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Lena [83]
3 years ago
14

In the year 1178. five monks at Canterbury Cathedral in England observed what appeared to be an asteroid colliding with the moon

, causing a red glow in and around it. It is hypothesized that this event created the crater Giordano Bruno, which is right on the edge of the area we can usually see from Earth. How long after the asteroid hit the Moon, which is 3.84 x 10^5 km away, would the light first arrive on Earth in seconds?
Physics
1 answer:
Lubov Fominskaja [6]3 years ago
8 0

Answer: 1.28 sec

Explanation:

Assuming that the glow following the collision was produced instantaneously, as the light propagates in a straight line from Moon to the Earth at a constant speed, we can get the time traveled by the light applying velocity definition as follows:

V = ∆x / ∆t

Solving for ∆t, we have:

∆t = ∆x/v = ∆x/c = 3.84 108 m / 3.8 108 m/s = 1.28 sec

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You observe a hockey puck of mass 0.12 kg, traveling across the ice at speed 18.3 m/sec. The interaction of the puck and the ice
Galina-37 [17]

The stopping distance is 143.1 m

Explanation:

First of all, we have to find the acceleration of the hockey puck. This can be done by using Newton's second law of motion:

\sum F =ma

where

\sum F = F_f = -0.14 N is the net force acting on the puck (the force of friction, negative because it acts in a direction opposite to the direction of motion)

m = 0.12 kg is the mass of the puck

a is the acceleration

Solving for a,

a=\frac{\sum F}{m}=\frac{-0.14}{0.12}=-1.17 m/s^2

The motion of the puck is a uniformly accelerated motion, therefore we can use the following suvat equation:

v^2-u^2=2as

where:

v = 0 is the final velocity (the puck comes to a stop)

u = 18.3 m/s is the initial velocity

a=-1.17 m/s^2 is the acceleration

s is the stopping distance

And solving for s, we find

s=\frac{v^2-u^2}{2a}=\frac{0-(18.3)^2}{2(-1.17)}=143.1 m

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

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#LearnwithBrainly

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3 years ago
PLEASE ANSWER ASAP FOR 75 POINTS!!!!
makkiz [27]

λ = c : f

λ = 3 x 10⁸ : 1.05 x 10⁸

λ = 2.86 m

E = hf

h = Planck's constant (6.626.10⁻³⁴ Js)

E = 6.626.10⁻³⁴ x 2.86

E = 1.896 x 10⁻³³ J

λ = 3 x 10⁸ : 1.011 x 10⁸

λ = 2.97 m

E = hf

h = Planck's constant (6.626.10⁻³⁴ Js)

E = 6.626.10⁻³⁴ x 2.97

E = 1.97 x 10⁻³³ J

λ = 3 x 10⁸ : 1.05 x 10⁸

λ = 2.96 m

E = hf

h = Planck's constant (6.626.10⁻³⁴ Js)

E = 6.626.10⁻³⁴ x 2.96

E = 1.96 x 10⁻³³ J

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There can be an experiment run using two or more cleansers at the same time as the manufacturers' one
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