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Ulleksa [173]
3 years ago
8

What type of properties would a scientist need to know to correctly place in new element into a group?

Physics
1 answer:
V125BC [204]3 years ago
3 0
Placing elements into columns, groups, rows and periods that share certain properties. properties such as gas, solid and liquid determine an element physical state at room temperature. 
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Heat is added to a substance, but its temperature does not rise. Which one of the following statements provides the best explana
GalinKa [24]

Answer:

Heat is added to a substance, but its temperature does not rise. Which one of the following statements provides the best explanation for this observation? the substance must be a gas. the substance must be a non-perfect solid. the substance undergoes a change of phase. the substance has unusual thermal properties. the substance must be cooler than its environment.

7 0
3 years ago
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A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft)
Phantasy [73]

Complete Question

A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft), where V0=110 V, f=60 Hz. The power supplied to the bulb is P=V2R J/s (joules per second) and the total energy expended over a time period [0,T] (in seconds) is U  =  \int\limits^T_0 {P(t)} \, dt

Compute U if the bulb remains on for 5h

Answer:

The value is  U  =  7.563 *10^{5} \  J

Explanation:

From the question we are told that

   The power rating of the bulb is P  =  100 \  W

   The resistance is   R =  143 \ \Omega

   The  voltage is  V  =  V_o  sin [2 \pi ft]

   The  energy expanded is U  =  \int\limits^T_0 {P(t)} \, dt

   The  voltage  V_o  =  110 \  V

   The frequency is  f =  60 \  Hz

    The  time considered is  t =  5 \  h  =  18000 \  s

Generally power is mathematically represented as

             P =  \frac{V^2}{ R}

=>          P =  \frac{( 110  sin [2 \pi * 60t])^2}{ 144}

=>           P =  \frac{ 110^2 [ sin [120 \pi t])^2}{ 144}

So  

     U  =  \int\limits^T_0 { \frac{ 110^2*  [sin [120 \pi t])^2}{ 144}} \, dt

=>  U  =  \frac{110^2}{144} \int\limits^T_0 { (   sin^2 [120 \pi t]} \, dt

=>  U =  \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 2 (120\pi t)}{2} } \, dt

=>  U =  \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 240 \pi t)}{2} } \, dt

=>  U =  \frac{110^2}{144} [\frac{t}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi t)}{240 \pi} ] ]\left  | T} \atop {0}} \right.

=>  U =  \frac{110^2}{144} [\frac{t}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi t)}{240 \pi} ] ]\left  | 18000} \atop {0}} \right.

U =  \frac{110^2}{144} [\frac{18000}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi (18000))}{240 \pi} ] ]

=>   U  =  7.563 *10^{5} \  J

7 0
3 years ago
An orbiting satellite can become charged by the photoelectric effect when sunlight ejects electrons from its outer surface. Sate
Rufina [12.5K]

Answer:

the longest wavelength of incident sunlight that can eject an electron from the platinum is 233 nm

Explanation:

Given data

Φ = 5.32 eV

to find out

the longest wavelength

solution

we know that

hf = k(maximum) +Ф   ...............1

here we consider k(maximum ) will be zero because photon wavelength max when low photon energy

so hf = 0

and hc/ λ = +Ф

so λ = hc/Ф  ................2

now put value hc = 1240 ev nm and Φ = 5.32 eV

so hc = 1240 / 5.32

hc = 233 nm

the longest wavelength of incident sunlight that can eject an electron from the platinum is 233 nm

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Can I still get 5 points bc u already figured it out
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