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3 years ago

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A car with a mass of 725 kg is moving at 32 m/s toward the east. What is the momentum of the car? A second car with a mass of 2,

175 kg has the same momentum. What is its velocity? Use details to support your answer.

1 answer:

Elden [556K]3 years ago

7 0

Answer:

A compact car, with mass 725 kg, ... at 115 km/h toward the east. ... b. A second car, with a mass of 2175 kg, has the same momentum. What is its ... Glisens : m = 2175 kg;. D 21,32 xrolka. anknown. r = ? 110.6 mis east

Explanation:

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Answer: f=150cm in water and f=60cm in air.

Explanation: Focal length is a measurement of how strong light is converged or diverged by a system. To find the variable, it can be used the formula:

$\frac{1}{f}$ = (nglass - ni)( $\frac{1}{R1}$ - $\frac{1}{R2}$ ).

nglass is the index of refraction of the glass;

ni is the index of refraction of the medium you want, water in this case;

R1 is the curvature through which light enters the lens;

R2 is the curvature of the surface which it exits the lens;

Substituting and calculating for water (nwater = 1.3):

$\frac{1}{f}$ = (1.5 - 1.3)( $\frac{1}{10}$ - $\frac{1}{15}$ )

$\frac{1}{f}$ = 0.2( $\frac{1}{30}$ )

f = $\frac{30}{0.2}$ = 150

For air (nair = 1):

$\frac{1}{f}$ = (1.5 - 1)( $\frac{1}{10}$ - $\frac{1}{15}$ )

f = $\frac{30}{0.5}$ = 60

In water, the focal length of the lens is f = 150cm.

In air, f = 60cm.

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A person jogs eight complete laps around a quarter-mile track in a total time of 12.5 min. Calculate (a) the average speed and (

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$\large\displaystyle\text{$\begin{gathered}\sf 1\ mile = 1609.34 \ m \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf 1/4 \ mile = 402.33 \ m \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf 12.5 \not{min}*\frac{60 \ s}{1\not{min}}=750 \ s \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf \bf{A) \ Calculate \ the \ average \ speed: } \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf 402.33 \ m*8 \ laps = 3218.64 \ m \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf d=3218.64 \ m \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf t=750 \ s \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf V=\frac{d}{t} \ \ \ \ \ \ V= \frac{3218.64 \ m }{750 \ s} \end{gathered}$}\\\\\\\large\displaystyle\text{$\begin{gathered}\sf V=4.29 \ m/s \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf \bf{B) \ Calculate \ the \ average \ speed \ in \ m/s} \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf V=402.33 \ m \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf t=750 \ s \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf V=\frac{D}{T} \ \ \ \ \ V=\frac{402.33 \ m}{750 \ s} \end{gathered}$}\\\\\\\large\displaystyle\text{$\begin{gathered}\sf V= 0.53 \ m/s \end{gathered}$}$

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Which of the following is the unit of momentum<br><br>a) ms-1 b) Kg ms-1<br><br>c) Ns d) Kgms-2

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Answer is b) I think

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If the airman had a mass of 80 kg, find the magnitude of the air drag acting on him when he reached terminal velocity of 54 m/s

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The magnitude of the air drag is 784 N

Explanation:

An object falling down reaches the terminal velocity when the magnitude of the air drag acting on it becomes equal to the weight of the object. Mathematically, this condition can be written as:

$F_D = mg$

where

$F_D$ is the magnitude of the air drag

m is the mass of the object

g is the acceleration of gravity

In this problem, we have

m = 80 kg is the mass of the airman

$g=9.8 m/s^2$ is the acceleration of gravity

Substituting into the formula, we find:

$F_D = (80)(9.8)=784 N$

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