Help asapp!!!!!!!!!!!!!!

Five 6 ohm resistors are connected in parallel. Which of the following is their effective resistance?

Ganezh [65]

1/Rt = 1/R1 + 1/R2 + 1/R3+ 1/R4+ 1/R5

1/Rt = 1/6+ 1/6+1/6+1/6+1/6

1/Rt = 5/6

Rt = 6/5

Rt = 1.2 ohm

so B is the answer

3 0

3 years ago

Electron spin: Radio astronomers can detect clouds of hydrogen too cool to radiate optical wavelengths of light by means of the

Ivan

Answer:

the magnetic field experienced by the electron is 0.0511 T

Explanation:

Given the data in the question;

Wavelength λ = 21 cm = 0.21 m

we know that Bohr magneton μ $_B$ is 9.27 × 10⁻²⁴ J/T

Plank's constant h is 6.626 × 10⁻³⁴ J.s

speed of light c = 3 × 10⁸ m/s

protein spin causes magnetic field in hydrogen atom.

so

Initial potential energy = -μ $_B$ B × cos0°

= -μ $_B$ B × 1

= -μ $_B$ B

Final potential energy = -μ $_B$ B × cos180°

= -μ $_B$ B × -1

= μ $_B$ B

so change in energy will be;

ΔE = μ $_B$ B - ( -μ $_B$ B )

ΔE = 2μ $_B$ B

now, difference in energy levels will be;

ΔE = hc/λ

2μ $_B$ B = hc/λ

2μ $_B$ Bλ = hc

B = hc / 2μ $_B$ λ

so we substitute

B = [(6.626 × 10⁻³⁴) × (3 × 10⁸)] / [2(9.27 × 10⁻²⁴) × 0.21 ]

B = [ 1.9878 × 10⁻²⁵ ] / [ 3.8934 × 10⁻²⁴ ]

B = 510556326.09

B = 0.0511 T

Therefore, the magnetic field experienced by the electron is 0.0511 T

7 0

3 years ago

Find the average velocity for the time interval beginning when t = 4 with duration 1 seconds, 0.5 seconds, and 0.05 seconds.

s344n2d4d5 [400]

This question is incomplete, the complete question is;

A student dropped a textbook from the top floor of his dorm and it fell according to the formula s(t) = -16t² + 8√t, where t is the time in seconds and s(t) is the distance in feet from the top of the building.

(a) Write a formula for the average velocity of the ball for t near 4.

(b) Find the average velocity for the time interval beginning when t = 4 with duration 1 seconds, 0.5 seconds, and 0.05 seconds

(c) What is your estimate for the instantaneous velocity of the ball at t = 4

Answer:

a)

Average velocity, (Vavg) of the ball for t near 4.

Vavg = [s(4) - s(0)] / (4 - 0)

Where s(4) = -16 × 4² + 8 × √4= - 240 m

s(0) = -16 × 0 + 8 * 0 = 0

b)

duration = 1 sec

Vavg = [s(5) - s(4)] / (5 - 4)

s(5) = -16 × 52 + 8 × √5 = - 382 m

s(4) = -16 × 42 + 8 √4 = - 240 m

Vavg = (-382 - (-240)) / (5 - 4)

Vavg = - 142.1 m/s

duration = 0.5 sec

Vavg = [s(4.5) - s(4)] / (4.5 - 4)

s(4.5) = -16 × 4.52 + 8 × √4.5 = - 307 m

s(4) = -16 × 42 + 8 × √4 = - 240 m

Vavg = (-307 - (-240)) / (4.5 - 4)

Vavg= - 134.1 m/s

duration = 0.05 sec

Vavg = [s(4.05) - s(4)] / (4.05 - 4)

s(4.05) = -16 × 4.052 + 8 × √4.05 = - 246 m

s(4) = -16 × 42 + 8 × √4 = - 240 m

Vavg = (-246 - (-240)) / (4.05 - 4)

Vavg= - 126.8 m/s

c)

Instantaneous velocity, v = ds/dt

= - 16 × 2 × t + 8 ×× (0.5 / √t )

= - 32 × t + 4/√t

ds/dt at t = 4 is,

v = - 32 × 4 + 4 / √4

= - 126 m/s

5 0

3 years ago

State the objects in the universe that reflect light from stars

stepan [7]

Everything in the universe that is not a star reflects light from Stars. Otherwise you can't see it.

7 0

3 years ago

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 2.20×106 N , one at an angle 19.0 ∘ west of north,

valina [46]

Answer:

$=2.99\times10^9J$

Explanation:

According to the question

net force F = 2.20×10^6 N

displacement $S = 0.72\times10^3m$

from figure , the horizontal forces are same in magnitude and opposite direction.

so , neglect these two forces.

we can take only vertical components of the force.

total force F' = F cos 19° + F cos 19°

= 2×F×cos 19° ................. (1

therefore , total work is

W = F'S

= (2F cos19)×S

$= (2)(2.20\times10^6 N)cos19° (0.720\times10^3 m)$

$=2.99\times10^9J$

6 0

3 years ago