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ipn [44]
2 years ago
13

A vw beetle goes from 0 to 60 mi/h with an acceleration of 2.35 m/s^2.

Physics
1 answer:
Vlad1618 [11]2 years ago
6 0
Part a.
u = 0, the initial velocity
v = 60 mi/h, the final velocity
a = 2.35 m/s², the acceleration.

Note that
1 m = 1609.34 m.
Therefore
v = (60 mi/h)*(1609.34 m/mi)*(1/3600 h/s) = 26.822 m/s
Use the formula
v = u + at
(26.822 m/s) = (2.35 m/s²)*(t s)
t = 26.822/2.35 = 11.4 s

Answer: 11.4 s

Part b.
We already determined that v = 60 mi/h = 26.822 m/s.
t = 0.6 s
Therefore
(26.822 m/s) = (a m/s²)*(0.6 s)
a = 26.822/0.6 = 44.7 m/s²

Answer:  44.7 m/s²
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Answer:

The answer is "effective stress at point B is 7382 ksi "

Explanation:

Calculating the value of Compressive Axial Stress:

\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}\\\\

       = [ (-50.9)^2 +3((63.7)^2 +(0.17)^2 )] \times \frac{1}{2}\\\\=[2590.81+ 3(4057.69)+0.0289]\times \frac{1}{2}\\\\=[2590.81+12,173.07+0.0289] \times \frac{1}{2}\\\\=14763.9089\times \frac{1}{2}\\\\ = 7381.95445 \ ksi\\\\ = 7382 \ ksi

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A hiker walks 11 km due north from camp and then turns and walks 11 km due east. What is the magnitude of the displacement (on a
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Answer:

16 km

Explanation:

Drawing a right triangle to model the problem helps. I started by drawing the lines of the triangle to model the hiker's journey- a vertical straight line for 11 km north and then a horizontal line connected to the top of it for 11 km east; I then drew the hypothenuse to connect the two lines.

The hypothenuse is what we have to solve for, so we will use the Pythagorean Theorem, a^2 + b^2 = c^2. Since both distances are 11 km both a and b in the equation are 11.

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