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Alex17521 [72]
3 years ago
13

Bromine vapour is heavier than air. even so it's spreads upwards in the experiment above. Why?

Physics
1 answer:
Andrews [41]3 years ago
8 0

Answer:

Bromine is heavier than air in terms of its atomic weight. Look at the periodic table bromine is lower than oxygen, nitrogen, etc. bromine in its gaseous phase will still float in air. <em>This is due to the fact that gaseous bromine takes up a wide volume of space while having little mass.</em>

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You've recently read about a chemical laser that generates a 20.0-cm-diameter, 26.0 MW laser beam. One day, after physics class,
aksik [14]

Answer :

(a). The speed of the block is 0.395 m/s.

(b). No

Explanation :

Given that,

Diameter = 20.0 cm

Power = 26.0 MW

Mass = 110 kg

diameter = 20.0 cm

Distance = 100 m

We need to calculate the pressure due to laser

Using formula of pressure

P_{r}=\dfrac{I}{c}

P_{r}=\dfrac{P}{Ac}Put the value into the formula[tex]P_{r}=\dfrac{26.0\times10^{6}}{\pi\times(10\times10^{-2})^2\times3\times10^{8}}

P_{r}=2.75\ N/m^2

We need to calculate the force

Using formula of force

F=P\times A

F=P\times \pi r^2

Put the value into the formula

F=2.75\times\pi (0.01)^2

F=0.086\ N

We need to calculate the acceleration

Using formula of force

F=ma

Put the value into the formula

0.086=110\times a

a=\dfrac{0.086}{110}

a=0.000781\ m/s^2

a=7.81\times10^{-4}\ m/s^2

(a). We need to calculate speed of the block

Using equation of motion

v^2=u^2+2ad

Put the value into the formula

v=\sqrt{2\times7.81\times10^{-4}\times100}

v=0.395\ m/s

(b). No because the velocity is very less.

Hence, (a). The speed of the block is 0.395 m/s.

(b). No

8 0
3 years ago
Why the change of the pressure and temperature affect the velocity of the sound ​
Marysya12 [62]

<h2>\large{\underbrace{\underline{\fcolorbox{White}{aqua}{\bf{ANSWER♥︎}}}}}</h2>

<u>Air pressure has no effect at all in an ideal gas approximation. This is because pressure and density both contribute to sound velocity equally, and in an ideal gas the two effects cancel out, leaving only the effect of temperature. Sound usually travels more slowly with greater altitude, due to reduced temperature.</u>

5 0
3 years ago
If a 20 N object has been lifted 5 meters above the ground, how much gravitational potential energy does it have?
Andreyy89

The gravitational potential energy of the object is 100 J.

Gravitational potential energy stored in an object is the work done in raising the object to a height <em>h</em> against the gravitational force acting on it.

The gravitational force acting on a body is its weight mg, where m is its mass and g, the acceleration due to gravity.

Work done by a force is equal to the product of the force and the displacement made by the point of application of the force.

GPE = mgh

The weight of the object is given as 20 J and it is raised to a height of 5 m.

GPE =(mg)*h = (20 N)*(5m)=100 J

The gravitational potential energy of the object is 100 J.

5 0
3 years ago
Read 2 more answers
The amount of energy necessary to remove an electron from an atom is a quantity called the ionization energy, Ei. This energy ca
Verizon [17]

Answer:

The value is E_i  =  1.5596 *10^{-18} \  J

Explanation:

From the question we are told that

The wavelength is \lambda  =  48.2 nm  =  48.2  *10^{- 9 }\  m

The velocity is v = 2.371*10^6 \ m/s

The mass of electron is m_e  =  9.109*10^{-31} \  kg

Generally the energy of the incident light is mathematically represented as

E =  \frac{h *  c}{\lambda}

Here c is the speed of light with value c =  3.0 *10^{8} \  m/s

h is the Planck constant with value h = 6.62607015 *  10^{-34 }  J\cdot s

So

E =  \frac{6.62607015 *  10^{-34 }* 3.0 *10^{8}}{48.2  *10^{- 9 }}

=> E = 4.12 *10^{-18} \  J

Generally the kinetic energy is mathematically represented as

E_k  =  \frac{1}{2} *  m_e * v^2

=> E_k  =  \frac{1}{2} *  9.109*10^{-31} * (2.371*10^6 )^2

=> E_k  =  2.56 *0^{-18} \  J

Generally the ionization energy is mathematically represented as

E_i  =  4.12 *10^{-18} -   2.56 *0^{-18}

=>     E_i  =  1.5596 *10^{-18} \  J

7 0
3 years 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
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