The atom that is the most electronegative is fluorine (F).
<h3>
What is electronegative?</h3>
Electronegativity, is the tendency for an atom of a given chemical element to attract shared electrons when forming a chemical bond.
Electronegativity increases across the groups from left to right of the periodic table and decreases down the group.
Examples of electronegative elements arranged in decreasing order;
- fluorine,
- oxygen,
- nitrogen,
- chlorine,
- bromine,
- iodine,
- sulfur,
- carbon, and
- hydrogen.
Thus, the atom that is the most electronegative is fluorine (F).
Learn more about electronegativity here: brainly.com/question/24977425
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Answer:
it b
Explanation:
bc A water droplet falling in the atmosphere is spherical
Answer:
Stars emit colors of many different wavelengths, but the wavelength of light where a star's emission is concentrated is related to the star's temperature - the hotter the star, the more blue it is; the cooler the star, the more red it is
Answer:
B. 2 m/s
B. Acceleration = 4.05 m/s² and Tension = 297.5 N.
Explanation:
A force is applied on a mass m whose acceleration is 4 m/s
Force = mass × acceleration
a = F/m = 4 m/s
4 m/s = F/m
F = 4 m/s (m)
If Force of 2F is applied on a mass of 4m ; it acceleration is as follows:
2F/4 m = F/ 2m
4m/s (m) / 2m = 2 m/s
a = 2 m/s
2.
Given that
mass 
= 30 kg
mass 
= 50 kg
= 0.1
From the question; we can arrive at two cases;
That :
----- equation (1)
---- equation (2)
50 a = 50 g - T
30 a = T - 30 g sin 30 - 4 × 30 g cos 30
By summation
80 a =![[ 50 - 30 * \frac{1}{2} - 0.1 *30* \frac{\sqrt{3}}{2}]](https://tex.z-dn.net/?f=%5B%2050%20%20-%2030%20%2A%20%5Cfrac%7B1%7D%7B2%7D%20-%200.1%20%2A30%2A%20%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%5D)
g
80 a = 32. 4 × 10 m/s ² (using g as 10m/s²)
80 a = 324 m/s ²
a = 324/80
a = 4.05 m/s²
From equation , replace a with 4.05
50 × 4.05 = 50 × 10 - T
T = 500 -202.5
T =297.5 N
Answer:
a) α = 0.338 rad / s² b) θ = 21.9 rev
Explanation:
a) To solve this exercise we will use Newton's second law for rotational movement, that is, torque
τ = I α
fr r = I α
Now we write the translational Newton equation in the radial direction
N- F = 0
N = F
The friction force equation is
fr = μ N
fr = μ F
The moment of inertia of a saying is
I = ½ m r²
Let's replace in the torque equation
(μ F) r = (½ m r²) α
α = 2 μ F / (m r)
α = 2 0.2 24 / (86 0.33)
α = 0.338 rad / s²
b) let's use the relationship of rotational kinematics
w² = w₀² - 2 α θ
0 = w₀² - 2 α θ
θ = w₀² / 2 α
Let's reduce the angular velocity
w₀ = 92 rpm (2π rad / 1 rev) (1 min / 60s) = 9.634 rad / s
θ = 9.634 2 / (2 0.338)
θ = 137.3 rad
Let's reduce radians to revolutions
θ = 137.3 rad (1 rev / 2π rad)
θ = 21.9 rev
