Using scientific information from chemistry and biology to devise a plan to clean up a lake and make it healthy again describes

A large storage tank, open to the atmosphere at the top and filled with water, develops a small hole in its side at a point 15.6

VladimirAG [237]

To solve this problem it is necessary to take into account the kinematic equations of motion and the change that exists in the volume flow.

By definition the change in speed is given by

$v_f^2-v_i^2 = 2ax$

Where,

x= distance

$v_f =$ final velocity

$v_i =$ initial velocity

a = acceleration

On the other hand we know that the flow of a fluid is given by

$\dot{V} = Av$

Where,

A = Area

v = Velocity

PART A )

Applying this equation to the previously given values we have to

$v_f^2-v_i^2 = 2ax$

$v_f^2-0 = 2*(9.8)(15.6)$

$v_f^2=305.76$

$v_f = 17.48$

Therefore the velocity of the water leaving the hole is 17.48m/s

PART B )

In the case of the hole we take the area of a circle, therefore replacing in the flow equation we have to,

$\dot{V} = \pi r^2 v$

$r = \sqrt{\frac{\dot{V}}{\pi v}}$

$r = \sqrt{\frac{3*10^{-3}*\frac{1}{60}}{\pi (17.48)}$

$r = \sqrt{9.10*10^{-7}}$

$r = 0.54*10^{-4}$

The diameter is 2 times the radius, then is $1.91*10^{-3}$ m or 1.91mm

Note: The rate flow was converted from minutes to seconds.

8 0

3 years ago

If an ideal monatomic gas undergoes an adiabatic expansion, in which the volume increases by a factor of 4.0, by what factor doe

julsineya [31]

Answer:

e) 0.099

Explanation:

For an adiabatic expansion:

$P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$

where

P1 is the initial pressure

P2 is the final pressure

V1 is the initial volume

V2 is the final volume

$\gamma$ is the adiabatic index (which is $\frac{5}{3}$ for an ideal monoatomic gas

In this problem, we have

$V_2 = 4 V_1$ since the volume increases by a factor 4

We can re-write the equation to find by what factor the pressure changes:

$\frac{P_2}{P_1}=(\frac{V_1}{V_2})^{\gamma}=(\frac{V_1}{4V_1})^{5/3}=(\frac{1}{4})^{5/3}=0.099$

8 0

3 years ago

If a particle with a charge of +4.3 × 10−18 C is attracted to another particle by a force of 6.5 × 10−8 N, what is the magnitude

AveGali [126]

Answer: 1.5×10^10 N/C

Explanation:

E= F/q

Where E= magnitude of the electric field

F= force of attraction

q= charge of the given body

Given F= 6.5×10^-8 N

q= 4.3× 10^-18 C

Therefore, E = 6.5×10 ^-8/ 4.3×10^-18

E = 1.5×10^10 N/C

7 0

3 years ago

PLeAsE hElp What is the kinetic energy of 14 Kg traveling at a velocity of 3m/s east

harkovskaia [24]

Answer:

Please see the given attachment.

Explanation:

Stay safe, stay healthy and be blessed.

Thank you.

<h2>PLEASE MARK ME AS BRAINLEST.</h2>

4 0

3 years ago

1.A bullet is fired with a velocity of 50m/s at an angle tita to the

VladimirAG [237]

Answer:

b

Explanation:

see how far the bullet goes

7 0

4 years ago