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anyanavicka [17]
2 years ago
8

Gasoline is an example of an accelerant. True False

Physics
2 answers:
Kay [80]2 years ago
7 0

Answer:

true

Explanation:

const2013 [10]2 years ago
4 0

Answer:

I believe its true

Explanation:

Gasoline is the most common fire accelerant used but it could also be present at a scene as an ignitable liquid due to gasoline being a common fuel. Although ignitable liquids are the most common fire accelerants it is possible to have other chemicals being used as a fire accelerant.

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You drive a bumper car into another bumper car whos driver has a much larger body mass than you do? Who experience more of a jol
pav-90 [236]
The other driver unexpectedly
4 0
3 years ago
In the steady state 1.2 ✕ 1018 electrons per second enter bulb 1. There are 6.3 ✕ 1028 mobile electrons per cubic meter in tungs
bekas [8.4K]

Answer:

E=12.2V/m

Explanation:

To solve this problem we must address the concepts of drift velocity. A drift velocity is the average velocity attained by charged particles, such as electrons, in a material due to an electric field.

The equation is given by,

V=\frac{I}{nAq}

Where,

V= Drift Velocity

I= Flow of current

n= number of electrons

q = charge of electron

A = cross-section area.

For this problem we know that there is a rate of 1.8*10^{18} electrons per second, that is

\frac{I}{q} = 1.2*10^{18}

A= 1.3*10^{-8}m^2

n=6.3*10^{28} e/m^3

\omicron{O} = 1.2*10^{-4}(m/s)(N/c) Mobility

We can find the drift velocity replacing,

V = \frac{1.2*10^{18}}{(1.3*10^{-8})(6.3*10^{28})}

V= 1.465*10^-3m/s

The electric field is given by,

E= \frac{V}{\omicron{O}}

E=\frac{1.465*10^-3}{1.2*10^{-4}}

E=12.2V/m

7 0
3 years ago
How often is water added to the Earth system?
rosijanka [135]
<span>Water is never added to earth system. Water forever remains in the water cycle on earth, so it goes from the ground, to the air, to the rain, to the sea, and round and round continuously. This cycle means that there does not need to be new water added to the earth, because it recycles any water that already exists of its own accord.</span>
4 0
3 years ago
A transverse wave on a string is described by the wave function
svp [43]

The period of the transverse wave from what we have here is 0.5

<h3>How to find the period of the transverse wave</h3>

The period of a wave can be defined as the time that it would take for the wave to complete one complete vibrational cycle.

The formula with which to get the period is

w = 4π

where w = 4 x 22/7

2π/T = 4π

6.2857/T = 12.57

From here we would have to cross multiply

6.2857 = 12.57T

divide through by 12.57

6.2857/12.57 = T

0.500 = T

Hence we can conclude that the value of T that can determine the period based on the question is 0.500.

Read more on transverse wave here

brainly.com/question/2516098

#SPJ4

5 0
1 year ago
A tuning fork vibrating at 508 Hz falls from rest and accelerates at 9.80 m/s^2. How far below the point of release is the tunin
JulijaS [17]

Answer:

Explanation:

given,

tuning fork vibration = 508 Hz

accelerates = 9.80 m/s²

speed of sound = 343 m/s

observed frequency = 490 Hz

f_s = f(\dfrac{v}{v-(-v_s)})

f_s = f(\dfrac{v}{v+v_s})

v_s = v[\dfrac{f_s}{f_o}-1]

      = 343[\dfrac{508}{490}-1]

      v_s=12.6 m/s

distance the tunning fork has fallen

y_1=\dfrac{v^2}{2a_y}

     =\dfrac{12.6^2}{2\times 9.8}

     =8.1 m

now, time required for the observed will be

t = \dfrac{8.1}{343} = 0.023 s

now, for the distance calculation

y_2 = u\ t + \dfrac{1}{2}at^2

  = 12.6\times 0.023 +\dfrac{1}{2}\times 9.8 \times 0.023^2

  =0.293 m

total distance

 = 8.1 + 0.293 = 8.392 m

3 0
2 years ago
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