Need physics help<br> ASAP please!

A small balloon is released at a point 150 feet away from an observer, who is on level ground. If the balloon goes straight up a

Elza [17]

Answer:

$\dfrac{dz}{dt}=0.65\ ft/s$

Explanation:

Given that

x= 150 ft

$\dfrac{dy}{dt}= 7\ ft/s$

y= 14 ft

From the diagram

$z^2=x^2+y^2$

When ,x= 150 ft and y= 14 ft

$z^2=150^2+14^2$

$z=\sqrt{150^2+15^2}$

z=150.74 ft

$z^2=x^2+y^2$

By differentiating with respect to time t

$2z\dfrac{dz}{dt}= 2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}$

$z\dfrac{dz}{dt}= x\dfrac{dx}{dt}+y\dfrac{dy}{dt}$

Here x is constant that is why

$\dfrac{dx}{dt}=0$

$z\dfrac{dz}{dt}= y\dfrac{dy}{dt}$

Now by putting the values in the above equation we get

$150.74\times \dfrac{dz}{dt}=14\times 7$

$\dfrac{dz}{dt}=\dfrac{14\times 7}{150.74}\ ft/s$

$\dfrac{dz}{dt}=0.65\ ft/s$

Therefore the distance between balloon and observer increasing with 0.65 ft/s.

5 0

3 years ago

A) what is the light source for a plant growing in the shade?

Ulleksa [173]

a.) Plants that thrive in the shade are often able to hold on to sunlight for extensive periods of time; they're in a sense like the camels of the plaNt WoRld.

b.) Though artificial lights are not nearly as beneficial as the sun, one could invest in one of them plant growing light thingies, but sun-loving plants might be sad if u do this instead of letting them soak in ePic rays from the sun.

6 0

3 years ago

The pressure drop needed to force water through a horizontal 1-in diameter pipe if 0.60 psi for every 12-ft length of pipe. Dete

oksian1 [2.3K]

Answer:

The shear stress at a distance 0.3-in away from the pipe wall is 0.06012lb/ft²

The shear stress at a distance 0.5-in away from the pipe wall is 0

Explanation:

Given;

pressure drop per unit length of pipe = 0.6 psi/ft

length of the pipe = 12 feet

diameter of the pipe = 1 -in

Pressure drop per unit length in a circular pipe is given as;

$\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\$

make shear stress (τ) the subject of the formula

$\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\\tau = \frac{\delta P *r}{2L}$

Where;

τ is the shear stress on the pipe wall.

ΔP is the pressure drop

L is the length of the pipe

r is the distance from the pipe wall

Part (a) shear stress at a distance of 0.3-in away from the pipe wall

Radius of the pipe = 0.5 -in

r = 0.5 - 0.3 = 0.2-in = 0.0167 ft

ΔP = 0.6 psi/ft

ΔP, in lb/ft² = 0.6 x 144 = 86.4 lb/ft²

$\tau = \frac{\delta P *r}{2L} = \frac{86.4 *0.0167}{2*12} =0.06012 \ lb/ft^2$

Part (b) shear stress at a distance of 0.5-in away from the pipe wall

r = 0.5 - 0.5 = 0

$\tau = \frac{\delta P *r}{2L} = \frac{86.4 *0}{2*12} =0$

3 0

3 years ago

What is a likely consequence of continued human population growth

quester [9]

Answer:

Rapid human population growth has a number of consequences. Population grows fastest in the world's poorest countries. High fertility rates have historically had a strong relation with poverty, and high childhood mortality rates

Explanation:

i got it from goo.gle but i hope that helps:)

have a good day bro

7 0

3 years ago

Need help please and thank you

Scilla [17]

C
B
A
C
C

Hope this helps you!!!

8 0

3 years ago

Need physics help ASAP please!