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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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A Hall probe, consisting of a rectangular slab of current-carrying material, is calibrated by placing it in a known magnetic fie
Citrus2011 [14]

Answer:

(a) 0.345 T

(b) 0.389 T

Solution:

As per the question:

Hall emf, V_{Hall} = 20\ mV = 0.02\ V

Magnetic Field, B = 0.10 T

Hall emf, V'_{Hall} = 69\ mV = 0.069\ V

Now,

Drift velocity, v_{d} = \frac{V_{Hall}}{B}

v_{d} = \frac{0.02}{0.10} = 0.2\ m/s

Now, the expression for the electric field is given by:

E_{Hall} = Bv_{d}sin\theta                            (1)

And

E_{Hall} = V_{Hall}d

Thus eqn (1) becomes

V_{Hall}d = dBv_{d}sin\theta

where

d = distance

B = \frac{V_{Hall}}{v_{d}sin\theta}                      (2)

(a) When \theta = 90^{\circ}

B = \frac{0.069}{0.2\times sin90} = 0.345\ T

(b) When \theta = 60^{\circ}

B = \frac{0.069}{0.2\times sin60} = 0.398\ T

5 0
3 years ago
What is the kinetic energy of a an 80kg football player running at 8 m/s?
Ugo [173]
In the question it is already given that the football player is 80 kg.
Then the mass of the football player = 80 kg
Velocity at which the football player is running = 8 m/s
<span>Kinetic Energy = 0.5 • mass • square of velocity
Now we have to put the known data in this equation to find the actual velocity of the footballer.
</span> <span></span>So
Kinetic Energy of the footballer = 0.5 * 80 * (8 * 8)
                                                 = 0.5 * 80 * 64
                                                 = 2560
So the Kinetic energy of the footballer is 2560 joules


4 0
3 years ago
A client experiences difficulty in performing the prone iso-abs exercise. Which of the following should be suggested as an regre
butalik [34]

Answer:

a. Quadruped arm and opposite leg raise

Explanation:

Quadruped arm and opposite leg lift

- Kneel on the floor, lean forward and place your hands down.

- Keep your knees in line with your hips and hands directly under your shoulders.

- Simultaneously raise one arm and extend the opposite leg, so that they are in line with the spine.

- Go back to the starting position.

This method is usually used as an alternative to iso-abs exercise or also known as a bridge, which allows you to exercise the abdominal and spinal area at the same time.

It is also used together with other exercises for the treatment of hyperlordosis.

8 0
3 years ago
An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the
andrew-mc [135]

Answer:

50 W

Explanation:

Case 1

Power = V * I

100 = 220 * I

I = \frac{100}{220} A

Case 2

P = V * I

P = 110 * \frac{100}{220}

P = 50 W

I think the answer is 50 W

Hope it helps

8 0
3 years ago
What is the effective resistance of a car’s starter motor when 150 A flows through it as the car battery applies 12.0 V to the m
QveST [7]

Answer:

From ohms law,

V=IR

R=V/I =12.0/150 =0.08 ohm.

8 0
4 years ago
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