Through which element would someone most likely receive a shock

Using parallax, what is the distance to the<br> farthest stars you can measure?

aev [14]

Parallax angles of less than 0.01 arcsec are very difficult to measure from Earth because of the effects of the Earth's atmosphere. This limits Earth based telescopes to measuring the distances to stars about 1/0.01 or 100 parsecs away.

6 0

3 years ago

An FM radio consists of a series RLC circuit with a 6.57 pF (peco – 10-12) capacitor. If a person is dialed into a station broad

faust18 [17]

Answer:

3.6 x 10⁻⁷ H

Explanation:

C = capacitance of the capacitor = 6.57 x 10⁻¹² F

L = inductance of the inductor = ?

f = frequency of broadcasting system = 103.9 MHz = 103.9 x 10⁶ Hz

frequency is given as

$f = \frac{1}{2\pi \sqrt{LC}}$

$103.6\times 10^{6} = \frac{1}{2(3.14)\sqrt{(6.57\times 10^{-12})L}}$

Inserting the values

L = 3.6 x 10⁻⁷ H

4 0

3 years ago

023 (part 1 of 2) 10.0 points

Annette [7]

Answer:

Part 1

The angular speed is approximately 1.31947 rad/s

Part 2

The change in kinetic energy due to the movement is approximately 675.65 J

Explanation:

The given parameters are;

The rotation rate of the merry-go-round, n = 0.21 rev/s

The mass of the man on the merry-go-round = 99 kg

The distance of the point the man stands from the axis of rotation = 2.8 m

Part 1

The angular speed, ω = 2·π·n = 2·π × 0.21 rev/s ≈ 1.31947 rad/s

The angular speed is constant through out the axis of rotation

Therefore, when the man walks to a point 0 m from the center, the angular speed ≈ 1.31947 rad/s

Part 2

Given that the kinetic energy of the merry-go-round is constant, the change in kinetic energy, for a change from a radius of of the man from 2.8 m to 0 m, is given as follows;

$\Delta KE_{rotational} = \dfrac{1}{2} \cdot I \cdot \omega ^2 = \dfrac{1}{2} \cdot m \cdot v ^2$

I = m·r²

Where;

m = The mass of the man alone = 99 kg

r = The distance of the point the man stands from the axis, r = 2.8 m

v = The tangential velocity = ω/r

ω ≈ 1.31947 rad/s

Therefore, we have;

I = 99 × 2.8² = 776.16 kg·m²

$\Delta KE_{rotational}$ = 1/2 × 776.16 kg·m² × (1.31947)² ≈ 675.65 J

3 0

3 years ago

I'm doing a presentation over real-world applications of physics and the group I am in wants to do "the physics of video games"

artcher [175]

Most modern games have a sense of real-world physics, but not exactly perfect. In a video game, the realistic movement or action greatly depends on the precision of coding. In real life, movement isn't done or programmed by a strand of code.

Your presentation sounds interesting, being a gamer myself, I would look forward to it. But the choice lies in your hands. If you do reconsider, be sure you have a backup plan. Good luck to you.

3 0

3 years ago

and an ideal spring of unknown force constant. The oscillator is found to have a period of 0.157 s and a maximum speed of 2 m/s

DENIUS [597]

Answer:

k = 1073.09 N/m

A = 0.05 m

Explanation:

Given:

- Time period T = 0.147 s

- maximum speed V_max = 2 m/s

- mass of the block m = 0.67 kg

Find:

- The spring constant k

- The amplitude of the motion A.

Solution:

- A general simple harmonic motion is modeled by:

x (t) = A*sin(w*t)

- The velocity of the above modeled SHM is:

v = dx / dt

v(t) = A*w*cos(w*t)

- Where A is the amplitude in meters, w is the angular speed rad/s and time t is in seconds.

- We can see that maximum velocity occurs when (cos(w*t)) maximizes i.e it is equal to 1 or -1. Hence,

- V_max = A*w

- Where w is related to mass of the object and spring constant k as follows,

w = sqrt ( k / m )

- The relationship between w angular speed and Time period T is:

w = 2*pi / T

- Equating the above two equations we have,

m*(2*pi / T)^2 = k

- Hence, k = 0.67*(2*pi / 0.157)^2

k = 1073.09 N / m

- So, amplitude A is:

A = V_max*sqrt ( m / k )

A = 2*sqrt ( 0.67 / 1073.09 )

A = 0.05 m

6 0

4 years ago