For an object to orbit another object

elena-s [515]

Roughly 1609 meters in one mile

6 0

3 years ago

The intensity of a sound wave at a fixed distance from a speaker vibrating at 1.00 kHz is 0.750 W/m2. (a) Determine the intensit

sveticcg [70]

Answer:

a) I = 3.63 W / m² , b) I = 0.750 W / m²

Explanation:

The intensity of a sound wave is given by the relation

I = P / A = ½ ρ v (2π f $s_{max}$ )²

I = (½ ρ v 4π² s_{max}²) f²

a) with the initial condition let's call the intensity Io

cte = (½ ρ v 4π² s_{max}²)

I₀ = cte s² f₀²

I₀ = cte 10 6

If frequency is increase f = 2.20 10³ Hz

I = constant (2.20 10³) 2

I = cte 4.84 10⁶

let's find the relationship of the two quantities

I / Io = 4.84

I = 4.84 Io

I = 4.84 0.750

I = 3.63 W / m²

b) in this case the frequency is reduced to f = 0.250 10³ Hz and the displacement s = 4 s or

I = cte (f s)²

I = constant (0.250 10³ 4)²

I = cte 1 10⁶

the relationship

I / Io = 1

I = Io

I = 0.750 W / m²

6 0

2 years ago

Spacetime interval: What is the interval between two events if in some given inertial reference frame the events are separated b

shepuryov [24]

Answer:

a. $\Delta s ^2 = 8.0888 \ 10^{17} m^2$

b. $\Delta s ^2 = 3.0234 \ 10^{16} m^2$

c. $\Delta s ^2 = 3.0234 \ 10^{20} m^2$

Explanation:

The spacetime interval $\Delta s^2$ is given by

$\Delta s ^2 = \Delta (c t) ^ 2 - \Delta \vec{x}^2$

please, be aware this is the definition for the signature ( + - - - ), for the signature (- + + + ) the spacetime interval is given by:

$\Delta s ^2 = - \Delta (c t) ^ 2 + \Delta \vec{x}^2$ .

Lets work with the signature ( + - - - ), and, if needed in the other signature, we can multiply our interval by -1.

$\Delta \vec{x}^2 = (7.5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 5,625 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 3 \ s)^2$

$\Delta (c t) ^ 2 = (899,377,374 \ m)^2$

$\Delta (c t) ^ 2 = 8.0888 \ 10^{17} m^2$

so

$\Delta s ^2 = 8.0888 \ 10^{17} m^2 - 5,625 m^2$

$\Delta s ^2 = 8.0888 \ 10^{17} m^2$

$\Delta \vec{x}^2 = (5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 2,500 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 0.58 \ s)^2$

$\Delta (c t) ^ 2 = (173,879,625.6 \ m)^2$

$\Delta (c t) ^ 2 = 3.0234 \ 10^{16} m^2$

so

$\Delta s ^2 = 3.0234 \ 10^{16} m^2 - 2,500 m^2$

$\Delta s ^2 = 3.0234 \ 10^{16} m^2$

$\Delta \vec{x}^2 = (5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 2,500 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 58 \ s)^2$

$\Delta (c t) ^ 2 = (1.73879 \ 10^{10} \ m)^2$

$\Delta (c t) ^ 2 = 3.0234 \ 10^{20} m^2$

so

$\Delta s ^2 = 3.0234 \ 10^{20} m^2 - 2,500 m^2$

$\Delta s ^2 = 3.0234 \ 10^{20} m^2$

5 0

2 years ago

If the angle of refraction is 20 degrees what is the angle of incidence

Romashka-Z-Leto [24]

Answer:

13.1

Explanation:

that's what I'm gonna go with, but u can research more

5 0

2 years ago

What is the acceleration for an object moving with constant speed? Explain your answer

Oduvanchick [21]

Answer:

What is the acceleration of an object moving at a constant speed?

The Meaning of Constant Acceleration

The data table above show an object changing its velocity by 10 m/s in each consecutive second. This is referred to as a constant acceleration since the velocity is changing by a constant amount each second.

3 0

3 years ago

Read 2 more answers