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4 years ago

What element holds the most electrons? Li , B, Be

Physics

2 answers:

Svetach [21]4 years ago

8 0

The correct answer is B

Alisiya [41]4 years ago

3 0

Question is kinda vague.

"holds the most"

holds most OTHER electrons - Lithium, it has 1 valence
holds the most of its OWN - Boron, it has more electrons than the others

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Which feature of a heating curve indicates a change of state

ziro4ka [17]

Answer:

The diagonal or the inclined lines shows the changes in terms of temperature, and the horizontal lines shows the changing of phases.

Explanation:

hope it is useful

3 0

3 years ago

The next two questions refer to this situation: A rectangular loop with sides of length a= 1.00 cm and b= 2.70 cm is placed near

alisha [4.7K]

Answer:

a) $1.007 * 10^{-7}$ V.m , into the page

b) $-5.39 * 10^{-8}$ V, counterclockwise

Explanation:

a) $d\Phi = B.dA\\\\B_r = \frac{\mu_0I}{2\pi r}\\\\dA = b.dr\\\\\Phi = \int\limits^b_a {B.dA} = \int\limits^{a+d}_d {\frac{\mu_0I}{2\pi r}.b.dr} =\frac{\mu_0I}{2\pi }.b.ln(\frac{a+d}{d} )\\$

At t = 2.90, I = 3.62 + 1.49 * (2.90)^2 = 16.15 A

$\Phi = \frac{4\pi * 10^{-7}*16.15*0.027}{2\pi} ln(1.46/0.46) = 1.007 * 10^{-7}$ V.m

Direction is into the page.

b) $emf = -d\Phi/dt = -\frac{\mu_0}{2\pi }.b.ln(\frac{a+d}{d}).(2.98t)= \\=-\frac{4\pi * 10^{-7}*0.027}{2\pi} ln(1.46/0.46)*2.98*2.90= -5.39 * 10^{-8} V$

Direction is counterclockwise.

8 0

3 years ago

alexandr402 [8]

Answer:

Velocity Has vector Quantity

4 0

3 years ago

An electron is released from rest in a uniform electric field of 418 N/C near a particle detector. The electron arrives at the d

Oksanka [162]

Answer:

a) 7.35 x 10¹³ m/s²

b) 5.03 x 10⁻⁸ sec

c) 9.3 cm

d) 6.23 x 10⁻¹⁸ J

Explanation:

E = magnitude of electric field = 418 N/C

q = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C

m = mass of the electron = 9.1 x 10⁻³¹ kg

acceleration of the electron is given as

$a = \frac{qE}{m}$

$a = \frac{(1.6\times 10^{-19})(418)}{(9.1\times 10^{-31})}$

a = 7.35 x 10¹³ m/s²

v = final velocity of the electron = 3.70 x 10⁶ m/s

v₀ = initial velocity of the electron = 0 m/s

t = time taken

Using the equation

v = v₀ + at

3.70 x 10⁶ = 0 + (7.35 x 10¹³) t

t = 5.03 x 10⁻⁸ sec

d = distance traveled by the electron

using the equation

d = v₀ t + (0.5) at²

d = (0) (5.03 x 10⁻⁸) + (0.5) (7.35 x 10¹³) (5.03 x 10⁻⁸)²

d = 0.093 m

d = 9.3 cm

Kinetic energy of the electron is given as

KE = (0.5) m v²

KE = (0.5) (9.1 x 10⁻³¹) (3.70 x 10⁶)²

KE = 6.23 x 10⁻¹⁸ J

4 0

4 years ago

n isolated charged soap bubble of radius R0=7.45 cmR0=7.45 cm is at a potential of V0=307.0 volts.V0=307.0 volts. If the bubble

Gnesinka [82]

Complete Question

An isolated charged soap bubble of radius R0 = 7.45 cm is at a potential of V0=307.0 volts. V0=307.0 volts. If the bubble shrinks to a radius that is 19.0%19.0% of the initial radius, by how much does its electrostatic potential energy ????U change? Assume that the charge on the bubble is spread evenly over the surface, and that the total charge on the bubble r

Answer:

The difference is $U_f -U_i = 16 *10^{-7} J$