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Elenna [48]
2 years ago
6

Why is nuclear energy able to be used for practical purposes? the reactions are controlled to regulate energy output. the reacti

ons are uncontrolled for maximum energy output. all of the products are fed back into the reaction to keep it going. all of the products are immediately removed to inhibit more reactions.
Physics
1 answer:
olasank [31]2 years ago
7 0

Nuclear energy is able to be used for practical purposes because ; ( A ) The reactions are controlled to regulate energy output

<h3>Nuclear energy generation </h3>

Nuclear energy is a clean energy source which has a zero-emission, nuclear energy is used to generate electricity/power following a controlled reaction known as fission of uranium atoms to produce heat energy required to drive a turbine to produce electricity

Hence we can conclude that Nuclear energy is able to be used for practical purposes because ; ( A ) The reactions are controlled to regulate energy output.

Learn more about Nuclear energy : brainly.com/question/11963586

#SPJ4

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Mrs. Smith can walk 1.4 m/s. If it takes her 8.5 seconds to get to the teacher lounge, how far is the teacher lounge from her ro
AlexFokin [52]

Answer:

6 meters away

Explanation:

6*1.4= 8.4 which is pretty close

5 0
3 years ago
A result of chemical change is
Tcecarenko [31]
We're u can never put it back together
4 0
3 years ago
Can anyone help me? (physics)
Masja [62]

Answer:

The initial velocity of the golf is 15.7 m/s.

The direction of the golf is 57°.

Explanation:

The following data were obtained from the question:

Time of flight (T) = 2.7 secs

Range (R) = 23 m

Acceleration due to gravity (g) = 9.8 m/s²

Initial velocity (u) =.?

Direction (θ) =.?

T = 2U Sine θ /g

2.7 = 2 × U × Sine θ /9.8

Cross multiply

2.7 × 9.8 = 2 × U × Sine θ

26.46 = 2 × U × Sine θ

Divide both side by 2 × Sine θ

U = 26.46 /2 Sine θ

U = 13.23 / Sine θ ... (1)

R = U² Sine 2θ /g

23 = U² Sine 2θ / 9.8

U = 13.23 / Sine θ

23 = (13.23/ Sine θ)² Sine 2θ / 9.8

23 = (175.0329 / Sine² θ) × Sine 2θ / 9.8

23 = 17.8605/Sine² θ × Sine 2θ

Recall:

Sine 2θ = 2SineθCosθ

23 = 17.8605/ Sine² θ × 2SineθCosθ

23 = 17.8605/ Sine θ × 2Cosθ

23 = 35.721 Cos θ /Sine θ

Cross multiply

23 × Sine θ = 35.721 Cos θ

Divide both side by 23

Sine θ = 35.721 Cos θ /23

Sine θ = 1.5531 × Cos θ

Divide both side by Cos θ

Sine θ /Cos θ = 1.5531

Recall:

Sine θ /Cos θ = Tan θ

Sine θ /Cos θ = 1.5531

Tan θ = 1.5531

Take the inverse of Tan

θ = Tan¯¹ (1.5531)

θ = 57°

Therefore, the direction of the golf is 57°

Thus, the initial velocity can be obtained as follow:

U = 13.23 / Sine θ

θ = 57°

U = 13.23 / Sine 57

U = 13.23/0.8387

U = 15.7 m/s

Therefore, the initial velocity of the golf is 15.7 m/s

8 0
3 years ago
When light passes through an object unchanged, scientists call that process _____.
Nataly [62]

<u>Answer</u>:

When light passes through an object unchanged, scientists call that process Transmission.

<u>Explanation</u>:

Transmission is the process where all the light that is passed through the material moves via the material without being absorbed. The Transmission depends on the affected radiation.The Transmittance of the medium is defined as the ratio between transmitted radiant power and incident radiant power. The light that is passed through the medium and not reflected will be either scattered or reflected. The light can be transmitted only through transparent or translucent material. Opaque object does not allows transmission of light.

5 0
4 years ago
Two point charges are fixed on the y axis: a negative point charge q1 = -25 μC at y1 = +0.18 m and a positive point charge q2 at
dedylja [7]

Answer:

50.91 \mu C

Explanation:

The magnitude of the net force exerted on q is known, we have the values and positions for q_{1} and q. So, making use of coulomb's law, we can calculate the magnitude of the force exerted byq_{1} on q. Then we can know the magnitude of the force exerted by q_{2} about q, finally this will allow us to know the magnitude of q_{2}

q_{1} exerts a force on q in +y direction, and q_{2} exerts a force on q in -y direction.

F_{1}=\frac{kq_{1} q }{d^2}\\F_{1}=\frac{(8.99*10^9)(25*10^{-6}C)(8.4*10^{-6}C)}{(0.18m)^2}=58.26 N\\

The net force on q is:

F_{T}=F_{1} - F_{2}\\25N=58.26N-F_{2}\\F_{2}=58.26N-25N=33.26N\\\mid F_{2} \mid=\frac{kq_{2}q}{d^2}

Rewriting for q_{2}:

q_{2}=\frac{F_{2}d^2}{kq}\\q_{2}=\frac{33.26N(0.34m)^2}{8.99*10^9\frac{Nm^2}{C^2}(8.4*10^{-6}C)}=50.91*10^{-6}C=50.91 \mu C

8 0
3 years ago
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