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anyanavicka [17]
3 years ago
9

Enter the expression 2gΔym−−−−−√, where Δ is the uppercase Greek letter Delta.

Physics
1 answer:
Anon25 [30]3 years ago
5 0
This doesnt make sense. 2g<span>Δym is simply an equation. </span>
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Regions around the nucleus where electrons are most likely to be found are called
ryzh [129]

Answer:

orbitals

Explanation:

The region where an electron is most likely to be is called an orbital. Each orbital can have at most two electrons. Some orbitals, called S orbitals, are shaped like spheres, with the nucleus in the center.

3 0
3 years ago
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Each tire on a car has a radius of 0.330 m and is rotating with an angular speed of 11.7 revolutions/s. Find the linear speed v
Eduardwww [97]

Answer:

The linear speed of the car, v, is 24.26 m/s

Explanation:

Given;

radius of the car's tire, r = 0.330 m

angular speed of the car, ω = 11.7 revolutions/s

The angular speed of the car in radian per second:

\omega = 11.7 \ \frac{rev}{s} \times \frac{2\pi \ rad}{1 \ rev} \\\\\omega = 73.523 \ rad/s

The linear speed of the car, v, is calculated as;

v = ωr

v = 73.523 rad/s  x  0.33 m

v = 24.26 m/s

Therefore, the linear speed of the car, v, is 24.26 m/s

7 0
3 years ago
n alpha particle (q = +2e, m = 4.00 u) travels in a circular path of radius 5.94 cm in a uniform magnetic field with B = 1.10 T.
9966 [12]

Answer:

a). V = 3.13*10⁶ m/s

b). T = 1.19*10^-7s

c). K.E = 2.04*10⁵

d). V = 1.02*10⁵V

Explanation:

q = +2e

M = 4.0u

r = 5.94cm = 0.0594m

B = 1.10T

1u = 1.67 * 10^-27kg

M = 4.0 * 1.67*10^-27 = 6.68*10^-27kg

a). Centripetal force = magnetic force

Mv / r = qB

V = qBr / m

V = [(2 * 1.60*10^-19) * 1.10 * 0.0594] / 6.68*10^-27

V = 2.09088 * 10^-20 / 6.68 * 10^-27

V = 3.13*10⁶ m/s

b). Period of revolution.

T = 2Πr / v

T = (2*π*0.0594) / 3.13*10⁶

T = 1.19*10⁻⁷s

c). kinetic energy = ½mv²

K.E = ½ * 6.68*10^-27 * (3.13*10⁶)²

K.E = 3.27*10^-14J

1ev = 1.60*10^-19J

xeV = 3.27*10^-14J

X = 2.04*10⁵eV

K.E = 2.04*10⁵eV

d). K.E = qV

V = K / q

V = 2.04*10⁵ / (2eV).....2e-

V = 1.02*10⁵V

7 0
3 years ago
HELP!!
raketka [301]
Hello!

Since the two weights are <em>off</em> the table, the block will move towards letter F. 

I hope this helps :))
5 0
3 years ago
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How long does it take for 4 coulombs of charge to pass through a cross
Galina-37 [17]

Answer: 2 seconds

Explanation:

Given that,

Time (T) = ?

Charge (Q) = 4 coulombs

current (I) = 2 Amps

Since charge depends on the amount of current flowing through the wire in a given time, hence

Charge = Current x Time

Q = IT

4 coulombs = 2 Amps x Time

Time = 4 coulombs / 2 Amps

Time = 2 seconds

Thus, it takes 2 seconds for the current to flow through the wire

4 0
3 years ago
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