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Makovka662 [10]
2 years ago
15

A car travels north at 30 m/s for one half hour. It then travels south at 40 m/s for 15 minutes. The total distance the car has

traveled and its displacement are: Group of answer choices 36 km; 36 km N. 90 km; 18 km N. 90 km; 36 km N. 36 km; 36 km S. 18 km; 18 km S.
Physics
1 answer:
kari74 [83]2 years ago
5 0

Answer:

xtotal = 90km

displacement = 18km N

Explanation:

To find the total distance traveled by the car, you first calculate the distance traveled by the car when it travels to north. You use the following formula:

x=vt    (1)

x: distance

v: speed of the car = 30 m/s

t: time = one half hour

In order to calculate the distance you convert the time from hours to seconds:

t=0.5\ h*\frac{3600s}{1\ h}=1800s

Then, you replace the values of t and v in the equation (1):

x=(30m/s)(1800s)=54000m     (2)

Next, you calculate the distance traveled by the car when it travels to south:

x'=v't'\\\\v'=40\frac{m}{s}\\\\t'=15\ min

You convert the time from minutes to seconds:

t'=15\ min*\frac{60s}{1min}=900s

x'=(40m/s)(900s)=36000m

Finally, you sum both distances x and x':

x_{total}=x+x'=54000m+36000m=90000m=90km

The total distance traveled by the car is 90km

The total displacement is the final distance of the car respect to the starting point of the motion. This is calculated by subtracting x' to x:

d=x-x'=54000m-36000m=18000m=18km

The total displacement of the car is 18km to the north from its starting point of motion.

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A force Fof 40000 lbf is applied to rod AC the negative Y-direction. The rod is 1000 inches tall. A Force P of 25 lbf is applied
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Explanation:

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\to \sigma y  =\frac{F}{A} = \frac{4 F}{( p d ^2 )} = \frac{(4 x ( - 40000 \ lbf))}{[ p \times (1 \ in)^2 ]} = - 50.9 \ ksi \\

Calculating Shear Transverse:

\to \frac{4v}{ 3 A} = \frac{4 (75 \ lbf + 25 \ lbf)}{ \frac{3 ( lni)^2}{4}}

        = \frac{4 (100 \ lbf)}{ \frac{3 ( lni)^2}{4}} \\\\ = \frac{400 \ lbf)}{ \frac{3 ( lni)^2}{4}} \\\\= 0.17 \ ksi

= R \times 200 \ in - P \times 100 \ in = 12500 \ lbf \times\  in

\to \sigma' =[ s y^2 +3( t \times y^2 + t yz^2 )] \times \frac{1}{2}\\\\

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