Trying to manage stress during which phase of the ABCs of Stress generally provides the most favorable results?

An 80 kg skydiver is falling at terminal velocity. What is the value of air resistance acting on his body? Consider, what are th

Mrac [35]

Answer:

800 N

Explanation:

During terminal velocity, air resistance=weight

5 0

3 years ago

Consider the electronic elements that are cooled by forced convection in Problem 6.31. The cooling system is designed and tested

Stella [2.4K]

Answer:

surface temperature of the chip located 120 mm Ts=42.5°C

surface temperature of the chip in Mexico Ts=46.9°C

Explanation:

from the energy balance equation we have to:

q=E=30W

from Newton´s law:

Ts=Tα+(q/(h*A)), where A=l^2

N=h/k=0.04*(Vl/V)^0.85*Pr^1/3

data given:

l=0.12 m

v=10 m/s

k=0.0269 W/(m*K)

Pr=0.703

Replacing:

h=0.04*(0.0269/0.12)*(10*0.12)/((16*69x10^-6))^0.85*(0.703^1/3) = 107 W/m^2*K

The surface temperature at sea level is equal to:

Ts=25+(30x10^-3/107*0.004^2)=42.5°C

h=0.04*(0.0269/0.12)*((10*0.12)/(21*81x10^-6))^0.85*(0.705^1/3)=85.32 W/(m*K)

the surface temperature at Mexico City is equal to:

Ts=25+(30x10^-3/85.32*0.004^2)=46.9°C

8 0

3 years ago

As important as it is to plan ahead,sometimes you

Natali5045456 [20]

Answer:

B. Cant stop things from going wrong.

Explanation:

To me it's the only reasonable answer...

5 0

3 years ago

Read 2 more answers

Three charges are arranged as shown in the picture above. Find the magnitude and direction of the electrostatic force on the 6 n

bagirrra123 [75]

Answer:

Bot Nm

Explanation:

jdjdmxjd mdjdcj jdsdj jedidj jddj

6 0

3 years ago

If an object is thrown in an upward direction from the top of a building 160 ft. High at an initial speed of 21.82 mi/h what is

viktelen [127]

To solve this problem we are going to use tow kinematic equations for falling objects.
1. Kinematic equation for final velocity: $V_{f}=V_{i}+gt$
where
$V_{f}$ is the final velocity
$V_{i}$ is the initial velocity
$g$ is the acceleration due to gravity $32 \frac{ft}{s^2}$
$t$ is the time
2. Kinematic equation for distance: $d=V_{i}t+ \frac{1}{2} gt^2$
where
$d$ is the distance
$V_{i}$ is the initial velocity
$V_{f}$ is the final velocity
$g$ is the acceleration due to gravity $32 \frac{ft}{s^2}$
$t$ is the time

First, we are going to convert 21.82 mi/h to ft/s:
$21.82 \frac{mi}{h} =31.21 \frac{ft}{s}$

Next, we are going to use the first equation to find how long it takes for the rock to reach its maximum height.
We know for our problem that the object is thrown in upward direction, so its velocity at its maximum height (before falling again) will be zero; therefore: $V_{f}=0$ . We also know that it initial speed is 31.21 ft/s, so $V_{i}=31.21$ . Lets replace those values in our formula to find $t$ :
$V_{f}=V_{i}+gt$
$0=31.21+(-32)t$
$-32t=-31.21$
$t= \frac{-31.21}{-32}$
$t=0.98$ seconds

Next, we are going to use that time in our second kinematic equation to find the distance the object reach at its maximum height:
$d=V_{i}t+ \frac{1}{2} gt^2$
$d=31.21(0.98)+ \frac{1}{2} (-32)(0.98)^2$
$d=15.22$ ft

Now we can add the height of the building and the maximum height of the object:
$d=160+15.22=175.22$ ft

Next, we are going to use that height (distance) in our second kinematic equation one more time to fin how long it takes for the object to fall from its maximum height to the ground:
$d=V_{i}t+ \frac{1}{2} gt^2$
$175.22=31.21t+ \frac{1}{2} (32)t^2$
$16t^2+31.21t-175.22=0$
$t=2.47$ or $t=-4.43$
Since time cannot be negative, $t=2.47$ is the time it takes the object to fall to the ground.

Finally, we can use that time in our first kinematic equation to find the final speed of the object when it hits the ground:
$V_{f}=V_{i}+gt$
$V_{f}=31.21+(32)(2.47)$
$V_{f}=110.25$ ft/s

We can conclude that the speed of the object when it hits the ground is 110.25 ft/s

5 0

3 years ago