Ask question

Ask question

All categories

Montano1993 [528]

3 years ago

7

The aircraft link is made from an a992 steel rod. determine the smallest diameter of the rod, to the nearest 1 /16 in., that wil

l support the load of 4 kip without buckling. the ends are pin connected.

1 answer:

Bas_tet [7]3 years ago

3 0

Answer:

hjhjhjhj

Explanation:

jhjhjhjhjh

You might be interested in

Careful measurements have been made of Olympic sprinters in the 100-meter dash. A quite realistic model is that the sprinter's v

mihalych1998 [28]

Answer:

a.

$\displaystyle a(0 )=8.133\ m/s^2$

$\displaystyle a(2)=2.05\ m/s^2$

$\displaystyle a(4)=0.52\ m/s^2$

b. $\displaystyle X(t)=11.81(t+1.45\ e^{-0.6887t})-17.15$

c. $t=9.9 \ sec$

Explanation:

Modeling With Functions

Careful measurements have produced a model of one sprinter's velocity at a given t, and it's is given by

$\displaystyle V(t)=a(1-e^{bt})$

For Carl Lewis's run at the 1987 World Championships, the values of a and b are

$\displaystyle a=11.81\ ,\ b=-0.6887$

Please note we changed the value of b to negative to make the model have sense. Thus, the equation for the velocity is

$\displaystyle V(t)=11.81(1-e^{-0.6887t})$

a. What was Lewis's acceleration at t = 0 s, 2.00 s, and 4.00 s?

To compute the accelerations, we must find the function for a as the derivative of v

$\displaystyle a(t)=\frac{dv}{dt}=11.81(0.6887\ e^{0.6887t})$

$\displaystyle a(t)=8.133547\ e^{-0.6887t}$

For t=0

$\displaystyle a(0)=8.133547\ e^o$

$\displaystyle a(0 )=8.133\ m/s^2$

For t=2

$\displaystyle a(2)=8.133547\ e^{-0.6887\times 2}$

$\displaystyle a(2)=2.05\ m/s^2$

$\displaystyle a(4)=8.133547\ e^{-0.6887\times 4}$

$\displaystyle a(4)=0.52\ m/s^2$

b. Find an expression for the distance traveled at time t.

The distance is the integral of the velocity, thus

$\displaystyle X(t)=\int v(t)dt \int 11.81(1-e^{-0.6887t})dt=11.81(t+\frac{e^{-0.6887t}}{0.6887})+C$

$\displaystyle X(t)=11.81(t+1.45201\ e^{-0.6887t})+C$

To find the value of C, we set X(0)=0, the sprinter starts from the origin of coordinates

$\displaystyle x(0)=0=>11.81\times1.45201+C=0$

Solving for C

$\displaystyle c=-17.1482\approx -17.15$

Now we complete the equation for the distance

$\displaystyle X(t)=11.81(t+1.45\ e^{-0.6887t})-17.15$

c. Find the time Lewis needed to sprint 100.0 m.

The equation for the distance cannot be solved by algebraic procedures, but we can use approximations until we find a close value.

We are required to find the time at which the distance is 100 m, thus

$\displaystyle X(t)=100=>11.81(t+1.45\ e^{-0.6887t})-17.15=100$

Rearranging

$\displaystyle t+1.45\ e^{-0.6887t}=9.92$

We define an auxiliary function f(t) to help us find the value of t.

$\displaystyle f(t)=t+1.45\ e^{-0.687t}-9.92$

Let's try for t=9 sec

$\displaystyle f(9)=9+1.45\ e^{-0.687\times 9}-9.92=-0.92$

Now with t=9.9 sec

$\displaystyle f(9.9)=9.9+1.45\ e^{-0.687\times 9.9}-9.92=-0.0184$

That was a real close guess. One more to be sure for t=10 sec

$\displaystyle f(10)=10+1.45\ e^{-0.687\times 10}-9.92=0.081$

The change of sign tells us we are close enough to the solution. We choose the time that produces a smaller magnitude for f(t).

$At t\approx 9.9\ sec, \text{ Lewis sprinted 100 m}$

7 0

3 years ago

A roller coaster car is going over the top of a 18-mm-radius circular rise. At the top of the hill, the passengers "feel light,"

Andre45 [30]

Answer:

0.29713 m/s

Explanation:

m = Mass of person

g = Acceleration due to gravity = 9.81 m/s²

v = Velocity

r = Radius = 18 mm

By balancing the forces in the system we have

$mg-N=\dfrac{mv^2}{r}\\\Rightarrow mg-\dfrac{mg}{2}=\dfrac{mv^2}{r}\\\Rightarrow v=\sqrt{r(g-\dfrac{g}{2})}\\\Rightarrow v=\sqrt{0.018\times (9.81-\dfrac{9.81}{2})}\\\Rightarrow v=0.29713\ m/s$

The velocity of the coaster is 0.29713 m/s

7 0

2 years ago

Air temperature in a desert can reach 58.0°C (about 136°F). What is the speed of sound (in m/s) in air at that temperature?

Veronika [31]

The approximate speed of sound in dry (0% humidity) air, in meters per second, at temperatures near 0 °C, can be calculated from

$c_{air} = (331.3+0.606 \upsilon)$

Here

$\upsilon =$ Temperature in Celsius

Replacing with our values we have that

$\upsilon=58\° C$

$c_{air} = (331.3+0.606*58)$

$c_{air} = 366.1m/s$

Therefore the speed of sound in air at that temperature is 366.1m/s

3 0

3 years ago

temperatura en la ciudad de la página madrugada y 2 grados bajo cero y a las 8 de la mañana sube a 6 grados y luego ya a las 10

vivado [14]

Answer:

kjkjnklhklsehfiodxlfngioduherl

Explanation:

8 0

2 years ago

Read 2 more answers

Where is the resistor located on the diagram?

Andrej [43]

The answer is B hope this helps

5 0

3 years ago

Read 2 more answers