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2 years ago

There are 4.2 cups in a liter. of a bottle holds 2.0 cups what does it hold in liters

Physics

2 answers:

xenn [34]2 years ago

6 0

(1 liter / 4.2 cups) x (2.0 cups)

= (1 · 2.0 / 4.2) (liter · cups / cups)

= 0.476 liter

SOVA2 [1]2 years ago

3 0

2.0 I'm guessing is the answer

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Two sections, A and B, are 0.5 km apart along a 0.05 m diameter rough concrete pipe. A is 115 m higher than B, the water tempera

KonstantinChe [14]

Answer:

Q = 178.41 m^3 / s

Explanation:

Given:

Length of the pipe L = 0.5 km
Diameter D = 0.05 m
Pressure head @ A (P_a / γ )= 21.7 m
Pressure head @ B (P_b / γ )= 76.1 m
Elevation head Z_a = 115 m
Elevation head Z_b = 0 m
Minor Losses = 0 m
Major Losses = f*L*V^2 / 2*D*g = (500/2*0.05*9.81) *f*V^2 = 509.684*f*V^2
Velocity at cross section A and B: V_a = V_b m/s
Roughness e = 2.5 mm
Dynamic viscosity of water u = 8.9*10^-4 Pa-s
Density of water p = 997 kg/m^3

Find:

Flow Rate Q = pi*V*D^2/4 m^3/s ??

Solution:

We will use the Head Balance as derived from Energy Balance:

(P_a / γ ) - (P_b / γ ) + (V_a^2 - V_b^2) / 2*g + (Z_a - Z_b) = Major Losses

21.7 - 76.1 + 0 + 115-0 = 509.684*f*V^2

f*V^2 = 0.18897199

To find correction factor f which is a function of e / D = 0.05, and Reynold's number which is unknown. In such cases we will guess a value of f and perform iterations as follows:

Guess: f_o = 0.072 (Moody's Chart @ e / D = 0.05 and most turbulent function).

V_o = sqrt(0.18897199 / 0.072) = 1.28504 m/s

Re_o = p*V_o*D / u = 997*1.28504*0.05 / 8.9*10^-4 = 71976.6

1st iteration

f_1 = g (Re_o , e/d) = 0.0718702 (Moody's Chart)

V_1 = sqrt(0.18897199 / 0.0718702) = 1.621528 m/s

Re_1 = p*V_1*D / u = 997*1.621528*0.05 / 8.9*10^-4 = 90823.81698

2nd iteration

f_2 = g (Re_1 , e/d) = 0.0718041 (Moody's Chart)

V_2 = sqrt(0.18897199 / 0.0718041) = 1.662273585 m/s

Re_2 = p*V_2*D / u = 997*1.662273585*0.05 / 8.9*10^-4 = 90865.54854

3rd iteration

f_3 = g (Re_2 , e/d) = 0.0718040 (Moody's Chart)

V_3 = sqrt(0.18897199 / 0.0718040) = 1.622274714 m/s

Re_3 = p*V_3*D / u = 997*1.622274714*0.05 / 8.9*10^-4 = 90865.61182

We can observe the convergence of V to 1.6222 m /s. Hence, the required velocity will be used to calculate the Flow rate Q:

Q = pi*V*D^2/4 = pi*1.6222*0.05^2 / 4

Q = 178.41 m^3 / s

3 0

3 years ago

A pencil rolls horizontally of a 1 meter high desk and lands .25 meters from the base of the desk. How fast was the pencil rolli

vovikov84 [41]

Answer: 0.55 m/s

Explanation:

This situation is related to projectile motion (also called parabolic motion), where the main equations are as follows:

$x=V_{o} cos\theta t$ (1)

$y=y_{o}+Vo sin \theta t + \frac{g}{2}t^{2}$ (2)

Where:

$x=0.25 m$ is the horizontal displacement of the pencil

$V_{o}$ is the pencil's initial velocity

$\theta=0\°$ since we are told the pencil rolls <u>horizontally</u> before falling

$t$ is the time since the pencil falls until it hits the ground

$y_{o}=1 m$ is the initial height of the pencil

$y=0$ is the final height of the pencil (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity, always acting vertically downwards

Begining with (1):

$x=V_{o} cos(0\°) t$ (3)

$x=V_{o}t$ (4)

Finding $t$ from (2):

$0=1 m+ \frac{-9.8m/s^{2}}{2}t^{2}$ (5)

$t=\sqrt{\frac{-2y_{o}}{g}}$ (6)

Substituting (6) in (4):

$x=V_{o}\sqrt{\frac{-2y_{o}}{g}}$ (7)

Isolating $V_{o}$ :

$V_{o}=\frac{x}{\sqrt{\frac{-2y_{o}}{g}}}$ (8)

$V_{o}=\frac{0.25 m}{\sqrt{\frac{-2(1 m)}{-9.8m/s^{2}}}}$ (9)

Finally:

$V_{o}=0.55 m/s$

4 0

3 years ago

Hey! Can someone help with this question? Thx :)

Schach [20]

<span>A. Chemical energy to chemical energy</span>

4 0

2 years ago

Pls help A car starts from rest and gains a velocity of 20m/s in 10 seconds calculate its acceleration and the distance covered

Soloha48 [4]

Answer:

$\boxed{\sf Acceleration \ (a) = 2 \ m/s^{2}}$

$\boxed{\sf Distance \ covered \ (s) = 100 \ m}$

Given:

Initial velocity (u) = 0 m/s

Final velocity (v) = 20 m/s

Time taken (t) = 10 sec

To Find:

(i) Acceleration (a)

(ii) Distance covered (s)

Explanation:

$\sf (i) \ From \ 1^{st} \ equation \ of \ motion:$

$\sf \implies v = u + at$

$\sf \implies 20 = 0 + a(10)$

$\sf \implies 10a = 20$

$\sf \implies \frac{10a}{10} = \frac{20}{10}$

$\sf \implies a = 2 \: m/ {s}^{2}$

$\sf (ii) \ From \ 2^{nd} \ equation \ of \ motion:$

$\sf \implies s = ut + \frac{1}{2} a {t}^{2}$

$\sf \implies s = (0)(10) + \frac{1}{2} \times 2 \times {(10)}^{2}$

$\sf \implies s = \frac{1}{ \cancel{2}} \times \cancel{2} \times {(10)}^{2}$

$\sf \implies s = {10}^{2}$

$\sf \implies s = 100 \: m$

6 0

3 years ago

What happens when the thermal energy of a substance increases?

Sonja [21]

When thermal energy of a substance increases, it's entropy(randomness) & Kinetic energy increases.

For more appropriate answer, you should put the options 'cause there could be more than one answer for this question.

4 0

2 years ago