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ohaa [14]
2 years ago
14

In a nuclear power plant, uranium atoms ______

Physics
1 answer:
motikmotik2 years ago
4 0

Answer:

All nuclear power plants use nuclear fission, and most nuclear power plants use uranium atoms. During nuclear fission, a neutron collides with a uranium atom and splits it, releasing a large amount of energy in the form of heat and radiation. More neutrons are also released when a uranium atom splits.

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Two resistors of 15 and 30 ω are connected in parallel. if the combination is connected in series with a 9. 0-v battery and a 20
Stels [109]

Resistors Working Together.

Resistors are shown coupled in parallel to a voltage source in Figure 10.3.4. When all of the resistors' ends are connected to one another by a continuous wire of minimal resistance and their other ends are also connected to one another by a continuous wire of minimal resistance, the resistors are said to be in parallel. There is a constant potential drop across all resistors. Ohm's law, I=V/R, can be used to determine the current flowing through each resistor while the voltage is constant across each resistor. For instance, the headlights, radio, and other components of an automobile are linked in parallel so that each subsystem can use the entire voltage of the source and function independently. The wiring in your home or any other structure shares the same

The original circuit is shown in part a with two parallel resistors linked to a voltage source, and the equivalent circuit is shown in part b with one equivalent resistor connected to the voltage source.

learn more about  resistors brainly.com/question/22259983

#4159

3 0
1 year ago
Help with this physics task pls
cupoosta [38]

Answer:

Answers can be seen below

Explanation:

First we must explain the essential when we clear equations, and that is that if the term we need to clear is accompanied by other terms that are being added up, then those terms go to the other side of the equation to subtract if those terms are subtracting, then they go to the other side to add, if those terms are found multiplying then they go to the other side of the equation to divide and if those other terms are found dividing then they go to the other side of the equation to multiply.

(Primero debemos explicar lo esencial cuando despejamos ecuaciones, y es que si el término que necesitamos despejar va acompañado de otros términos que se están sumando, entonces esos términos van al otro lado de la ecuación para restar si esos términos están restando, luego van al otro lado para sumar, si esos términos se encuentran multiplicando luego van al otro lado de la ecuación a dividir, y si esos términos se encuentran dividiendo, pasan al otro lado de la ecuación a multiplicar.)

1 )  

t=\frac{v}{a} ; d=s*(t-t_{0} )

2)

k=\frac{2*U}{x^{2} }; T_{2}=\frac{P_{2}*V_{2}*T_{1}  }{P_{1}*V_{1}  }  \\

3)

L=\frac{F}{\pi*r*P}; d=\frac{w}{F*cos(o)}

4)

t^{2}=\frac{2*x}{g}  ; V_{2}=\frac{A_{1}*V_{1} }{A_{2} }  \\

5)

h=\frac{V}{\pi *r^{2} } ; r=\frac{t}{F*sin(o)}

6)

h=\frac{m}{(1/2)*\pi *r^{2} }  ; h_{2}=\frac{F_{2}*(1/2)*b_{1} *h_{1} }{F_{1}*(1/2)*b_{2}*h_{2}   }

7)

b=\frac{mg-ma}{v}; m=\frac{F+kx}{g*cos(o)}

8)

a=\frac{v-v_{o} }{t} ; u=\frac{m_{1}+m_{2}  }{M}

9)

v_{o}=\frac{x-\frac{1}{2}*a*t^{2}  }{t}  ; F=\frac{W+uNd}{d*cos(o)}

10)

h=\frac{E-\frac{1}{2}*m*v^{2}  }{mg} ; v_{2} ^{2} = \frac{Dk-\frac{1}{2} m*v_{1}^{2}  }{\frac{1}{2}m }

11)

N=\frac{mg*sin(o)-F}{u} ; x^{2}=\frac{W+\frac{1}{2}k*x_{1}^{2}   }{\frac{1}{2}*k }

12)

x=x_{o} +\frac{v^{2-v_{o}^{2}  } }{2a}  ;  m=\frac{P*A-F_{1}-F_{2} }{g}

13)

x_{o} = x-\frac{F}{k} ;  u=\frac{cos(o)-\frac{a}{g} }{sin(o)}

14)

t=\frac{d}{v} +t_{o} ; t_{o} = t-(\frac{v-v_{o} }{a} )

15)

F_{2}=\frac{W-F_{1} *d}{d}+F_{3}   ;  v_{2}^{2}=v_{1}^{2}+\frac{2*Dk}{m}

16)

y_{1}=y-\frac{u}{mg}  ; x^{2} = \frac{2W}{k}+x_{o} ^{2}

7 0
3 years ago
The mass luminosity relation L  M 3.5 describes the mathematical relationship between luminosity and mass for main sequence sta
ivanzaharov [21]

Answer:

(a) <u>11.3 L</u>

(b) <u>10 M</u>

Explanation:

The mass-luminosity relationship states that:

Luminosity ∝ Mass^3.5

Luminosity = (Constant)(Mass)^3.5

So, in order to find the values of luminosity or mass of different stars, we take the luminosity or mass of sun as reference. Therefore, write the equation for a star and Sun, and divide them to get:

Luminosity of a star/L = (Mass of Star/M)^3.5 ______ eqn(1)

where,

L = Luminosity of Sun

M = mass of Sun

(a)

It is given that:

Mass of Star = 2M

Therefore, eqn (1) implies that:

Luminosity of star/L = (2M/M)^3.5

Luminosity of Star = (2)^3.5 L

<u>Luminosity of Star =  11.3 L</u>

(b)

It is given that:

Luminosity of star = 3160 L

Therefore, eqn (1) implies that:

3160L/L = (Mass of Star/M)^3.5

taking ln on both sides:

ln (3160) = 3.5 ln(Mass of Star/M)

8.0583/3.5 = ln(Mass of Star/M)

Mass of Star/M = e^2.302

<u>Mass of Star = 10 M</u>

3 0
3 years ago
Chemical properties of an acid include...
Cerrena [4.2K]

and C. corrosive, increases the concentration of hydrogen ions when added to water, forms hydrogen gas when it comes in contact with a metal, and formssalt and water when added to a base.

5 0
3 years ago
Two objects are dropped from rest from the same height. Object A falls through a distance during a time t, and object B falls th
UNO [17]

Answer:

Distance covered by B is 4 times distance covered by A

Explanation:

For an object in free fall starting from rest, the distance covered by the object in a time t is

s=\frac{1}{2}gt^2

where

s is the distance covered

g is the acceleration due to gravity

t is the time elapsed

In this problem:

- Object A falls through a distance s_A during a time t, so the distance covered by object A is

s_A=\frac{1}{2}gt^2

- Object B falls through a distance s_B during a time 2t, so the distance covered by object B is

s_B=\frac{1}{2}g(2t)^2 = 4(\frac{1}{2}gt^2)=4s_A

So, the distance covered by object B is 4 times the distance covered by object A.

5 0
3 years ago
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