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castortr0y [4]
2 years ago
6

Which terrestrial planet exhibits retrograde rotation?.

Physics
1 answer:
Sati [7]2 years ago
3 0

Answer:

Planets that are farther from the sun than the earth (all but Mercury and Venus) will exhibit retrograde motion.

If the position of the planet is observed relative to the background stars, the planet will appear to move backward relative to the stars when the earth is moving in an Eastward direction faster than the planet, and the planet appears to move backwards relative to the stars

(The planet will be on the side of the earth that is opposite that of the sun)

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Two general points to remember for recovering from skids are: keep your eyes pointed in the direction you want to go and you wil
Semmy [17]
The answer is do not break, the key avoiding skids is to always smoothly apply your brakes and accelerator and to turn slowly and smoothly. Reducing of the speed before oncoming turns and once driving in possibly hazardous circumstances such as wet, icy or snow covered roadways or on roadways with loose gravel.  
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4 years ago
Small fragments of orbiting bodies that have fallen on Earth's surface are known as
kolezko [41]
There are known as B. meteorites: asteroids are much larger.  
6 0
3 years ago
A hoop and a solid disc are relased from rest
Murljashka [212]

Answer:

1) The hoop and a solid disc rolling without slipping down an incline plane.

Their final velocities are proportional to their moment of inertia.

The condition for moment of inertia: v = ωR

We will use conservation of energy.

<u>For the hoop:</u>

K_1 + U_1 = K_2 + U_2\\0 + m_hgh = \frac{1}{2}m_hv_h^2 + \frac{1}{2}I\omega_h^2 + 0

They are released from rest, so their initial kinetic energy is zero. And when they reach the bottom, their final potential energy is also zero.

The moment of inertia of a hoop is

I_h = m_hR^2

Let's continue with the energy equations:

m_h gh = \frac{1}{2}m_hv_h^2 + \frac{1}{2}(m_hR^2)(\frac{v_h^2}{R^2})\\m_hgh = \frac{1}{2}m_hv_h^2 + \frac{1}{2}m_hv_h^2\\m_hgh = m_hv_h^2\\v_h = \sqrt{gh}

Similarly <u>for the solid disk</u> with a moment of inertia of (1/2)mR^2:

K_1 + U_1 = K_2 + U_2\\m_dgh = \frac{1}{2}m_dv_d^2 + \frac{1}{2}I_d\omega_d^2\\m_dgh = \frac{1}{2}m_dv_d^2 + \frac{1}{2}(\frac{1}{2}m_dR^2)(\frac{v_d^2}{R^2})\\m_dgh = \frac{1}{2}m_dv_d^2 + \frac{1}{4}m_dv_d^2\\m_dgh = \frac{3}{4}m_dv_d^2\\v_d = \sqrt{\frac{4gh}{3}}

Comparing the final velocities, we can conclude that the solid disk reaches the bottom first.

2) The angular acceleration of the pebble is equal to the angular acceleration of the tire, since they stuck together. We can deduce the angular acceleration of the tire from the linear acceleration of the bicycle.

The kinematics equations states that

v = v_0 + at\\4.47 = 0 + 2a\\a = 2.235 ~m/s^2

where a is the linear acceleration.

The relation with the angular and linear acceleration is

a = \alpha R

where R is the radius of the tire. Since it is not given in the question, we will leave it as R.

The angular acceleration of the small pebble is

\alpha = 2.235/R ~m/s^2

4 0
4 years ago
Read 2 more answers
Lindsay is planning a flight from St. Catherines to Hamilton, which lies due west of St. Catharines.
Schach [20]

Lindsay should fly the plane in the direction [W 12.5° S] to get Hamilton.

Using Sine rule to solve this question

Sine rule => SinA/a = SinB/b = SinC/c = constant

The magnitude of wind is 50 with an angle of 60 degrees.

The magnitude of plane is 200 and the angle at which it should fly is unknown and should be θ.

One side is 50 km/hr at an angle of 60 degrees.

sin 60°/200 = sin θ / 50

50 × sin 60° = 200 × sin θ

√3/2 = 4 × sin θ

√3/8 = sin θ

sin θ = 0.2165

θ = sin⁻¹(0.2165)

θ = 12.5°

So Lindsay have to fly the plane in the direction of [W 12.5° S].

Learn more about Sine Rule here:

brainly.com/question/27174058

#SPJ10

3 0
2 years ago
G Railroad tracks are made from segments L = 79 m long at T = 20° C. When the tracks are laid, the engineers leave gaps of width
Andreyy89

Answer:

l=L\alpha(T_c-T)

Explanation:

L = Initial length of segment = 79 m

T = Normal temperature = 20^{\circ}\text{C}

l = Width to be left for expansion

\alpha = Coefficient of linear expansion of the material = 12\times 10^{-6}^{\circ}\text{C}^{-1}

T_c = Maximum temperature = 39.5^{\circ}\text{C}

\Delta T = Change in temperature = T_c-T

The expression of linear expansion is given by

l=L\alpha\DeltaT\\\Rightarrow l=L\alpha(T_c-T)

The expression for the minimum gap distance l the engineers must leave for a track rated at temperature T_c is l=L\alpha(T_c-T)

8 0
3 years ago
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