How does an annular eclipse differ from a total eclipse.

Physics

1 answer:

konstantin123 [22]2 years ago

7 0

Answer:

Solar eclipses may be classified as either total, in which the Moon completely covers the Sun, or annular, in which the Moon obscures all but an outer ring of the Sun.

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A rock with a mass of 8 kg falls straight down from a height of 7 m. What work is done?

fenix001 [56]

F=MA
F=(8 kg)(9.8 m/s)
F= 78.4 N
W=FD
W=(78.4 N)(7 m)
W=548.8 J
How this helps

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3 years ago

In which situation is a person doing work on an object?

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Answer : When a person applies force and covers distance then the person is doing work on an object.

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Find the coefficient of kinetic friction μk. express your answer in terms of some or all of the variables d1, d2, and θ.

Andre45 [30]

<span>internet tension = mass * acceleration internet tension = 23 – Friction tension = 14 * acceleration Friction tension = µ * 14 * 9.8 = µ * 137.2 23 – µ * 137.2 = 14 * acceleration Distance = undemanding speed * time undemanding speed = ½ * (preliminary speed + very final speed) Distance = ½ * (preliminary speed + very final speed) * time Distance = 8.a million m, preliminary speed = 0 m/s, very final speed = a million.8 m/s 8.a million = ½ * (0 + a million.8) * t Time = 8.a million ÷ 0.9 = 9 seconds Acceleration = (very final speed – preliminary speed) ÷ time Acceleration = (a million.8 – 0) ÷ 9 = 0.2 m/s^2 23 – µ * 137.2 = 14 * 0.2 resolve for µ</span>

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(d) Suppose you use a spring to launch a payload horizontally from the asteroid so that the payload ends up far from the asteroi

nydimaria [60]

Answer:

ks= 133.2 N/m

Explanation:

Assuming that we can neglect the gravitational potential energy of the mass, and that no other forces acting on the payload, total mechanical energy must be conserved.
This energy, at any time, is part elastic potential energy (stored in the spring) and part kinetic energy.
When the spring is initially compressed, the payload is at rest, so all energy is elastic potential.
Once the spring has returned to its natural state, all this elastic potential energy must have been turned into kinetic energy.
If the payload is launched horizontally, and no gravity is present,this means that its final speed will be horizontal only also, according to Newton's First Law.
So, we can write the following equation:

$\Delta U + \Delta K = 0 (1)$

where ΔU = -1/2*k*(Δx)² (2)
and ΔK = 1/2*m*v² (3)
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$k_{s} =\frac{m*v^{2}}{\Delta x^{2}} = \frac{29 kg*(3m/s)^{2}}{(1.4m)^{2}} = 133.2 N/M (4)$

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