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3 years ago

What is the ecliptic, and why is it tilted with respect to the celestial equator?

1 answer:

4 0

The ecliptic is the line in the celestial sphere traced out by the sun as the rotates (the path in the sky that the sun takes over head during one 24hr day). The celestial equator is just the extension of the regular equator on the surface of the earth.

This means that there will be a "tilt" between the ecliptic and the celestial equator that depends on your latitude of the observers location on the earths surface.

source: astrophysicist

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A fly travels along the x-axis. His starting point is x = 16 m and his ending point 15x = - 25 m. His flight lasts 4.0 seconds H

gogolik [260]

Explanation:

Starting position at x = 16m

Ending position at x = -25m

Time of flight = 4s

Unknown:

Distance flown = ?

Displacement = ?

Speed = ?

Velocity = ?

Solution:

To find the distance flown, we should understand that the body is moving on the x - plane;

So distance = 16 + 25 = 41m

Displacement is 41m to the left or -x axis

Speed is the distance divided by the time taken;

Speed = $\frac{distance}{time}$ = $\frac{41}{4}$ = 10.25m/s

Velocity is 10.25m/s along -x axis

5 0

2 years ago

When will the motion of an object be

mafiozo [28]

Answer:

C. The forces are balanced.

Explanation:

Every answer other than C means the object moved.

5 0

2 years ago

Please help me with this (with explanation)

Sergeeva-Olga [200]

Suppose the cyclist travels for a total time of t hours.

For 20 min = 1/3 hr, the cyclist does not move.

Over the remaining (t - 1/3) hr, the cyclist is moving at a constant speed of 22.0 km/hr, so that the cyclist would travel a distance of

x = (22.0 km/hr) • ((t - 1/3) hr) ≈ (22.0 km/hr) t - 7.33 km

If the cyclist's average speed over the total time t was 17.5 km/hr, then by the definition of average speed,

17.5 km/hr = x / t

Replace x with the distance expression from earlier:

17.5 km/hr = ((22.0 km/hr) t - 7.33 km) / t

Solve for t :

17.5 km/hr = 22.0 km/hr - (7.33 km) / t

(7.33 km) / t = 4.5 km/hr

t = (7.33 km) / (4.5 km/hr)

t ≈ 1.62963 hr

Then the distance the cyclist traveled must have been

x ≈ (22.0 km/hr) (1.62963 hr) - 7.33 km ≈ 28.5 km

and so the answer is A.

Alternatively, as soon as you arrive at

17.5 km/hr = x / t

you can instead solve for t in terms of x, then plug that into the distance equation.

t = x / (17.5 km/hr)

then

x ≈ (22.0 km/hr) (x / (17.5 km/hr)) - 7.33 km

x ≈ 1.25714 x - 7.33 km

0.25714x ≈ 7.33 km

x = (7.33 km) / 0.25714 ≈ 28.5 km

6 0

3 years ago

A 0.40-kg block is attached to the end of a horizontal ideal spring and rests on a frictionless surface. The block is pulled so

lubasha [3.4K]

Answer:

160N/m

Explanation:

According to Hooke's law which states that the extension of an elastic material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically,

F = ke where

F is the applied force

k is the spring constant

e is the extension

From the formula k = F/e

Since the body accelerates when the block is released, F = ma according to Newton's second law of motion.

The spring constant k = ma/e where

m is the mass of the block = 0.4kg

a is the acceleration = 8.0m/s²

e is the extension of the spring = 2.0cm = 0.02m

K = 0.4×8/0.02

K = 3.2/0.02

K = 160N/m

The spring constant of the spring is therefore 160N/m

4 0

3 years ago

At t=0, a train approaching a station begins decelerating from a speed of 80 mi/hr according to the acceleration function a(t)=−

vredina [299]

Answer:

a) Δx = 49.23 mi , b) Δx = 5.77 mi

Explanation:

As we have an acceleration function we must use the definition of kinematics

a = dv / dt

∫dv = ∫ a dt

we integrate and evaluators

v - vo = ∫ (-1280 (1 + 8t)⁻³ dt = -1280 ∫ (1+ 8t)⁻³ dt

We change variables

1+ 8t = u

8 dt = du

v - v₀ = -1280 ∫ u⁻³ du / 8

v -v₀ = -1280 / 8 (-u⁻²/2)

v - v₀ = 80 (1+ 8t)⁻²

We evaluate between the initial t = 0 v₀ = 80 and the final instant t and v

v- 80 = 80 [(1 + 8t)⁻² - 1]

v = 80 (1+ 8t)⁻²

We repeat the process for defining speed is

v = dx / dt

dx = vdt

x-x₀ = 80 ∫ (1-8t)⁻² dt

x-x₀ = 80 ∫ u⁻² dt / 8

x-x₀ = 80 (-1 / u)

x-x₀ = -80 (1 / (1 + 8t))

We evaluate for t = 0 and x₀ and the upper point t and x

x -x₀ = -80 [1 / (1 + 8t) - 1]

We already have the function of time displacement

a) let's calculate the position at the two points and be

t = 0 h

x = x₀

t = 0.2 h

x-x₀ = -80 [1 / (1 +8 02) -1]

x-x₀ = 49.23

displacement is

Δx = x (0.2) - x (0)

Δx = 49.23 mi

b) in the interval t = 0.2 h at t = 0.4 h

t = 0.4h

x- x₀ = -80 [1 / (1+ 8 0.4) -1]

x-x₀ = 55 mi

Δx = x (0.4) - x (0.2)

Δx = 55 - 49.23

Δx = 5.77 mi

3 0

3 years ago