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guapka [62]
3 years ago
13

the metal wire in an incandescent lightbulb glows when the light is switched on and stops glowing when it is switched off. the s

imple process in which kind of a change
Physics
1 answer:
Travka [436]3 years ago
3 0
When the metal wire in an incandescent lightbulb glows when the light is switched on and stops glowing when it is switched off, this is an example of resistance, which provides light and heat. 
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the coefficient of static friction between a 40 kg picnic table and the ground below is .43. what is the greatest horizontal for
Gre4nikov [31]
The force equals the coefficient of static friction times the weight. Use gravity g=9.8 m/s^2
0.43*40*9.8=16.856 N
4 0
4 years ago
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The number ocean waves that pass a buoy in one second is _ of the wave
mr_godi [17]
The number of ocean waves that pass a buoy in one second is the frequency of the <span>wave. The crest of a transverse wave is its highest point. </span>
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3 years ago
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Which elements are metalloids
zhannawk [14.2K]

The six commonly recognised metalloids are boron, silicon, germanium, arsenic, antimony, and tellurium....

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4 years ago
A spaceship moves radially away from Earth with acceleration 29.4 m/s 2 (about 3g). How much time does it take for sodium street
Gemiola [76]

Answer:

doppler shift's formula for source and receiver moving away from each other:

<em>λ'=λ°√(1+β/1-β)</em>

Explanation:

acceleration of spaceship=α=29.4m/s²

wavelength of sodium lamp=λ°=589nm

as the spaceship is moving away from earth so wavelength of earth should increase w.r.t increasing speed until it vanishes at λ'=700nm

using doppler shift's formula:

<em>λ'=λ°√(1+β/1-β)</em>

putting the values:

700nm=589nm√(1+β/1-β)

after simplifying:

<em>β=0.17</em>

by this we can say that speed at that time is: v=0.17c

to calculate velocity at an acceleration of a=29.4m/s²

we suppose that spaceship started from rest so,

<em>v=v₀+at</em>

where v₀=0

so<em> v=at</em>

as we want to calculate t so:-

<em>t=v/a</em>                                                v=0.17c      ,c=3x10⁸           ,a=29.4m/s²

putting values:

=0.17(3x10⁸m/s)/29.4m/s²

<em>t=1.73x10⁶</em>

7 0
3 years ago
The engine of a locomotive exerts a constant force of 8.1*10^5 N to accelerate a train to 68 km/h. Determine the time (in min) t
Bumek [7]

Answer:443.1 s

Explanation:

Given

Engine of a locomotive exerts a force of 8.1\times 10^5 N

Mass of train=1.9\times 10^7

Final speed (v)=68 km/h \approx 18.88 m/s

F=ma

so acceleration(a) =\frac{F}{m}=\frac{8.1\times 10^5}{1.9\times 10^7}

a=0.042631 m/s^2

and acceleration is

a=\frac{v-u}{t}

0.042631=\frac{18.88-0}{t}

t=443.089 \approx 443.1 s

8 0
3 years ago
Read 2 more answers
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