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3 years ago

15

Which two elements have similar characteristics?

2 answers:

joja [24]3 years ago

7 0

Lithium and calcium because they are in the same group.

Vanyuwa [196]3 years ago

5 0

<h2>Two Elements have similar Characteristics </h2>

Two elements have similar characteristics when they relate to the same family which indicates that they have placed in the same group in the periodic table. There are only those elements which have same properties that are present in the same group. This is because there are same properties along the same group.

Their valency also remains same along the same group. For instance, sodium (Na) and lithium (Li) are present in the same group. They have same valency which is 1 and other properties as well. Good example are electronegativity, density, melting point, boiling point, and many other features.

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The answer is C but man if you have leak or a meltdown good luck to anyone downstream.

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Write your opinion about achievement made by during rana rule

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Answer:

Your opinion about achievement made by during rana rule

Explanation:

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3 years ago

Brandon buys a new Seadoo. He goes 22 km north from the beach. He then goes 11km to the east. Then chases a boat 3 km north. Wha

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Answer:

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3 years ago

Calculate the charge that flows through the cell in 1 minute. Each filament lamp has a power of 3 W and a resistance of 12 Ω

lapo4ka [179]

Answer:

24 Coulumbs

Explanation:

Given data

time= 1 minute= 6 seconds

P=2 W

R= 12 ohm

We know that

P= I^2R

P/R= I^2

2/12= I^2

I^2= 0.166

I= √0.166

I= 0.4 amps

We know also that

Q= It

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3 years ago

At one point in a pipeline, the water's speed is 3.57 m/s and the gauge pressure is 68.7 kPa. Find the gauge pressure at a secon

ArbitrLikvidat [17]

Answer:

The pressure at point 2 is $P_2 = 254.01 kPa$

Explanation:

From the question we are told that

The speed at point 1 is $v_1 = 3.57 \ m/s$

The gauge pressure at point 1 is $P_1 = 68.7kPa = 68.7*10^{3}\ Pa$

The density of water is $\rho = 1000 \ kg/m^3$

Let the height at point 1 be $h_1$ then the height at point two will be

$h_2 = h_1 - 18.5$

Let the diameter at point 1 be $d_1$ then the diameter at point two will be

$d_2 = 2 * d_1$

Now the continuity equation is mathematically represented as

$A_1 v_1 = A_2 v_2$

Here $A_1 , A_2$ are the area at point 1 and 2

Now given that the are is directly proportional to the square of the diameter [i.e $A= \frac{\pi d^2}{4}$ ]

which can represent as

$A \ \ \alpha \ \ d^2$

=> $A = c d^2$

where c is a constant

so $\frac{A_1}{d_1^2} = \frac{A_2}{d_2^2}$

=> $\frac{A_1}{d_1^2} = \frac{A_2}{4d_1^2}$

=> $A_2 = 4 A_1$

Now from the continuity equation

$A_1 v_1 = 4 A_1 v_2$

=> $v_2 = \frac{v_1}{4}$

=> $v_2 = \frac{3.57}{4}$

$v_2 = 0.893 \ m/s$

Generally the Bernoulli equation is mathematically represented as

$P_1 + \frac{1}{2} \rho v_1^2 + \rho * g * h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho * g * h_2$

So

$P_2 = \rho * g (h_1 -h_2 )+P_1 + \frac{1}{2} * \rho (v_1^2 -v_2 ^2 )$

=> $P_2 = \rho * g (h_1 -(h_1 -18.3) + P_1 + \frac{1}{2} * \rho (v_1^2 -v_2 ^2 )$

substituting values

$P_2 = 1000 * 9.8 (18.3) )+ 68.7*10^{3} + \frac{1}{2} * 1000 ((3.57)^2 -0.893 ^2 )$

$P_2 = 254.01 kPa$

8 0

3 years ago