Unbalanced forces are not examples of Newton's third law because not all opposite reactions are unbalanced forces.

Explanation:

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The electron beam inside a television picture tube is 0.40 {\rm mm} in diameter and carries a current of 50 {\rm \mu A}. This el
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A.3.13x10^14 electrons

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A bullet starting from rest accelerates uniformly at a rate of 1,250 meters per square second. What is the bullet's speed after
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125,000

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8 0
3 years ago
(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
A car starts from rest, speeds up with constant acceleration, and travels 400 meters in 10 seconds. What was the acceleration of
PIT_PIT [208]

Answer:

Acceleration of the car will be a=8m/sec^2

Explanation:

We have given that car starts from rest so initial velocity of the car u = 0 m/sec

And car traveled 400 m in 10 sec

So distance traveled by car s = 400 m

Time taken to compete this distance t = 10 sec

We have to find the acceleration of the car

From second equation of motion we know that s=ut+\frac{1}{2}at^2

So 400=0\times 10+\frac{1}{2}\times a\times 10^2

a=8m/sec^2

So acceleration of the car will be a=8m/sec^2

7 0
3 years ago
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