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3 years ago

CAN SOME PLZ HELP I REALY NEED THIS PLZ!!!!

Physics

1 answer:

mamaluj [8]3 years ago

7 0

Answer:

you won't find then

Explanation:

every test is new for each year

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A town requiring 2.0 m3/s of drinking water has two sources, a local well with 15 g/m3 nitrate (as N) and a distant reservoir wi

kati45 [8]

Explanation:

Drinking water requirement in town 2.0 m^3/s of water per second

nitrate in local well nitrate per 15 $\mathrm{m}^{3}$ of water

nitrate in distant reservoir $=5 \mathrm{~g} / \mathrm{m}^{3}$

Let the flow rate of well

flow rate of reservoir $=y m^{3} / s$

Drinking water requirement is $45 \mathrm{ppm}$ or $45 \mathrm{~g} / \mathrm{m}^{3}$

therefore, the total flow of drinking water

8 0

3 years ago

A 1kg ball is dropped (from rest) 100m onto a spring with spring constant 125N/m. How much does the spring compress?

Anastasy [175]

Answer:

3.95 m

Explanation:

m = 1 kg, h = 100 m, k = 125 N/m

Let the spring is compressed by y.

Use the conservation of energy

potential energy of the mass is equal to the energy stored in the spring

m x g x h = 1/2 x ky^2

1 x 9.8 x 100 = 0.5 x 125 x y^2

y^2 = 15.68

y = 3.95 m

7 0

4 years ago

In an equation f = l^2-d^2/4l the intercept is<br>

DanielleElmas [232]

Answer:

the intercept is the orgin (0,0)

5 0

3 years ago

2.5 gram sample of a radioactive element was formed in a 1960 explosion of an atomic bomb at Johnson Island in the Pacific Test

kow [346]

Answer:

0.15625 grams

Explanation:

Half life: It is related to the decay of radioactive material. The duration in which half of the material will be degraded/decayed. That means after half life 50% of the radioactive material will be left. Here the half life is 28 years.

Initial quantity of the sample: 2.5 grams.

After 28 years, the leftover quantity = 1.25 grams

After 56 years, the leftover quantity = 0.625 grams

After 84 Years, the leftover quantity = 0.3125 grams

After 112 years, the leftover quantity = 0.15625 grams

5 0

3 years ago

The existence of the dwarf planet Pluto was proposed based on irregularities in Neptune's orbit. Pluto was subsequently discover

givi [52]

Answer:

Acceleration due to gravity, $a=4.61\times 10^{-14}\ m/s^2$

Explanation:

It is given that,

Mass of Pluto, $m=1.4\times 10^{22}\ kg$

Distance between Neptune and Pluto, $r=4.5\times 10^{12}\ m$

The force of gravity is balanced by the gravitational force between Neptune and Pluto. It is given by :

$a=\dfrac{Gm}{r^2}$

$a=\dfrac{6.67\times 10^{-11}\times 1.4\times 10^{22}}{(4.5\times 10^{12})^2}$

$a=4.61\times 10^{-14}\ m/s^2$

So, the acceleration due to gravity at Neptune due to Pluto is $4.61\times 10^{-14}\ m/s^2$ . Hence, this is the required solution.

4 0

4 years ago