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What are the products of a fusion reaction? Check all that apply.

Free_Kalibri [48]

Energy, heavier atoms atoms, neutron and a proton are the products of a fusion reaction.

<h3>What are the products of a fusion reaction?</h3>

Nuclear fusion is a type of reaction in which two or more atomic nuclei fuse or combine to form heavy nuclei and subatomic particles such as neutrons or protons and release of energy.

So we can conclude that Energy, heavier atoms, neutron and a proton are the products of a fusion reaction.

Learn more about reaction here: brainly.com/question/11231920

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8 0

2 years ago

Eric drops a 2.20 kg water balloon that falls a distance of 45.08 m off the top of a

Marianna [84]

Answer:

972 J

Explanation:

At the bottom, all the gravitational potential energy was converted into kinetic energy. If you calculate the GPE, its value will be the same that the KE at the bottom. The GPE can be calculated this way:

GPE = mass×gravity×heigth

GPE = 2.2×9.8×45.08 ≈ 972

4 0

2 years ago

"The drawing shows three layers of different materials, with air above and below the layers. The interfaces between the layers a

My name is Ann [436]

Answer:

Angle of incidence that entered material b= 63.1°

Angle of incidence between a and b = 55.9°

Explanation: Using the formular:

n1sintheta1= n2sintheta2

The light ray which enters material B will be

1.4Sin72.8° = 1.5Sin theta

1.3373= 1.5Sintheta

sintheta = 1.3373/1.5

Sin^-1 0.8916 = Theta

63.1 = theta

When the ray hits interface with material a

1.5Sin63.1 = 1.3 Sin theta

1.3374 = 1.3Sin theta

Sintheta= 1.3374/1.3

Sin theta = 1.0877

There will be total reflection off the boundary b c because sin theta exceeded 1 in value.

The equation should be

1.4sin63.1 = 1.4 sin theta

Sin theta=72.8°

When the ray hits air-c boundary:

1.4sin72.8=1.00sin theta

Sin theta=1.3374/1 =1.3374

There is total reflection.

In material a,the ray will:

1.3sin72.8° = 1.00sin theta

There will be total reflection when the ray hits a-b boundary.

1.3sin72.8= 1.5sintheta

Sin theta= 1.2419/ 1.5

Sin theta =0.8279

Theta= Sin^-10.8279= 55.88°

When ray hits c-air boundary

1.4sin63.1= 1.00sintheta

1.2485= sin theta = Toal reflection.

Therefore when the ray of light pass through the layers of material a, b and c the boundary with air on top and bottom will be total reflection.

7 0

3 years ago

As a space shuttle moves through the dilute ionized gas of Earth's ionosphere, the shuttle's potential is typically changed by -

postnew [5]

Answer:

-2.3 × 10^-9 Coulombs(C).

Explanation:

So, we are given the following data or information or parameters that is going to help us to solve the problem effectively and efficiently;

=> " the shuttle's potential is typically changed by -1.4 V during one revolution. "

=> " Assuming the shuttle is a conducting sphere of radius 15 m".

So, in order to estimate the value for the charge we will be making use of the equation below:

Charge, C =( radius × voltage or potential difference) ÷ Coulomb's law constant.

Note that the value of Coulomb's law constant = 9 x 10^9 Nm^2 / C^2.

So, charge = { 15 × (- 1.4)} / 9 x 10^9 Nm^2 / C^2.

= -2.3 × 10^-9 Coulombs(C).

4 0

3 years ago

A flywheel with a radius of 0.300 m starts from rest and accelerates with a constant angular acceleration of 0.600 rad/s2. Compu

boyakko [2]

Answer:

0.42 m/s²

Explanation:

r = radius of the flywheel = 0.300 m

w₀ = initial angular speed = 0 rad/s

w = final angular speed = ?

θ = angular displacement = 60 deg = 1.05 rad

α = angular acceleration = 0.6 rad/s²

Using the equation

w² = w₀² + 2 α θ

w² = 0² + 2 (0.6) (1.05)

w = 1.12 rad/s

Tangential acceleration is given as

$a_{t}$ = r α = (0.300) (0.6) = 0.18 m/s²

Radial acceleration is given as

$a_{r}$ = r w² = (0.300) (1.12)² = 0.38 m/s²

Magnitude of resultant acceleration is given as

$a = \sqrt{a_{t}^{2} + a_{r}^{2}}$

$a = \sqrt{0.18^{2} + 0.38^{2}}$

$a$ = 0.42 m/s²

8 0

3 years ago