All categories

3 years ago

During nuclear fission and fusion, matter that seems to disappear is actually converted into

Physics

2 answers:

iris [78.8K]3 years ago

8 0

Answer:

Energy (I need one more brainlist can i has?)

Explanation:

- Nuclear fusion occurs when two light nuclei fuse together into a heavier nucleus

- Nuclear fission occurs when a heavy, unstable nucleus breaks apart into two or more lighter nuclei

In both processes, the mass of the products is always smaller than the mass of the initial nuclei. This means that part of the initial mass has been converted into something else: into energy, which is released in the process.

The amount of energy released in the process can be calculated by using the famous Einstein's equivalence:

where m is the difference between the mass of the product and the initial mass of the nuclei, and c is the speed of light.

LenKa [72]3 years ago

6 0

Answer:

Energy

Explanation:

You might be interested in

What is the current in a 10V Circuit if the resistance is 2 ohms

VladimirAG [237]

V=IR can be changed to V/R=I so 10V/2 ohms = 5amps so 5 amps is your answer boss

8 0

3 years ago

During the first 6 years of its operation, the Hubble Space Telescope circled the Earth 37,000 times, for a total of 1,280,000,0

madreJ [45]

Answer:

$Kilometers\ in\ 1\ Orbit=\frac{1.28*10^9 Km}{3.7*10^4orbits}$

$Kilometers\ in\ 1\ Orbit=3.46*10^4 Km/Orbit$

Explanation:

Given Data:

Numbers of times Telescope cycled around the earth in 6 years=37,000 times

Total Distance traveled in 6 years by the Hubble Space Telescope=1,280,000,000 Km

Find:

Kilometers in one Orbit=?

Solution:

Kilometers in 37,000 Orbits=1,280,000,000 Km

Kilometers in 1 Orbit=1,280,000,000/37,000

In Scientific Notation:

$Kilometers\ in\ 3.7*10^4\ Orbits=1.28*10^9 Km$

$Kilometers\ in\ 1\ Orbit=\frac{1.28*10^9 Km}{3.7*10^4 orbits}$

Kilometers in 1 Orbit=34594.594 Km

Kilometers in 1 Orbit in Scientific notation:

$Kilometers\ in\ 1\ Orbit=3.46*10^4 Km$

8 0

3 years ago

If the sprinter from the previous problem accelerates at that rate for 20 m, and then maintains that velocity for the remainder

kakasveta [241]

Question:

A 63.0 kg sprinter starts a race with an acceleration of 4.20m/s square. What is the net external force on him? If the sprinter from the previous problem accelerates at that rate for 20m, and then maintains that velocity for the remainder for the 100-m dash, what will be his time for the race?

Answer:

Time for the race will be t = 9.26 s

Explanation:

Given data:

As the sprinter starts the race so initial velocity = v₁ = 0

Distance = s₁ = 20 m

Acceleration = a = 4.20 ms⁻²

Distance = s₂ = 100 m

We first need to find the final velocity (v₂) of sprinter at the end of the first 20 meters.

Using 3rd equation of motion

(v₂)² - (v₁)² = 2as₁ = 2(4.2)(20)

v₂ = 12.96 ms⁻¹

Time for 20 m distance = t₁ = (v₂ - v ₁)/a

t₁ = 12.96/4.2 = 3.09 s

He ran the rest of the race at this velocity (12.96 m/s). Since has had already covered 20 meters, he has to cover 80 meters more to complete the 100 meter dash. So the time required to cover the 80 meters will be

Time for 100 m distance = t₂ = s₂/v₂

t₂ = 80/12.96 = 6.17 s

Total time = T = t₁ + t₂ = 3.09 + 6.17 = 9.26 s

T = 9.26 s

5 0

3 years ago

As part

umka21 [38]

Answer:

Part a)

$a = 3.68 m/s^2$

Part b)

$a = 11.8 m/s^2$

Explanation:

Part a)

For force conditions of two blocks we will have

$m_1g - T = m_1 a$

$T - m_2g = m_2 a$

now from above equations we have

$(m_1 - m_2) g = (m_1 + m_2) a$

$a = \frac{m_1 - m_2}{m_1 + m_2} g$

now we know that

$m_1 = \frac{908}{9.8} = 92.65 kg$

$m_2 = \frac{412}{9.8} = 42 kg$

now from above equation we have

$a = \frac{92.65 - 42}{92.65 + 42}(9.8)$

$a = 3.68 m/s^2$

Part b)

When heavier block is removed and F = 908 N is applied at the end of the string then we have

$F - mg = ma$

$908 - 412 = 42 a$

$a = 11.8 m/s^2$

8 0

3 years ago

Which part of the electromagnetic spectrum is divided into seven ranges of wavelengths and is the part visible to the human eye

oksano4ka [1.4K]

In one of the most amazing coincidences in all of science,
the part of the electromagnetic spectrum that's visible to the
human eye is called "visible light".

Visible light is not 'divided' into anything. We mention the names
to seven of the colors in visible light. But all of the thousands of
OTHER colors that we can see are in there too, even though we
don't bother to list their names when we buzz through the rainbow
in the third grade.

6 0

3 years ago

Read 2 more answers