All categories

2 years ago

Create a list of 5 potential jobs that students of genealogy can obtain.

Physics

1 answer:

Kipish [7]2 years ago

4 0

Answer:

Private Investigator.

Investigative Genetic Genealogist.

Historic Preservationist.

Military Repatriation Expert.

Heir Searcher.

Explanation:

You might be interested in

Electrons in a particle beam each have a kinetic energy of 4.0 × 10 −17 J. What is the magnitude of the electric field that will

Dmitry_Shevchenko [17]

Answer:

-833.3 N/C

Explanation:

Kinetic energy, K, in terms of electric field, E, is given as:

K = qEr

q = charge = e = 1.6 × 10⁻¹⁹C

E = Electric field

r = distance = 0.3m

The electric field can be gotten by making E subject of formula:

E = K/(qr)

The electeic field needed to stop the electrons must be equal in magnitude to the electric field carried by these electrons:

E = (4.0 × 10⁻¹⁷)/(-1.6 × 10⁻¹⁹ * 0.3)

E = -833.3 N/C

This is the electric field needed to stop the electrons.

The negative sign means that the electric field must be in a direction opposite to the motion of the electrons.

3 0

3 years ago

Read 2 more answers

A space expedition discovers a planetary system consisting of a massive star and several spherical planets. The planets all have

Juliette [100K]

Answer:

T/√8

Explanation:

From Kepler's law, T² ∝ R³ where T = period of planet and R = radius of planet.

For planet A, period = T and radius = 2R.

For planet B, period = T' and radius = R.

So, T²/R³ = k

So, T²/(2R)³ = T'²/R³

T'² = T²R³/(2R)³

T'² = T²/8

T' = T/√8

So, the number of hours it takes Planet B to complete one revolution around the star is T/√8

7 0

3 years ago

12.51 A parallel RLC circuit, which is driven by a variable frequency 2-A current source, has the following values: R = 1 kΩ, L

Anastaziya [24]

Answer:

BW = 100 rad/s

wlow = 452.49 rad/s

whigh = 552.49 rad/s

V(jwlow) =1414.21 < 45°V

V(jwhigh) =1414.21 <-45°V

Explanation:

To calculate bandwidth we have formula

BW = 1/RC

BW = 1/ 1000x10x10^¯6

BW = 100 rad/s

We will first calculate resonant frequency and quality factor for half power frequencies.

For resonant frequency

wo = 1/(SQRT LC)

wo = 1/SQRT 400×10¯³ × 10×10^¯6

wo = 500 rad/s

For Quality

Q = wo / BW

Q = 500/100

Q = 5

wlow = wo [-1/2Q+ SQRT (1/2Q)² + 1]

wlow = 500 [-1/2×5 + SQRT (1/2×5)² + 1]

wlow = 452.49 rad/s

whigh = wo [1/2Q+ SQRT (1/2Q)² + 1]

whigh = 500 [1/2×5 + SQRT (1/2×5)² + 1]

whigh = 552.49 rad/s

We will start with admittance at lower half power frequency

Y(jwlow) = (1/R) + (1/jwlow L) + (jwlow C)

Y(jwlow) = (1/1000) + (1/j×452.49×400×10¯³) + (j×452.49×10×10^¯6)

Y(jwlow) = 0.001 - j5.525×10¯³ + j4.525×10¯³

Y(jwlow) = (1-j).10¯³ S

Voltage across the network is calculated by ohm's law

V(jwlow) = I/Y(jwlow)

V(jwlow) = 2/(1-j).10¯³

V(jwlow) = 1414.2 < 45°V

Now we will calculate the admittance at higher half power frequency

Y(jwhigh) = (1/R) + (1/jwhigh L) + (jwhigh C)

Y(jwhigh) = (1/1000) + (1/j×552.49×400×10¯³) + (j×552.49×10×10^¯6)

Y(jwhigh) = 0.001 - j4.525×10¯³ + j5.525×10¯³

Y(jwhigh) = (1+j).10¯³ S

Voltage across network will be calculated by ohm's law

V(jwhigh) = I/Y(jwhigh)

V(jwhigh) = 2/(1+j).10¯³

V(jwhigh) = 1414.2 < - 45°V

6 0

3 years ago

Having landed on a newly discovered planet, an astronaut sets up a simple pendulum of length 1.38 m and finds that it makes 441

Tasya [4]

The period of a simple pendulum is given by:
$T=2 \pi \sqrt{ \frac{L}{g} }$
where L is the pendulum length, and g is the gravitational acceleration of the planet. Re-arranging the formula, we get:
$g= \frac{4 \pi^2}{T^2}L$ (1)

We already know the length of the pendulum, L=1.38 m, however we need to find its period of oscillation.

We know it makes N=441 oscillations in t=1090 s, therefore its frequency is
$f= \frac{N}{t}= \frac{441}{1090 s}=0.40 Hz$
And its period is the reciprocal of its frequency:
$T= \frac{1}{f}= \frac{1}{0.40 Hz}=2.47 s$

So now we can use eq.(1) to find the gravitational acceleration of the planet:
$g= \frac{4 \pi^2}{T^2}L = \frac{4 \pi^2}{(2.47 s)^2} (1.38 m) =8.92 m/s^2$

3 0

3 years ago

Suppose the initial kinetic energy and final potential energy in an experiment are both zero. What can you conclude?

puteri [66]

Answer:

That an item is neither moving nor staying still in a position that is building up energy.

Explanation:

3 0

3 years ago

Read 2 more answers