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3 years ago

the brightness of stars as they appear from earth is measured by _______ magnitude. a. light b. apparent c. relative d. absolute

Physics

1 answer:

xxTIMURxx [149]3 years ago

8 0

How bright a star appears from Earth is the star's apparent magnitude.

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When would you use a compound microscope?

Vladimir [108]

a compound microscope is used for viewing samples at high magnification<span> 40 - 1000x, which is achieved by the combined effect of two sets of lenses: the ocular lens in the eyepiece and the objective lenses close to the sample.</span>

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3 years ago

Air at 400 kPa, 980 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occ

Angelina_Jolie [31]

Answer:

a). $\frac{\dot{W}}{m}= 311$ kJ/kg

b). $\frac{\dot{\sigma _{gen}}}{m}=0.9113$ kJ/kg-K

Explanation:

a). The energy rate balance equation in the control volume is given by

$\dot{Q} - \dot{W}+m(h_{1}-h_{2})=0$

$\frac{\dot{Q}}{m} = \frac{\dot{W}}{m}+m(h_{1}-h_{2})$

$\frac{\dot{W}}{m}= \frac{\dot{Q}}{m}+c_{p}(T_{1}-T_{2})$

$\frac{\dot{W}}{m}= -30+1.1(980-670)$

$\frac{\dot{W}}{m}= 311$ kJ/kg

b). Entropy produced from the entropy balance equation in a control volume is given by

$\frac{\dot{Q}}{T_{boundary}}+\dot{m}(s_{1}-s_{2})+\dot{\sigma _{gen}}=0$

$\frac{\dot{\sigma _{gen}}}{m}=\frac{-\frac{\dot{Q}}{m}}{T_{boundary}}+(s_{2}-s_{1})$

$\frac{\dot{\sigma _{gen}}}{m}=\frac{-\frac{\dot{Q}}{m}}{T_{boundary}}+c_{p}ln\frac{T_{2}}{T_{1}}-R.ln\frac{p_{2}}{p_{1}}$

$\frac{\dot{\sigma _{gen}}}{m}=\frac{-30}{315}+1.1ln\frac{670}{980}-0.287.ln\frac{100}{400}$

$\frac{\dot{\sigma _{gen}}}{m}=0.0952+0.4183+0.3978$

$\frac{\dot{\sigma _{gen}}}{m}=0.9113$ kJ/kg-K

5 0

3 years ago

The force on the wire is directed?

USPshnik [31]

Answer: A

Explanation:

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1 year ago

Mary and her younger brother Alex decide to ride the 17-foot-diameter carousel at the State Fair. Mary sits on one of the horses

lorasvet [3.4K]

Answer:

$\omege_A=\omega_M$

$v_M=1.6v_A$

Explanation:

A denotes Alex

M denotes Mary

r = Distance from center

Mary and Alex will have the equal displacements in equal interval of time as they are in uniform circular motion. So,

$\omega_A=\omega_M$

Tangential speed speed is given by

$\dfrac{v_M}{v_A}=\dfrac{r_M\omega_M}{r_A\omega_A}\\\Rightarrow \dfrac{v_M}{v_A}=\dfrac{r_M}{r_A}\\\Rightarrow \dfrac{v_M}{v_A}=\dfrac{8}{5}\\\Rightarrow \dfrac{v_M}{v_A}=1.6\\\Rightarrow v_M=1.6v_A$