List two types of deposist made by water please help

S= 1/2(V f + V I )t solve for V f show me how

vaieri [72.5K]

S= 1/2(V f + V I )t solve for V f
2s/t=Vf+Vi
2s/t-Vi=Vf

8 0

4 years ago

In the water cycle matter moves towards the pull of gravity during ??

Bumek [7]

infiltration and evaporation

7 0

4 years ago

Poorly treated municipal wastewater is discharged to a stream. The river flow rate upstream of the discharge point is Qu/s = 8.7

Musya8 [376]

Answer

given,

Q u = 8.7 m³/s

Q d= 0.9 m³/s

BOD concentration = 50 mg/L

a) BOD concentration at the down stream

$C_{down}=\dfrac{0.9\times 50}{8.7+0.9}$

$C_{down}=\dfrac{0.9\times 50}{9.6}$

= 4.69 mg/L

b) discharge = 9.6 m³/s

cross sectional area = 10 m²

velocity steam = $\dfrac{8.7+0.9}{10}$

= 0.96 m/s

time taken to move 50 km down stream = $\dfrac{50 \times 1000}{0.96}$

= 52083.3 s

= $\dfrac{52083.3}{3600\times 24}$

= 0.6 days

now,

$C_t=C_0e^{-kt}$

$C_t=4.6875\ e^{-0.2\times 0.6}$

$C_t = 4.16 mg/L$

5 0

4 years ago

Monochromatic light falling on two very narrow slits 0.048mm apart. Successive fringes on a screen 5.00m away are 6.5cm apart ne

tino4ka555 [31]

Answer:

λ = 5.85 x 10⁻⁷ m = 585 nm

f = 5.13 x 10¹⁴ Hz

Explanation:

We will use Young's Double Slit Experiment's Formula here:

$Y = \frac{\lambda L}{d}\\\\\lambda = \frac{Yd}{L}$

where,

λ = wavelength = ?

Y = Fringe Spacing = 6.5 cm = 0.065 m

d = slit separation = 0.048 mm = 4.8 x 10⁻⁵ m

L = screen distance = 5 m

Therefore,

$\lambda = \frac{(0.065\ m)(4.8\ x\ 10^{-5}\ m)}{5\ m}$

λ = 5.85 x 10⁻⁷ m = 585 nm

Now, the frequency can be given as:

$f = \frac{c}{\lambda}$

where,

f = frequency = ?

c = speed of light = 3 x 10⁸ m/s

Therefore,

$f = \frac{3\ x\ 10^8\ m/s}{5.85\ x\ 10^{-7}\ m}\\\\$

f = 5.13 x 10¹⁴ Hz

5 0

3 years ago

A girl (mass M) standing on the edge of a frictionless merry-go-round (radius R, rotational inertia I) that is not moving. She t

vladimir1956 [14]

a) $\omega=\frac{-mvR}{I+MR^2}$

b) $v=\frac{-mvR^2}{I+MR^2}$

Explanation:

a)

Since there are no external torques acting on the system, the total angular momentum must remain constant.

At the beginning, the merry-go-round and the girl are at rest, so the initial angular momentum is zero:

$L_1=0$

Later, after the girl throws the rock, the angular momentum will be:

$L_2=(I_M+I_g)\omega +L_r$

where:

$I$ is the moment of inertia of the merry-go-round

$I_g=MR^2$ is the moment of inertia of the girl, where

M is the mass of the girl

R is the distance of the girl from the axis of rotation

$\omega$ is the angular speed of the merry-go-round and the girl

$L_r=mvR$ is the angular momentum of the rock, where

m is the mass of the rock

v is its velocity

Since the total angular momentum is conserved,

$L_1=L_2$

So we find:

$0=(I+I_g)\omega +mvR\\\omega=\frac{-mvR}{I+MR^2}$

And the negative sign indicates that the disk rotates in the direction opposite to the motion of the rock.

b)

The linear speed of a body in rotational motion is given by

$v=\omega r$

where

$\omega$ is the angular speed

r is the distance of the body from the axis of rotation

In this problem, for the girl, we have:

$\omega=\frac{-mvR}{I+MR^2}$ is the angular speed

$r=R$ is the distance of the girl from the axis of rotation

Therefore, her linear speed is:

$v=\omega R=\frac{-mvR^2}{I+MR^2}$

5 0

3 years ago