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MAVERICK [17]
2 years ago
14

How was Mendeleev first periodic table like the periodic table today

Physics
1 answer:
Mila [183]2 years ago
5 0

Answer:

He began organizing the known elements according to their atomic weights and chemical properties.

Explanation:

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In a 100 mm diameter horizontal pipe, a venturimeter of 0.5 contraction ratio has been fitted. The head of water on the meter wh
horrorfan [7]

Answer:

the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

Explanation:

Given:

Diameter of the pipe = 100mm = 0.1m

Contraction ratio = 0.5

thus, diameter at the throat of venturimeter = 0.5×0.1m = 0.05m

The formula for discharge through a venturimeter is given as:

Q=C_d\frac{A_1A_2}{\sqrt{A_1^2-A_2^2}}\sqrt{2gh}

Where,

C_d is the coefficient of discharge = 0.97 (given)

A₁ = Area of the pipe

A₁ = \frac{\pi}{4}0.1^2 = 7.85\times 10^{-3}m^2

A₂ = Area at the throat

A₂ = \frac{\pi}{4}0.05^2 = 1.96\times 10^{-3}m^2

g = acceleration due to gravity = 9.8m/s²

Now,

The gauge pressure at throat = Absolute pressure - The atmospheric pressure

⇒The gauge pressure at throat = 2 - 10.3 = -8.3 m (Atmosphric pressure = 10.3 m of water)

Thus, the pressure difference at the throat and the pipe = 3- (-8.3) = 11.3m

Substituting the values in the discharge formula we get

Q=0.97\frac{7.85\times 10^{-3}\times 1.96\times 10^{-3}}{\sqrt{7.85\times 10^{-3}^2-1.96\times 10^{-3}^2}}\sqrt{2\times 9.8\times 11.3}

or

Q=\frac{0.97\times15.42\times 10^{-6}\times 14.88}{7.605\times 10^{-3}}

or

Q = 29.28 ×10⁻³ m³/s

Hence, the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

5 0
3 years ago
If a 7 kg bowling ball is lifted 2 m into the air and dropped, what speed will it strike the ground? (
Naily [24]

Answer:

6.32m/s

Explanation:

note:Now these calculations are based in the fact that acc. due to gravity is 10m/s²

okay so I'm thinking you think the speed of a body depends on the mass of the body also,umh... well it doesn't at all!

when two bodies of different masses fall from the same height,they fall at the same time( this is just to say)

now enough of the talking let solve....

so the ball was dropped .ie from rest to the ground through a distance of 2m,

the formula for calculating the distance if a body moving in a straight line is given by:

S=ut + ½at² where u is initial velocity, a is acceleration ( of the body or due to gravity, but since its falling freely under the influence of gravity its " we use the acceleration due to gravity ,which is 10m/s²) and t is the time taken to cover the distance.

from our question the ball was dropped from rest thus its u is 0 therefore we use this equation to find the time it took to touch ground (S=½at²)

solving ....

we get t to be 0.632s

to find the speed we substitute t in the equation below:

V=u+at ,but since u=0

V=at =10•0.632=6.32m/s

therefore the speed the body uses to strike the ground is 6.32m/s

4 0
2 years ago
The general formula for an acid is
Umnica [9.8K]

Answer:

To write the general formula for an acid, we fix one atom which is hydrogen because this atom is common to all the acids. General formula for acid is written by HX. where H represents Hydrogen atom.

Explanation:

4 0
2 years ago
The velocity of a bob on a simple pendulum at the lowest position is 10.56 m/s. What is the maximum vertical height it is able t
kondaur [170]

um I do not know sorry

Explanation:

ummmmmmjjhfgfffff

7 0
2 years ago
During the first 6 years of its operation, the Hubble Space Telescope circled the Earth 37,000 times, for a total of 1,280,000,0
oksian1 [2.3K]

Answer:

v = 384km/min

Explanation:

In order to calculate the speed of the Hubble space telescope, you first calculate the distance that Hubble travels for one orbit.

You know that 37000 times the orbit of Hubble are 1,280,000,000 km. Then, for one orbit you have:

d=\frac{1,280,000,000km}{37,000}=34,594.59km

You know that one orbit is completed by Hubble on 90 min. You use the following formula to calculate the speed:

v=\frac{d}{t}=\frac{34,594.59km}{90min}=384.38\frac{km}{min}\approx384\frac{km}{min}

hence, the speed of the Hubble is approximately 384km/min

5 0
2 years ago
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