If light travels 300,000 km/s how long does light reflected from mars take to reach earth when mars is 65,000,000 km away

Calculate the angular velocity of the earth in its orbit around the sun.

Orlov [11]

Answer:

$0.0172rad/day=1.99x10^{-7}rad/second$

Explanation:

The definition of angular velocity is as follows:

$\omega=2\pi f$

where $\omega$ is the angular velocity, and $f$ is the frequency.

Frequency can also be represented as:

$f=\frac{1}{T}$

where $T$ is the period, (the time it takes to conclude a cycle)

with this, the angular velocity is:

$\omega=\frac{2\pi}{T}$

The period T of rotation around the sun 365 days, thus, the angular velocity:

$\omega=\frac{2\pi}{365days}=0.0172rad/day$

if we want the angular velocity in rad/second, we need to convert the 365 days to seconds:

Firt conveting to hous

$365days(\frac{24hours}{1day} )=8760hours$

then to minutes

$8760hours(\frac{60minutes}{1hour} )=525,600minutes$

and finally to seconds

$525,600minutes(\frac{60seconds}{1minute})=31,536x10^3seconds$

thus, angular velocity in rad/second is:

$\omega=\frac{2\pi}{31,536x10^3seconds}=1.99x10^{-7}rad/second$

3 0

3 years ago

Using dimensional analysis, construct a constant, with units of length only, out of all three of the following fundamental const

ludmilkaskok [199]

The quantity with units of length only is $\sqrt{\frac{hG}{c^3}}$

Explanation:

We have to combine the following constants:

- $h$ , Planck constant, with units $[m^2][kg][s^{-1}]$

- G, the Newton's gravitational constant, with units $[m^3][kg^{-1}][s^{-2}]$

- $c$ , the speed of light, with units $[m][s^{-1}]$

The combination of these constant should have units of length only, so with meters (m).

First, we notice that $h$ has [kg] in its units, while G has $[kg^{-1}]$ in its units, so in order to make the [kg] disappear, we have to multiply them and they should have same power, so:

$hG = [m^{2+3}][kg^{1-1}][s^{-1-2}]=[m^5][s^{-3}]$

Now we have to make the seconds, [s], disappear. We do that by dividing the new quantity by $c^3$ , so that the new units are:

$\frac{hG}{c^3}=\frac{[m^5][s^{-3}]}{([m][s^{-1}])^3}=\frac{[m^5][s^{-3}]}{[m^3][s^{-3}]}=[m^2]$

We are almost done: now the quantity has units of an area, squared meters. Therefore, in order to make it have it units of length, we just take its square root:

$\sqrt{\frac{hG}{c^3}}=\sqrt{[m^2]}=[m]$

Learn more about gravitational constant:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

4 0

3 years ago

Which describes the movement of a fluid during convection?

andre [41]

The hot molecules around the heat source expands, becomes less dense, then rises. When it rises, the cooler molecules moves down to take its place. This can occur in fluid, which include gas or liquid.

6 0

3 years ago

Read 2 more answers

Interactive Solution 8.29 offers a model for this problem. The drive propeller of a ship starts from rest and accelerates at 2.3

MAXImum [283]

Answer:

Δθ = 15747.37 rad.

Explanation:

The total angular displacement is the sum of three partial displacements: one while accelerating from rest to a certain angular speed, a second one rotating at this same angular speed, and a third one while decelerating to a final angular speed.
Applying the definition of angular acceleration, we can find the final angular speed for this first part as follows:

$\omega_{f1} = \alpha * \Delta t = 2.38*e-3rad/s2*2.04e3s = 4.9 rad/sec (1)$

Since the angular acceleration is constant, and the propeller starts from rest, we can use the following kinematic equation in order to find the first angular displacement θ₁:

$\omega_{f1}^{2} = 2* \alpha *\Delta\theta (2)$

Solving for Δθ in (2):

$\theta_{1} = \frac{\omega_{f1}^{2}}{2*\alpha } = \frac{(4.9rad/sec)^{2}}{2*2.38*e-3rad/sec2} = 5044.12 rad (3)$

The second displacement θ₂, (since along it the propeller rotates at a constant angular speed equal to (1), can be found just applying the definition of average angular velocity, as follows:

$\theta_{2} =\omega_{f1} * \Delta_{t2} = 4.9 rad/s * 1.48*e3 s = 7252 rad (4)$

Finally we can find the third displacement θ₃, applying the same kinematic equation as in (2), taking into account that the angular initial speed is not zero anymore:

$\omega_{f2}^{2} - \omega_{o2}^{2} = 2* \alpha *\Delta\theta (5)$

Replacing by the givens (α, ωf₂) and ω₀₂ from (1) we can solve for Δθ as follows:

$\theta_{3} = \frac{(\omega_{f2})^{2}- (\omega_{f1}) ^{2} }{2*\alpha } = \frac{(2.42rad/s^{2}) -(4.9rad/sec)^{2}}{2*(-2.63*e-3rad/sec2)} = 3451.25 rad (6)$

The total angular displacement is just the sum of (3), (4) and (6):
Δθ = θ₁ + θ₂ + θ₃ = 5044.12 rad + 7252 rad + 3451.25 rad
⇒ Δθ = 15747.37 rad.

4 0

3 years ago

during the course of songraphic exam, you notice lateral splaying of echoes in the far field. what can you do to improve the ima

horrorfan [7]

Answer:

lateral splaying of echoes in the far field can be improved by Increasing the maximum number of transmit focal zones and optimize their location.

8 0

3 years ago