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3 years ago

A motorist travels a distance of 406 km during a 7 hr period. What was the average speed in (a) km/hr and (b) m/sec?

Physics

2 answers:

Jobisdone [24]3 years ago

8 0

Answer:

The average speed was 58 $\frac{km}{hr}$ or 16.11 $\frac{m}{sec}$

Explanation:

Speed is a physical quantity that expresses the relationship between the space traveled by an object, the time used for it and its direction. Its unit of measurement in the International System (S.I.) is the meter per second $\frac{m}{s}$ .

So:

$speed=\frac{distance}{time}$

In this case you know:

distante= 506 km
time=7 hr

Replacing:

$speed=\frac{406 km}{7 hr}$

Solving:

$speed=58 \frac{km}{hr}$

Taking into account that 1 km is 1000 m and 1 hour is 60 minutes, which in turn is 60 seconds, the unit conversion is performed:

$speed= 58\frac{km}{hr} * 1 000\frac{m}{km} *\frac{1}{60} \frac{hr}{minutes} *\frac{1}{60} \frac{minute}{sec}$

$speed=16.11 \frac{m}{sec}$

The average speed was 58 $\frac{km}{hr}$ or 16.11 $\frac{m}{sec}$

Sholpan [36]3 years ago

4 0

The motorist travels (a) 58 km/h and (b) ~16.1 m/sec

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As shown in the diagram, two forces act on an object. The forces have magnitudes F1 = 5.7 N and F2 = 1.9 N. What third force wil

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F1->

Explanation:

Observe the attached image. We must calculate the sum of all the forces in the direction x and in the direction y and equal the sum of the forces to 0.

For the address x we have:

$-F_3sin(b) + F_1 = 0$

For the address and we have:

$-F_3cos(b) + F_2 = 0$

The forces $F_1$ and $F_2$ are known

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Now we clear $F_3$ from the second equation and introduce it into the first equation.

$F_3 = \frac{F_2}{cos (b)}$

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$-\frac{F_2}{cos (b)}sin(b)+F_1 = 0\\\\F_1 = \frac{F_2}{cos (b)}sin(b)\\\\F_1 = F_2tan(b)\\\\tan(b) = \frac{F_1}{F_2}\\\\tan(b) = \frac{5.7}{1.9}\\\\tan^{-1}(\frac{5.7}{1.9}) = b\\\\b= 72\°\\\\m = b +90\\\\\m= 162\°$

Then we find the value of $F_3$

$F_3 = \frac{F_1}{sin(b)}\\\\F_3 =\frac{5.7}{sin(72\°)}\\\\F_3 = 6.01 N$

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3 years ago