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3 years ago

If a proton emits a photon in a spin-flipping event in a magnetic field, which is true?

Physics

1 answer:

vodomira [7]3 years ago

5 0

Answer:

Explanation:

We know that the magnetic moment of electron is in same direction as it's spin so it's spin flip from down to up.

The correct answer is C that is . It flips from spin down to spin up.

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Answer all these and i'll mark brainliest.

Luda [366]

Formula for net force: force= mass x acceleration

6 0

3 years ago

A solid cylinder is released from the top of an inclined plane of height 0.50 m. From what height, in meters, on the incline sho

dangina [55]

Answer:

Explanation:

for rolling motion down the plane acceleration is given by the following expression

a = g sinθ / (1 + k² / R²)

here k is radius of gyration and R is radius of the object rolling down .

for cylinder I = 1/2 m R²

so k² = R² / 2

k² / R² = 1/2

a = g sinθ /( 1 + 1 / 2 )

= 2 / 3 x g sinθ

v = √ 2 a s

= √ (2 x 2 / 3 x g sinθ s )

= √ (4 / 3 x g h )

= √ (4 / 3 x g x .5 )

= √ 2g / 3

for sphere I = 2/5 m R²

so k² = 2/5 R²

k² / R² = 2 / 5

a = g sinθ / (1 + 2 / 5)

= 5 / 7 x g sinθ

v = √ 2 a s

= √ (2 x 5 / 7 x g sinθ s )

= √ (10/7 x g h )

Given

√ (10/7 x g h ) = √ 2g / 3

10/7 x g h = 2g / 3

h = 14 / 30 m

= .47 m .

5 0

4 years ago

Scientists want to place a 3400.0 kg satellite in orbit around Mars. They plan to have the satellite orbit at a speed of 2480.0

BaLLatris [955]

Answer:

Part a)

$r = 6.96 \times 10^6 m$

Part b)

$F_g = 3.004 \times 10^3 N$

Part c)

$a = 0.88 m/s^2$

Part d)

$v = \frac{2480}{2} = 1240 m/s$

Explanation:

Part a)

As we know that the orbital speed of the satellite is given as

$v = 2480 m/s$

now we will have

$v = \sqrt{\frac{GM_{mars}}{r}}$

now we have

$M_{mars} = 6.4191 \times 10^{23} kg$

$R_{mars} = 3.397 \times 10^6 m$

now we have

$2480 = \sqrt{\frac{(6.67 \times 10^{-11})(6.4191 \times 10^{23})}{r}}$

$r = 6.96 \times 10^6 m$

Part b)

Here force between mars and satellite is the gravitational attraction force which is given as

$F_g = \frac{GM_{mars} m}{r^2}$

$F_g = \frac{(6.67\times 10^{-11})(6.4191 \times 10^{23})(3400)}{(6.96\times 10^6)^2}$

$F_g = 3.004 \times 10^3 N$

Part c)

Acceleration of satellite is the ratio of gravitational force and mass of the satellite

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

$a = \frac{3004}{3400}$

$a = 0.88 m/s^2$

Part d)

As we know by III law of kepler

$\frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3}$

here we know that T2 = 8 T1

$(\frac{1}{8})^2= \frac{r_1^3}{r_2^3}$

$\frac{r_1}{r_2} = (\frac{1}{2})^2$

so we have

$r_2 = 4r_1$

as we know that speed is given as

$v = \sqrt{\frac{GM}{r}}$

so we can say since radius is orbit becomes 4 times so the orbital speed must be half

$v = \frac{2480}{2} = 1240 m/s$

7 0

4 years ago

Help m-e-e people -_-!

xxTIMURxx [149]

Answer:

the answer is

Explanation:For equilibrium

Weight = Tension

mg=T

∴T=4×3.1π=12.4πN (as can be inferred from the question)

△l/l

T/A

1000

0.031

/20

12.4π/π(

1000

)

4×0.031

12.4×20×1000×(1000)

=2×10

N/m

6 0

3 years ago

Use the image below to answer the following question (ruler not to scale).

Svetradugi [14.3K]

Answer:

it depends on wether the + and - are facing eachother

or away from eachother

Explanation:

4 0

3 years ago

Read 2 more answers