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3 years ago

The first periodic table of elements Mendeleev created contained only 70 elements. True or false?

Physics

2 answers:

mr Goodwill [35]3 years ago

7 0

Answer: True

Explanation: Mendeleev created the first periodic table containing 70 elements.

melamori03 [73]3 years ago

5 0

Explanation:

FALSE

The first periodic table of elements Mendeleev created contained 63 elements.

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In the text "White House Press Release Announcing the Bombing of Hiroshima August 6, 1945" by President Truman, in which country

stepladder [879]

D
Japan is the answer

7 0

2 years ago

Ở nhiệt độ không đổi, dưới áp suất 10^4 Pa l, một lượng khí có thể tích 10l. Tính thể tích lượng khí đó dưới áp suất 5.10^4 Pa

gtnhenbr [62]

Answer:

Thể tích cuối cùng, V2 = 2 L

Explanation:

Cho các dữ liệu sau đây;

Khối lượng ban đầu = 10 L

Áp suất ban đầu = 10⁴ = 10000 Pa

Áp suất cuối cùng = 5 * 10⁴ = 50000 Pa

Để tìm tập mới hoặc tập cuối cùng, chúng ta sẽ sử dụng định luật Boyle;

Để tìm tập mới V2, chúng tôi sẽ sử dụng định luật Boyles.

Boyles phát biểu rằng khi nhiệt độ của khí lý tưởng được giữ không đổi, áp suất của khí tỉ lệ nghịch với thể tích mà khí chiếm giữ.

Về mặt toán học, định luật Boyles được đưa ra bởi;

$PV = K$

$P_{1}V_{1} = P_{2}V_{2}$

Thay vào phương trình, ta có;

$10000 * 10 = 50000V_{2}$

$100000 = 50000V_{2}$

$V_{2} = \frac {100000}{50000}$

$V_{2} = 2$

Thể tích cuối cùng, V2 = 2 L

6 0

3 years ago

Suppose you drop a 10-pound weight and a 5-pound weight on the Moon, both from the same height at the same time. What will happe

Murljashka [212]

Answer:

Both will hit the ground at the same time.

3 0

2 years ago

A 4 cm diameter "bobber" with a mass of 3 grams floats on a pond. A thin, light fishing line is tied to the bottom of the bobber

Tasya [4]

Answer:

Explanation:

Calculate the volume of the lead

$V=\frac{m}{d}\\\\=\frac{10g}{11.3g'cm^3}$

Now calculate the bouyant force acting on the lead

$F_L = Vpg$

$F_L=(\frac{10g}{11.3g/cm^3} )(1g/cm^3)(9.8m/s^2)\\\\=8.673\times 10^{-3}N$

This force will act in upward direction

Gravitational force on the lead due to its mass will act in downward direction

Hence the difference of this two force

$T=mg-F_L\\\\=(10\times10^{-3}kg(9.8m/s^2)-8.673\times 10^{-3}\\\\=8.933\times10^{-3}N$

If V is the volume submerged in the water then bouyant force on the bobber is

$F_B=V'pg$

Equate bouyant force with the tension and gravitational force

$F_B=T_mg\\\\V'pg=\frac{(8.933\times10^{-2}N)+mg}{pg} \\\\V'=\frac{(8.933\times10^{-2}N)+mg}{pg}$

Now Total volume of bobble is

$\frac{V'}{V^B} =\frac{\frac{(8.933\times10^{-2})+Mg}{pg} }{\frac{4}{3} \pi R^3 }\times100\\\\=\frac{\frac{(8.933\times10^{-2})+(3)(9.8)}{(1000)(9.8)} }{\frac{4}{3} \pi (4.0\times10^{-2})^3 }\times100\\\\$

= $\large\boxed{4.52 \%}$

7 0

3 years ago

A 0.9 kg ball attached to a cord is whirled in a vertical circle of radius 2.5 m. Find the minimum speed needed at the top of th

lapo4ka [179]

Answer:

The minimum speed needed at the top of the circle so that the cord remains tensioned and the ball's path remains circular is approximately is 9.903 meters per second.

Explanation:

By the Principle of Energy Conservation we understand that the minimum speed needed by the ball is that speed such that maximum height reached is equal to the diameter of the vertical circle, that is:

$K =U_{g}$ (1)

Where:

$K$ - Translational kinetic energy, measured in joules.

$U_{g}$ - Gravitational potential energy, measured in joules.

By definitions of translational kinetic and gravitational potential energies, we expand the equation above and clear the initial speed of the ball:

$\frac{1}{2}\cdot m \cdot v^{2} = m\cdot g\cdot h$

$v = \sqrt{2\cdot g\cdot h}$ (2)

Where:

$m$ - Mass, measured in kilograms.

$v$ - Initial speed, measured in meters per second.

$g$ - Gravitational acceleration, measured in meters per square second.

$h$ - Maximum height of the ball, measured in meters.

If we know that $g = 9.807\,\frac{m}{s^{2}}$ and $h = 5\,m$ , then the initial speed of the ball is:

$v = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (5\,m)}$

$v\approx 9.903\,\frac{m}{s}$

The minimum speed needed at the top of the circle so that the cord remains tensioned and the ball's path remains circular is approximately is 9.903 meters per second.

3 0

3 years ago