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3 years ago

PLEASEE HELP WILLL GIVE 15+ POINTS AND BRAINLIEST PLEASEEEEEEEEEEEEEEEEEEEEEEE

Physics

1 answer:

Dovator [93]3 years ago

4 0

um you can't really get help on this it's mostly you sorry

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Which are the following are pure substance?

Jobisdone [24]

A pure substance is chemically combined

Therefore a compound is a pure substance because it cant be broken down physically

All elements are pure substances, along with tin, diamond, and etc...

6 0

4 years ago

Read 2 more answers

A carpenter builds an exterior house wall with a layer of wood 3.0 cm thick on the outside and a layer of Styrofoam insulation 2

Leno4ka [110]

Answer:

A. T=15.54 °C

B. Q/A= 0.119 W/m2

Explanation:

To solve this problem we need to use the Fourier's law for thermal conduction:

$Q= kA\frac{dT}{dx}$

Here, the rate of flow per square meter must be the same through the complete wall. Therefore, we can use it to find the temperature at the plane where the wood meets the Styrofoam as follows:

$\frac{Q}{A} =\frac{T_1-T_0}{d_w}k_w=\frac{T_2-T_1}{d_s}k_s\\T_1(\frac{k_w}{d_w}+\frac{k_s}{d_s})=T_2\frac{k_s}{d_s}+T_0\frac{k_w}{d_w}\\T_1=\frac{T_2\frac{k_s}{d_s}+T_0\frac{k_w}{d_w}}{\frac{k_w}{d_w}+\frac{k_s}{d_s}}\\T_1= 15.54 \°C$

Then, to find the rate of heat flow per square meter, we have:

$\frac{Q}{A}=\frac{T_1-T_0}{d_w}k_w=0.119 \frac{W}{m^2}\\\frac{Q}{A}=\frac{T_2-T_1}{d_s}k_s= 0.119 \frac{W}{m^2}$

$T_0: Temperature \ in \ the \ house\\T_1: Temperature \ at \ the \ plane \ between \ wood \ and \ styrofoam\\T_2: Temperature \ outside\\k_w: k \ for \ wood\\d_w: wood \ thickness\\k_s: k \ for \ styrofoam\\d_s: styrofoam \ thickness$

7 0

3 years ago

Equation of uniformly accelerated motion

Romashka [77]

This means acceleration a is constant.

Let

a) vo be the initial speed, at t=0

b) v be the final speed after time t

c) d distance travelled in time t

Then we have:

a) v=vo+a×t

b) v²=vo²+2×a×d (Galilei's equation)

c) d=vo×t+a×t²/2

d) average speed vm=(vo+v)/2

3 0

4 years ago

The Ha line of the Balmer series is emitted in the transition from n-3 to n 2. Compute the wavelength of this line for H and 2H.

Citrus2011 [14]

Explanation:

According to Rydberg's formula, the wavelength of the balmer series is given by:

$\frac{1}{\lambda}=R(\frac{1}{2^2}-\frac{1}{3^2})$

R is Rydberg constant for an especific hydrogen-like atom, we may calculate R for hydrogen and deuterium atoms from:

$R=\frac{R_{\infty}}{(1+\frac{m_e}{M})}$

Here, $R_{\infty}$ is the "general" Rydberg constant, $m_e$ is electron's mass and M is the mass of the atom nucleus

For hydrogen, we have, $M=1.67*10^{-27}kg$ :

$R_H=\frac{1.09737*10^7m^{-1}}{(1+\frac{9.11*10^{-31}kg}{1.67*10^{-27}kg})}\\R_H=1.09677*10^7m^{-1}$

Now, we calculate the wavelength for hydrogen:

$\frac{1}{\lambda}=R_H(\frac{1}{2^2}-\frac{1}{3^2})\\\lambda=[R_H(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=[1.0967*10^7m^{-1}(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=6.5646*10^{-7}m=656.46nm$

For deuterium, we have $M=2(1.67*10^{-27}kg)$ :

$R_D=\frac{1.09737*10^7m^{-1}}{(1+\frac{9.11*10^{-31}kg}{2*1.67*10^{-27}kg})}\\R_D=1.09707*10^7m^{-1}\\\\\lambda=[R_D(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=[1.09707*10^7m^{-1}(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=6.5629*10^{-7}=656.29nm$

5 0

3 years ago

Consider an experimental setup where charged particles (electrons or protons) are first accelerated by an electric field and the

MissTica

Answer:

V = 10581.59 V

Explanation:

To solve this problem, we have to take into the relation between the kinetic energy of an electron and the energy generated by the potential difference of an electric field. This relation is given by

$E_{k} = E_{V}$

$\frac{1}{2}mv^{2} = eV$

where m is the mass of the electron, v is the velocity of the electron, e is the electric charge and V is the potential difference that accelerated the electron.

By taking V from the last expression we have:

$V = \frac{mv^{2}}{2e} = \frac{(9.1*10^{-31} )(6.1*10^{7})^{2}}{2*(-1.6*10^{-19})}} = 10581.59 V$

where we have taken that

me = 9.1*10^(-31) Kg

e = -1.6*10^(-19) C

v = 6.1*10^(7)

I hope this is useful for you

Regards

6 0

4 years ago