The answer is D, all responses are correct
<u><em>Both solvent and solute are parts of a solution.</em></u> Solutions are mixtures of two or more substances, and the substance that dissolves into the solution is a solute. Meanwhile, the solute dissolves into a s Meanwhile, the solute dissolves into a substance called the solvent.
Solutes and solvents are substances not used only in chemical laboratories, but they are part of the day to day life. A solution contains only two components, which are solute and solvent. Solvent has the capability of dissolving the solute in a homogenous solution.
Answer:
262 kN/C
Explanation:
If the electrons is moving parallel, thus it has a retiline movement, and because the velocity is varing, it's a retiline variated movement. Thus, the acceleration can be calculated by:
v² = v0² + 2aΔS
Where v0 is the initial velocity (2.0x10⁷ m/s), v is the final velocity (4.0x10⁷ m/s), and ΔS is the distance (1.3 cm = 0.013 m), so:
(4.0x10⁷)² = (2.0x10⁷)² + 2*a*0.013
16x10¹⁴ = 4x10¹⁴ + 0.026a
0.026a = 12x10¹⁴
a = 4.61x10¹⁶ m/s²
The electric force due to the electric field (E) is:
F = Eq
Where q is the charge of the electron (-1.602x10⁻¹⁹C). By Newton's second law:
F = m*a
Where m is the mass, so:
E*q = m*a
The mass of one electrons is 9.1x10⁻³¹ kg, thus, the module of electric field strenght (without the minus signal of the electron charge) is:
E*(1.602x10⁻¹⁹) = 9.1x10⁻³¹ * 4.61x10¹⁶
E = 261,866.42 N/C
E = 262 kN/C
Hey there!
The periodic table is arranged by number of protons/electrons which is the same for all isotopes of an element. <u>A different isotope will have a different number of neutrons only, which does not qualify it for a separate space on the periodic table.</u>
I hope this helps!
Answer:
d_{b} = 2 d_{a}
Explanation:
The electrical resistance for a cylindrical wire is described by the expression
R = ρ L / A
The area of a circle is
A = π r²
r = d / 2
A = π d²/4
We substitute
R = ρ L 4 /π d²
Let's apply this expression to our case, they indicate that the resistance of wire A is 4 times the resistance of wire B
= 4 R_{b}
We substitute
ρ 4/π 
² = 4 (ρ 4/π d_{b}²)
1 / d_{a}² = 4 / d_{b}²
d_{a} = d_{b} / 2
