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Elden [556K]
3 years ago
7

A fountain sends a stream of water straight up into the air to a maximum height of 5.60 m. The effective area of the pipe feedin

g the fountain is 4.77 x 10-4 m2. Neglecting air resistance and any viscous effects, determine how many gallons per minute are being used by the fountain. (1 gal = 3.79 x 10-3 m3)
Physics
1 answer:
kifflom [539]3 years ago
5 0

Answer:It increases confidence because the more times you conduct the same experiment over and over should either prove your hypothesis right and wrong and eliminate any random occurrences that might affect your results.

Read more on Brainly.com - brainly.com/question/1492542#readmore

Explanation:

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A particle is moving along the x-axis. Its position as a function of time is given as x=bt-ct^2a) What must be the units of the
lisabon 2012 [21]

Answer:

We are given x= bt +ct²

So

A. bxt= m

Because m/s*s= m

So b= m/s and c= m/s²

B.

x= bt-ct²

So at x=0 t=0

x=0 t= 2

We have

bt = ct² so t = b/c at x= 0

So b-2ct= 0

B. To find velocity we use

dx / dt = b - 2 Ct

C. At rest wen V= 0

We have t= b/2c

D. To find acceleration we use

dv / dt = - 2C

3 0
3 years ago
A skier (m=59.0 kg) starts sliding down from the top of a ski jump with negligible friction and takes off horizontally. If h = 3
marissa [1.9K]

Answer:

35.20 m

Explanation:

By the law of conservation of energy we have,

mg(H-h)=\frac{1}{2}mv^2

g(H-h)=\frac{1}{2}v^2

\Rightarrow H=\frac{v^2}{2g}+h

where m= mass of the skier, h= 3.00 m

D= horizontal distance=13.9 m

H= maximum height attained

Also, the horizontal distance covered by the skier is

D=vt

=v\frac{2g}{h}

\Rightarrow v^2=\frac{gD^2}{2h}

thus, height H in terms of D  is given by

H=\frac{D^2}{2h}+h

H=\frac{13.9^2}{2\times3}+3

H=35.20 m

4 0
3 years ago
Calculate the pressure on a man’s foot when a woman who weighs 520 N steps on his foot with her heel which has an area of 0.001
Vinvika [58]

Answer:

520000  or 520000 pa

Force = 520N

Area of contact = 0.001

Pressure: 520000 or 520000

6 0
3 years ago
The half-life of cobalt-60 is 5.26 years. If 50 g are left after 15.8 years, how many
PtichkaEL [24]

Answer:

400 g

Explanation:

The computation of the number of grams in the original sample is shown below:

Given that

half-life = 5.26 years

total time of decay = 15.8 years

final amount = 50.0 g

Now based on the above information  

number of half-lives past is

=  15.8 ÷ 5.26

= 3 half-lives

Now

3 half-lives = 1 ÷ 8 remains = 50.0 g

So, the number of grams would be

= 50.0 g × 8

= 400 g

4 0
2 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
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