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lubasha [3.4K]
3 years ago
11

What element atomic number is in Group 17 and Period 6?

Physics
1 answer:
notsponge [240]3 years ago
7 0

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The halogen elements are the six elements in Group 17 of the periodic table. Group 17 is the second column from the right in the periodic table and contains six elements: fluorine (F), chlorine (Cl), bromine (Br), iodine (I), astatine (As), and tennessine (Ts).

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How are chargeable cells different from ordinary dry cells​
topjm [15]
Ordinary cells can convert chemical energy to electrical energy only, but rechargeable cells can also store electrical energy into chemical energy and vice versa. You will study more about it in your higher classes. secondary cells can be recharged and used again but dry cells cannot be recharged.
6 0
3 years ago
In the unsaturated zone, the pores of the soil are totally filled with water. T or F
creativ13 [48]







the answer is clearly false

6 0
3 years ago
Dos cargas puntuales q1 = −50μC y q2 = +30μC se encuentran
alina1380 [7]

Answer:

Datos:

q1 = -50 μC = 50*10^{-6}

q2 = +30 μC = 30*10^{-6}

F = 10 N

a) x si la <em>F = 10N</em>

Aplicando la Ley de Coulomb:

x = \sqrt\frac{k0 * q1 *q2}{F} = \sqrt\frac{(9*10^{9} )*(50*10^{-6})*(30*10^{-6})}{10} = 1,162m

b) x si la <em>F = 20 N</em>

x=<em> </em>\sqrt\frac{(9*10^{9} )*(50*10^{-6})*(30*10^{-6})}{20}<em> </em>= 0,822m

c)x si la <em>F = 50 N</em>

x = \sqrt\frac{(9*10^{9} )*(50*10^{-6})*(30*10^{-6})}{50} = 0,520m

3 0
3 years ago
One problem for humans living in outer space is that they are apparently weightless. One way around this problem is to design a
Maurinko [17]

This question is not complete.

The complete question is as follows:

One problem for humans living in outer space is that they are apparently weightless. One way around this problem is to design a space station that spins about its center at a constant rate. This creates “artificial gravity” at the outside rim of the station. (a) If the diameter of the space station is 800 m, how many revolutions per minute are needed for the “artificial gravity” acceleration to be 9.80m/s2?

Explanation:

a. Using the expression;

T = 2π√R/g

where R = radius of the space = diameter/2

R = 800/2 = 400m

g= acceleration due to gravity = 9.8m/s^2

1/T = number of revolutions per second

T = 2π√R/g

T = 2 x 3.14 x √400/9.8

T = 6.28 x 6.39 = 40.13

1/T = 1/40.13 = 0.025 x 60 = 1.5 revolution/minute

6 0
3 years ago
A skateboarder has an acceleration of −1.9 m/s2. If her initial speed is 6 m/s, how long does it take her to stop?
SSSSS [86.1K]
You said that she's losing 1.9 m/s of her speed every second.

So it'll take

             (6 m/s) / (1.9 m/s²)  =  3.158 seconds  (rounded)

to lose all of her initial speed, and stop.
6 0
3 years ago
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