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3 years ago

Obliquity describes:

Physics

1 answer:

Pachacha [2.7K]3 years ago

4 0

Answer:

c. the tilt of the axis of rotation with respect to the Plane of the Ecliptic

Explanation:

The inclination of the ecliptic (or known only as obliqueness) refers to the angle of the axis of rotation with respect to a perpendicular to the plane of the eclipse. He is responsible for the seasons of the year that the planet Earth lends. It is not constant but changes through the movement of nutation. The terrestrial plane of Ecuador and the ecliptic intersect in a line that has an end at the point of Aries and at the diametrically opposite point of Libra.

When the Sun crosses the Aries, the spring equation occurs (between March 20 and 21, the beginning of spring in the northern hemisphere and the early autumn of the southern hemisphere), and from which the Sun is in the North Hemisphere; Pound until you reach the point of the autumn equinox (around September 22-23, beginning fall in the northern hemisphere and spring in the southern hemisphere).

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A hill is 132 m long and makes an angle of 12.0 degrees with the horizontal. As a 54 kg jogger runs up the hill, how much work d

TEA [102]

Answer:

14523.55J

Explanation:

The work done by the jogger against gravity is given by the following equation;

$W=mgh.................(1)$

where m is the mass, g is acceleration due to gravity taken as $9.8m/s^2$ and h is the height of the hill.

Since the length of the hill is 132m and it is inclined at 12 degrees to the horizontal, the height is thus given as follows;

$h=132sin12^o\\h=27.44m$

Substituting this into equation (1) with all other necessary parameters, we obtain the following;

$W=54*9.8*27.44\\W=14523.55J$

4 0

3 years ago

How long would it take for a sound impulse to travel through a copper Rod 20km

Nana76 [90]

Not sure.can you give me a clue?

8 0

3 years ago

A uniform rod is 2.0 m long. The rod is pivoted about a horizontal, frictionless pin through one end. The moment of inertia of t

emmasim [6.3K]

Answer:

Angular acceleration = 6.37rad/sec²

Approximately, Angular acceleration =

6.4 rad/sec²

Explanation:

Length of the rod = 2.0m long

Inclination of the rod (horizontal) = 30°

Mass of the rod is not given so we would refer to it as = M

Rotational Inertia of the Rod(I) = 1/3ML²

Angular Acceleration = ?

There is an equation that shows us the relationship between Torque and Angular acceleration.

The equation is :

Torque(T) = Inertia × Angular Acceleration

Angular acceleration = Torque ÷ Inertia

Where:

Torque = L/2(MgCosθ)

Where M = Mass

L = Length = 2.0m

θ = Inclination of the rod (horizontal) = 30°

g = Acceleration due to gravity = 9.81m/s²

Inertia = 1/3ML²

Angular Acceleration = (Mass × g × Cos (30°) × (L÷2)) ÷ 1/3ML²

Angular Acceleration =

(3 × g × cos 30°) ÷ 2× L

Angular Acceleration = (3 × 9.81m/s² × cos 30°) ÷ 2× L

Angular Acceleration = 3 × 9.81m/s² × cos 30°) ÷ 2× 2.0m

Angular Acceleration = 6.37rad/sec²

Approximately Angular Acceleration =

6.4rad/sec²

5 0

3 years ago

At what phase would you expect to find extremely high and low tides?

creativ13 [48]

Answer:

full moon and new moon

4 0

3 years ago

Determine the amount of work done by the engine of a car with a mass of 2500 kg when it accelerates from 45 mph to 65 mph. (Use

Y_Kistochka [10]

Answer:

amount of work done, W = 549.36 kJ

Given:

mass of a car engine, m = 2500 kg

initial velocity, u = 45 mph

final velocity, v = 65 mph

1 mile = 1609

Solution:

We know that 1 hour = 3600 s

Now, velocities in m/s are given as:

u = 45 mph = $\frac{45\times 1609}{3600}$ = 20.11 m/s

v = 65 mph = $\frac{65\times 1609}{3600}$ = 29.05 m/s

Now, the amount of work done, W is given by the change in kinetic energy of the car and is given by:

W = $\frac{1}{2}m\Delta v^{2}$

W = $\frac{1}{2}m\times (v^{2} - u^{2})$

W = $\frac{1}{2}2500\times (29.05^{2} - 20.11^{2})$

W = 549.36 kJ

3 0

3 years ago