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Korolek [52]
3 years ago
6

Who theorized that electromagnetic radiation is emitted from a black hole due to quantum effects near the event horizon. Hint: A

famous scientist is named after this radiation who unfortunately passed away in March of last year. You will be missed :(
Physics
1 answer:
wolverine [178]3 years ago
6 0

Answer:

Stephen William Hawking

Explanation:

Stephen William Hawking was a popular theoretical physicist born on  8th January 1942.

He served as the professor of mathematics at the Cambridge university

He was director at the center for theoretical cosmology.

He has theorized that the black holes will emit radiations and the radiations is known by his name Hawking's radiation.

His book 'The Brief History of Time' appeared as one of the best seller according to the Sunday Times.

Hawking's was diagnosed for a rare disease named as motor neuron disease.

He received numerous awards and elected as the fellow of royal society.

He passed away on 14th March 2018.

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Between which two points did they travel fastest?
marusya05 [52]

Answer:

During the section CD , the speed is fastest.

Explanation:

The rate of change of distance is called speed.

Speed = distance / time

Its SI unit ism/s. It is a scalar quantity.

The slope of the distance time graph is given by the speed of the object.

Here, the speed of AB is 30/3= 10 m/s .

The speed of BC is = 0 m/s

The speed of CD is (50 - 30)/(6 - 5) = 20 m/s

So, the speed is maximum during the section CD.

7 0
3 years ago
Jet aircraft maintenance crews are required to wear protective earplugs. Members of a particular crew wear earplugs that reduce
Taya2010 [7]

Answer:

So the sound intensity level they would experience without the earplugs is 110.32dB.

Explanation:

Given data

Sound intensity by factor =215

Sound intensity level =87 dB

To find

Sound intensity level they would experience without the earplugs

Solution

First we need to find the new sound intensity level

So

I_{n}=215(10^{\frac{87}{10} } )\\I_{n}=1.08*10^{11})

The dB can be calculated as:

dB=10log(I_{n})\\

Substitute the given values

dB=10log(1.08*10^{11})\\dB=110.32dB

So the sound intensity level they would experience without the earplugs is 110.32dB.

7 0
3 years ago
If the potential across two parallel plates, separated by 4.0 cm, is 15.0 V, what is the electric field strength in volts per me
abruzzese [7]

Field strength = (15 V) / (4 cm)

Field strength = (15 V) / (0.04 meter)

Field strength = (15/0.04) (volts/meter)

<em>Field strength = 375 volts/meter </em>

3 0
3 years ago
Read 2 more answers
What type of system would allow light to enter and exit, but would keep any
andrew11 [14]
An isolated system , it does not allow any matter or energy to be exchanged
4 0
3 years ago
Read 2 more answers
An asteroid is on a collision course with Earth. An astronaut lands on the rock to bury explosive charges that will blow the ast
forsale [732]

Answer:

The maximum radius the asteroid can have for her to be able to leave it entirely simply by jumping straight up is approximately 1782.45 meters

Explanation:

Whereby the height the astronaut can jump on Earth = 0.500 m, we have the following kinematic equation;

v² = u² - 2·g·h

Where;

v = The final velocity

u = The initial velocity

g = The acceleration due to gravity ≈ 9.8 m/s²

h = The height she jumps

At the maximum height, h_{max} = 0.500 m, she jumps, v = 0, therefore, we have;

0² = u² - 2·g·h_{max}

u² = 2 × 9.8 × 0.5 = 9.8

u = √9.8 ≈ 3.13

u = 3.13 m/s

Her initial jumping velocity ≈ 3.13 m/s

Escape velocity, v_e = \sqrt{\dfrac{2 \cdot G \cdot M}{r} }

Where;

M = The mass of the asteroid

G = The Universal gravitational constant = 6.67408 × 10⁻¹¹ m³/(kg·s²)

r = The radius of the asteroid

The average density of the Earth = 5515 kg/m³

The mass of the asteroid, M = Density × Volume = 5515 kg/m³× 4/3 × π × r³

The escape velocity, she has, v_e ≈ 3.13 m/s is therefore;

3.13 = \sqrt{\dfrac{2 \times 6.67408 \times 10^{-11} \times 5515 \times \frac{4}{3} \times \pi \times r^3}{r} } = r \times \sqrt{3.084 \times 10^{-6}}

r = \dfrac{3.13}{ \sqrt{3.084 \times 10^{-6}}} \approx 1782.45

Therefore, the maximum radius of the asteroid can have for her jumping velocity to be equal to the escape velocity for her to be able to leave it entirely simply by jumping straight up = r ≈ 1782.45 meters.

7 0
3 years ago
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