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TiliK225 [7]
3 years ago
12

Consider four atoms from the second period: lithium, beryllium, boron, carbon, and nitrogen. Which of these emblems has the lowe

st electronegativity value?
•Lithium
•Beryllium
•Boron
•Carbon
•Nitrogen
Physics
1 answer:
xz_007 [3.2K]3 years ago
6 0
If you go to the right along the periodic table, electronegativity increases.
So the larger the column number, the greater the electronegativity.

-Lithium has lowest as it is in the 1st column



-Beryllium (2nd column)
-Boron s (13th column)
-Nitrogen (15th column)


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If the strings have the same thickness butdifferent lengths, which of the following parameters, if any, willbe different in the
gizmo_the_mogwai [7]

Answer:

1.Length is one of the four factors on which the wave frequency depends. So if the length of the string changes then there will be a change in the vibration of string. So in this case if the lengths are different then the wave frequency of both will be different.

2. Wave speed will be the same as it depends on tension and linear density of it.

3.  Wavelength itself is find out by the length of string so it depends on length and it will vary with the lengths of strings.

Explanation:

7 0
3 years ago
The asteroid belt circles the sun between the orbits of Mars and Jupiter. One asteroid has a period of 5.4 earth years.
Ad libitum [116K]

Answer:

(a) Radius = 4.6 x 10^11 m

(b) speed = 16.96 km/s

Explanation:

Time period, T = 5.4 earth years

mass of sun, M = 1.989 x 10^30 kg

(a) Let the orbital radius is R.

use the formula of period

T^2 = \frac{4 \pi^2 R^3}{G M}\\\\\left ( 5.4\times 365\times 24\times 3600 \right )^2 = \frac{4\times3.14\times 3.14\times R^3}{6.67\times10^{-11}\times 1.989\times 10^{30}}\\\\R = 4.6\times 10^{11} m

(b) Let the speed is v.

v=\frac{2 \pi\times R}{T}\\\\v=\frac{2\times 3.14\times 4.6\times 10^{11}}{5.4\times 365\times 24\times 3600}\\\\v = 16963.6 m/s =16.96 km/s

3 0
3 years ago
Point charges q1=+2.00μC and q2=−2.00μC are placed at adjacent corners of a square for which the length of each side is 5.00 cm
mihalych1998 [28]

Point charges q1=+2.00μC and q2=−2.00μC are placed at adjacent corners of a square for which the length of each side is 5.00 cm.?

Point a is at the center of the square, and point b is at the empty corner closest to q2. Take the electric potential to be zero at a distance far from both charges.  

(a) What is the electric potential at point a due to q1 and q2?  

(b) What is the electric potential at point b?

(c) A point charge q3 = -6.00 μC moves from point a to point b. How much work is done on q3 by the electric forces exerted by q1 and q2?

Answer:

a) the potential is zero at the center .

Explanation:

a) since the two equal-magnitude and oppositely charged particles are equidistant

b)(b) Electric potential at point b, v = Σ kQ/r

r = 5cm = 0.05m

k = 8.99*10^9 N·m²/C²

Q = -2 microcoulomb

v= (8.99*10^9) * (2*10^-6) * (1/√2m - 1) / 0.0500m

v =  -105 324 V

c)workdone = charge * potential

work = -6.00µC * -105324V

work = 0.632 J

6 0
3 years ago
PLS HELP Do you believe this relationship between incident and reflected angles would occur even if the medium interface were cu
Anna35 [415]
The Law of reflection would still hold even off a curved surface. Since the angles are measured from the normal, which is perpendicular to the surface, curved surfaces don't matter. This is basis of curved mirrors such as concave and convex 
4 0
3 years ago
A ball of silly putty hits andsticks to a bowling ball that was initially at rest. After thecollision, the total kinetic energy
Jobisdone [24]

Answer:

3. less than the kinetic energy of thesilly putty before the collision.

Explanation:

This is because kinetic energy is dependent on the mass and velocity of an object. Mathematically, it is given as:

K. E. = ½*m*v²

Where m = mass

v = velocity

In the case of the silly putty, we know that the masses of the ball of silly putty and the bowling ball are conserved, hence, the kinetic energy depends solely on the velocity at which the object moves.

After the collision with the bowling ball, because of how heavy a bowling ball is, the speed of the silly putty and bowling ball will definitely be less than the speed of the silly putty before collision, i. e. u > v.

Hence, the kinetic energy after collision will be less than the kinetic energy before collision.

7 0
3 years ago
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