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3 years ago

What is sludge dumping?HELP ASAP PLEASE

Physics

1 answer:

yulyashka [42]3 years ago

5 0

Answer:

What is sludge dumping? Sludge is the solid waste in raw sewage. Sludge dumping is discharging that waste into the ocean.

Explanation:

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Given that the concentration of bovine carbonic anhydrase is 3.3 pmol ⋅ L − 1 and R max ( V max ) = 222 μmol ⋅ L − 1 ⋅ s − 1 , d

LuckyWell [14K]

Answer:

The turnover number of the enzyme molecule bovine carbonic anhydrase = 67,272,727.27 s^–1.

Explanation:

Given:

The concentration of bovine carbonic anhydrase = total enzyme concentration = Et = 3.3 pmol⋅L^–1 = 3.3 × 10^–12 mol.L^–1

The maximum rate of reaction = Rmax (Vmax) = 222 μmol⋅L^–1⋅s^–1 = 222 × 10^–6 mol.L^–1⋅s^–1

The formula for the turnover number of an enzyme (kcat, or catalytic rate constant) = Rmax ÷ Et = 222 × 10^–6 mol.L^–1⋅s^–1 ÷ 3.3 × 10^–12 mol.L^–1 = 67,272,727.27 s^–1

Therefore, the turnover number of the enzyme molecule bovine carbonic anhydrase = 67,272,727.27 s^–1

3 0

3 years ago

How much work must be done on a 20-kg go-cart to increase its speed from 5 m/s to 10 m/s?

allochka39001 [22]

b 250 j

hope this helps

6 0

2 years ago

Read 2 more answers

A jogger runs 4.0 km [W] in 0.50 h, then turns and runs 1.0 km [E] in 0.20 h, then 1.5 km [N] in 0.25 h, then 3.0 km [E] in 0.75

GrogVix [38]

This is a sneaky trick question, to help you discover whether you know
one of the differences between velocity and speed.
=======================================
If you make a list of the distances and directions, and ignore the times,
you find these:

4 - west, (3 + 1) - east . . . . . zero in the east/west direction

1.5 - north, 1.5 - south . . . . . zero in the north/south direction

This jogger went out, had a nice jog around the neighborhood,and ended up exactly where he started.

Average velocity = (distance between start point and end point) / (time)

IF the question asked for average SPEED, then you would need the total distance, and divide it by the total time. But it asks for VELOCITY, and that only involves the straight distance between the start point and the end point, regardless of the route taken in between.

The jogger ended up exactly where he started. The distance between start and end points was zero. Average velocity is (zero) / (time) . And that fraction is going to be Zero, no matter how long or how short the trip was, and no matter how much time it took.

3 0

2 years ago

Read 2 more answers

Jupiter's moon Io has active volcanoes (in fact, it is the most volcanically active body in the solar system) that eject materia

kramer

Answer:

The height reached by the material on Earth is 91 km.

Explanation:

Given that,

Mass $M_{Io}=8.93\times10^{22}\ kg$

Radius = 1821 km

Height $h_{Io}=500\ km$

Suppose we need to find that how high would this material go on earth if it were ejected with the same speed as on Io?

We need to calculate the acceleration due to gravity on Io

Using formula of gravity

$g =\dfrac{GM_{Io}}{(R_{Io})^2}$

Put the value into the formula

$g=\dfrac{6.67\times10^{-11}\times8.93\times10^{22}}{(1821\times10^{3})^2}$

$g=1.79\ m/s^2$

Let v be the speed at which the material is ejected.

We need to calculate the height

Using the formula of height

$H=\dfrac{v^2}{2g}$

Using ratio of height of earth and height of Io

$\dfrac{H_{e}}{H_{Io}}=\dfrac{\dfrac{v^2}{2g_{e}}}{\dfrac{v^2}{2g_{Io}}}$

$\dfrac{H_{e}}{H_{Io}}=\dfrac{g_{Io}}{g_{e}}$

Put the value into the formula

$\dfrac{H_{e}}{H_{Io}}=\dfrac{1.79}{9.8}$

$\dfrac{H_{e}}{H_{Io}}=0.182$

$H_{e}=0.182\times H_{Io}$

$H_{e}=0.182\times500$

$H_{e}=91\ km$

Hence, The height reached by the material on Earth is 91 km.

3 0

2 years ago

ASAP please!! The force vectors on an aircraft are as shown. Find the net (resultant force).

ioda

Answer:

The magnitude of the net force is 5430N

Explanation:

I suggest to define the axes as aligned to the axis of the plane. This will require you to decompose only one vector, namely the Weight. We need two components of the W force: one in horizontal direction of the plane, the other perpendicular to it. Through a simple triangle argument you will se that the plane-horizontal component of W is

$W_D=3600 N\cdot\sin 27^\circ=1634N$

acting in the direction of the Drag, and the plane-perpendicular component is:

$W_L=-3600N\cdot\cos 27^\circ=-3208N$

with negative sign since it counteracts the Lift.

So the components of the netforce F are:

$F_h=T-D-W_D=(8000-1000-1634)N=5366N\\F_v=L+W_L=(4100-3208)N=829N$

The magnitude of the net force is:

$|F|=\sqrt{5366^2+829^2}N = 5430N$

6 0

3 years ago