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2 years ago

14

What happens during the apparent retrograde motion of a planet?

1 answer:

anygoal [31]2 years ago

6 0

Answer:

The planet will move from east to west for a couple of months in the night sky.

Explanation:

Retrograde motion is an optical effect due to the fact that Earth rotates more quickly than the planet that apparently has a retrograde motion in the sky.

For example, Saturn has a slower speed in its orbit around the Sun. That means that the Earth will pass it, and that will give the effect that the planet is moving backward. That same scenario can be seen between two cars on a highway, the faster car will see the slower car when it passes as it is moving for a short fragment of time in backward.

Remember that the planets in the night sky move from west to east, in the case of a planet with retrograde motion, it will move from east to west for a couple of months.

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A disk between vertebrae in the spine is subjected to a shearing force of 600 N. Find its shear deformation, taking it to have a

Verizon [17]

Answer:

The shear deformation is $\Delta x=3.34\times10^{-6}\ m$ .

Explanation:

Given that,

Shearing force F = 600 N

Shear modulus $S = 1\times10^{9}\ N/m^2$

length = 0.700 cm

diameter = 4.00 cm

We need to find the shear deformation

Using formula of shear modulus

$S=\dfrac{Fl_{0}}{A\Delta x}$

$\Delta x=\dfrac{Fl_{0}}{(\dfrac{\pi d^2}{4})S}$

$\Delta x=\dfrac{4Fl_{0}}{\pi d^2 S}$

Put the value into the formula

$\Delta x=\dfrac{4\times600\times0.700\times10^{-2}}{3.14\times1\times10^{9}\times(4.00\times10^{-2})^2}$

$\Delta x=3.34\times10^{-6}\ m$

Hence, The shear deformation is $\Delta x=3.34\times10^{-6}\ m$ .

7 0

2 years ago

A 12-kg projectile is launched with an initial vertical speed of 20 m/s. It rises to a maximum height of 18 m above the launch p

kirill115 [55]

Answer:

Change in mechanical energy, $\Delta E=283.2\ J$

Explanation:

It is given that,

Mass of the projectile, m = 12 kg

Speed of the projectile, v = 20 m/s

Maximum height, h = 18 m

Initially, the projectile have only kinetic energy. it is given by :

$K=\dfrac{1}{2}mv^2$

$K=\dfrac{1}{2}\times 12\ kg\times (20\ m/s)^2$

K = 2400 J

Finally, it have only potential energy. it is given by :

P = mgh

$P=12\ kg\times 9.8\ m/s^2\times 18\ m$

P =2116.8 J

The change in mechanical energy is given by :

$\Delta E=K-P$

$\Delta E=2400-2116.8$

$\Delta E=283.2\ J$

So, the change in mechanical energy is 283.2 J. Hence, this is the required solution.

8 0

2 years ago

15 POINTS!!! QUICK PLEASE

Margaret [11]

Answer:

The net force is 15 newtons

The direction is to the right

Explanation:

Hope this helps

6 0

2 years ago

Which processes cause rocks to be exposed<br> at Earth's surface? SC.7..6.2

seropon [69]

Answer:

weathering

Explanation:

5 0

2 years ago

Anna Litical analyzes the force between a planet and its moon, varying the mass of

Travka [436]

Answer:

Trial 1 is the largest, trial 3 is the smallest

Explanation:

Given:

<em>Trial 1</em>

M₁ = 6·10²² kg

d₁ = 3 500 km = 3.5·10⁶ м

<em>Trial 2</em>

M₂ = 6·10²² kg

d₂ = 7 000 km = 7·10⁶ м

<em>Trial 3</em>

M₃ = 3·10²² kg

d₃ = 7 000 km = 7·10⁶ м

___________

F - ?

Gravitational force:

F₁ = G·m·M₁ / d₁² = m·6.67·10⁻¹¹·6·10²² / (3.5·10⁶)² = 0.37·m (N)

F₂ = G·m·M₂ / d₂² = m·6.67·10⁻¹¹·6·10²² / (7·10⁶)² = 0.08·m (N)

F₃ = G·m·M₃ / d₃² = m·6.67·10⁻¹¹·3·10²² / (7·10⁶)² = 0.04·m (N)

Trial 1 is the largest, trial 3 is the smallest

5 0

6 months ago