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3 years ago

12

Choose two objects from the illustration. Based on the scene, how might these objects be in motion and a reference point at the

same time? Please answer Assignment is due in 30 minutes

1 answer:

Tanya [424]3 years ago

8 0

Answer:the sun and the person

Explanation:

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What is the (magnitude of the) centripetal acceleration (as a multiple of g=9.8~\mathrm{m/s^2}g=9.8 m/s 2 ) towards the Eart

Wittaler [7]

Answer:

The centripetal acceleration as a multiple of $g=9.8 m/s^{2}$ is $1.020x10^{-3}m/s^{2}$

Explanation:

The centripetal acceleration is defined as:

$a = \frac{v^{2}}{r}$ (1)

Where v is the velocity and r is the radius

Since the person is standing in the Earth surfaces, their velocity will be the same of the Earth. That one can be determined by means of the orbital velocity:

$v = \frac{2 \pi r}{T}$ (2)

Where r is the radius and T is the period.

For this case the person is standing at a latitude $71.9^{\circ}$ . Remember that the latitude is given from the equator. The configuration of this system is shown in the image below.

It is necessary to use the radius at the latitude given. That radius can be found by means of trigonometric.

$\cos \theta = \frac{adjacent}{hypotenuse}$

$\cos \theta = \frac{r_{71.9^{\circ}}}{r_{e}}$ (3)

Where $r_{71.9^{\circ}}$ is the radius at the latitude of $71.9^{\circ}$ and $r_{e}$ is the radius at the equator ( $6.37x10^{6}m$ ).

$r_{71.9^{\circ}}$ can be isolated from equation 3:

$r_{71.9^{\circ}} = r_{e} \cos \theta$ (4)

$r_{71.9^{\circ}} = (6.37x10^{6}m) \cos (71.9^{\circ})$

$r_{71.9^{\circ}} = 1.97x10^{6} m$

Then, equation 2 can be used

$v = \frac{2 \pi (1.97x10^{6} m)}{24h}$

Notice that the period is the time that the Earth takes to give a complete revolution (24 hours), this period will be expressed in seconds for a better representation of the velocity.

$T = 24h . \frac{3600s}{1h}$ ⇒ $84600s$

$v = \frac{2 \pi (1.97x10^{6} m)}{84600s}$

$v = 146.31m/s$

Finally, equation 1 can be used:

$a = \frac{(146.31m/s)^{2}}{(1.97x10^{6}m)}$

$a = 0.010m/s^{2}$

Hence, the centripetal acceleration is $0.010m/s^{2}$

To given the centripetal acceleration as a multiple of $g=9.8 m/s^{2}$ it is gotten:

$\frac{0.010m/s^{2}}{9.8 m/s^{2}} = 1.020x10^{-3}m/s^{2}$

6 0

3 years ago

Consider a sound wave moving through the air modeled with the equation s(x, t) = 5.00 nm cos(60.00 m−1x − 18.00 ✕ 103 s−1t). Wha

GaryK [48]

Answer:

Shortest time = 58.18 × 10^(-6) s

Explanation:

We are given;

s(x,t) = 5.00 nm cos((60.00 m^(−1)x) − (18.00 X 10³ s^(−1)t))

Let us set x = 0 as origin.

Now, for us to find the time difference, we need to solve 2 equations which are;

s(x,t) = 5.00 nm cos((60.00 m^(−1)x) − (18.00 X 10³ s^(−1)t1))

And

s(x,t) = 5.00 nm cos((60.00 m^(−1)x) − (18.00 X 10³ s^(−1)t2))

Now, since the wave starts from maxima at time at t = 0, the required time would be the difference (t2 - t1)

Thus, the solutions are;

t1 = (1/(18 × 10³)) cos^(-1) (2.5/5)

And

t2 = (1/(18 × 10³)) cos^(-1) (-2.5/5)

Angle of the cos function is in radians, thus;

t1 = 58.18 × 10^(-6) s

t2 = 116.36 × 10^(-6) s

So,

Required time = t2 - t1 = (116.36 × 10^(-6) s) - (58.18 × 10^(-6) s) = 58.18 × 10^(-6) s

4 0

3 years ago

Lelu [443]

The frequency of the wave will not change

5 0

3 years ago

Read 2 more answers

When the wind kicks up dust and sand, the dust grains are charged. The small grains tend to get a negative charge, and the large

diamong [38]

Answer:

Explanation:

Small grains are negatively charged by the wind while big grains is positively charged and remains at the ground . This process creates an electric field due to the presence of oppositely charged particles.

When ever electric field exists it is directed from a positive charge to a negative charge so the here electric field is towards an upwards direction.

4 0

3 years ago

We know we have exerted of force even when we have done no work this is called _____

zheka24 [161]

Answer: The correct answer is zero work done.

Explanation:

Work is said to be done when the object moves through a distance when the force is applied to the object.

If the object does not move a distance even the force is exerted on the object then the work done is zero in this case.

Therefore, when the force is exerted even when no work is done then this is called zero work done.

5 0

3 years ago