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soldier1979 [14.2K]
3 years ago
13

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.
Physics
2 answers:
Contact [7]3 years ago
7 0

Complete Question

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 *10^5 km away ,would the light first arrive on the earth in seconds

Answer:

The time it would take is  t = 1.28 \ sec

Explanation:

The objective of this question is to obtain the time the light would arrive earth

      We are told that the distance from moon to  earth is  

                  D = 3.84*10^5km = 3.84*10^5 * 1000 = 3.84*10^8m

Now generally time is mathematically represented as

                          t = \frac{D}{c}

The c here is the speed of light which has a value of c = 3*10^8 m/s

        Now substituting values

                        t = \frac{3.84 * 10^8 }{3*10^8}

                           = 1.28 \ sec

antiseptic1488 [7]3 years ago
4 0

Answer:

Incomplete question: The question is: How logn after the asteroid hit the Moon, which is 3.84x10⁵km away, would the light first arrive on Earth in seconds?

The answer is 1.28 s

Explanation:

Data given:

d = distance = 3.84x10⁵km = 3.84x10⁸m

v = velocity of light = 3x10⁸ m/s

The time will be:

t=\frac{d}{v} =\frac{3.84x10^{8} }{3x10^{8} } =1.28s

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3 years ago
A ball rolls down the hill which has a vertical height of 15 m. Ignoring friction what would be the gravitational potential ener
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a) Potential energy: 147 m [J]

The gravitational potential energy of an object is given by

U=mgh

where

m is its mass

g=9.8 m/s^2 is the acceleration of gravity

h is the height of the object above the ground

In this problem,

h = 15 m

We call 'm' the mass of the ball, since we don't know it

So, the potential energy of the ball at the top of the hill is

U=(m)(9.8)(15)=147 m (J)

b) Velocity of the ball at the bottom of the hill: 17.1 m/s

According to the law of conservation of energy, in absence of friction all the potential energy of the ball is converted into kinetic energy as the ball reaches the bottom of the hill. Therefore we can write:

U=K=\frac{1}{2}mv^2

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v is the final velocity of the ball

We know from part a) that

U = 147 m

Substituting into the equation above,

147 m = \frac{1}{2}mv^2

And re-arranging for v, we find the velocity:

v=\sqrt{2\cdot 147}=17.1 m/s

5 0
4 years ago
A ball is thrown horizontally from a window that is 15.4 meters high at a speed of 3.01 m/s. How far will the ball go before hit
tia_tia [17]

The distance travelled by the ball that is thrown horizontally from a window that is 15.4 meters high at a speed of 3.01 m/s is 5.34 m

s = ut + 1 / 2 at²

s = Distance

u = Initial velocity

t = Time

a = Acceleration

Vertically,

s = 15.4 m

u = 0

a = 9.8 m / s²

15.4 = 0 + ( 1 / 2 * 9.8 * t² )

t² = 3.14

t = 1.77 s

Horizontally,

u = 3.01 m / s

a = 0 ( Since there is no external force )

s = ( 3.01 * 1.77 ) + 0

s = 5.34 m

Therefore, the distance travelled by the ball before hitting the ground is 5.34 m

To know more about distance travelled

brainly.com/question/12696792

#SPJ1

7 0
1 year ago
A 2.98-kg object oscillates on a spring with an amplitude of 8.05 cm. Its maximum acceleration is 3.55 m/s2. Calculate the total
Aloiza [94]

Answer:

a = ω^2 A      formula for max acceleration (ignoring sign)

V = ω A         formula for max velocity

V^2 = ω^2 A^2 = a A   from first equation

E = 1/2 M V^2 = 1/2 * 2.98 * 3.55 * .0805 = .426 J

(kg * m/sec^2 * m = kg m^2 / sec^2 = Joule

6 0
2 years ago
Which of these is
Airida [17]

Answer:c

Explanation:work is force x distance. The wall doesn’t move any distance in A so that’s not work. B involves tasks that do work but I don’t think this is what they’re going for. For a short instance while kicking a ball, your foot is applying a force to the ball which moves a distance before following its own trajectory after contact with your foot. Because the contact has a short duration and isn’t instantaneous, this is work

8 0
2 years ago
Read 2 more answers
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