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taurus [48]
2 years ago
6

A sample of a substance burns more rapidly in pure oxygen than in air. which factor is most responsible for this high rate of re

action?
Physics
1 answer:
zmey [24]2 years ago
3 0
I had to look for the options and here is my answer. Based on the given substance above that burns faster in pure oxygen than in air, the factor that is most responsible for this kind of reaction is the concentration of the reactants. This answer is based on the actual options attached to this question.
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Ahmed fills a basket with shopping weighing 10 kg. How much work is being done on the shopping basket when he lifts it verticall
NISA [10]

The work done by the shopping basket is 147 J.

<h3>When is work said to be done?</h3>

Work is said to be done whenever a force moves an object through a certain distance.

The amount of work done on the shopping basket can be calculated using the formula below.

Formula:

  • W = mgh

Where:

  • W = Amount of work done by the basket
  • m = mass of the shopping basket
  • h = height of the shopping basket
  • g = acceleration due to gravity.

Form the question,

Given:

  • m = 10 kg
  • h = 1.5 m
  • g = 9.8 m/s²

Substitute these values into equation 2

  • W = 10(1.5)(9.8)
  • W = 147 J.

Hence, The work done by the shopping basket is 147 J.

Learn more about work done here: brainly.com/question/18762601

6 0
2 years ago
Two isolated, concentric, conducting spherical shells have radii R1 = 0.500 m and R2 = 1.00 m, uniform charges q1=+2.00 µC and q
scZoUnD [109]

Complete Question

The diagram for this question is shown on the first uploaded image  

Answer:

a E =1.685*10^3 N/C

b E =36.69*10^3 N/C

c E = 0 N/C

d V = 6.7*10^3 V

e   V = 26.79*10^3V

f   V = 34.67 *10^3 V

g   V= 44.95*10^3 V

h    V= 44.95*10^3 V

i    V= 44.95*10^3 V

Explanation:

From the question we are given that

       The first charge q_1 = 2.00 \mu C = 2.00*10^{-6} C

       The second charge q_2 =1.00 \muC = 1.00*10^{-6}

      The first radius R_1 = 0.500m

      The second radius R_2 = 1.00m

 Generally \ Electric \ field = \frac{1}{4\pi\epsilon_0}\frac{q_1+\ q_2}{r^2}

And Potential \ Difference = \frac{1}{4\pi \epsilon_0}   [\frac{q_1 }{r}+\frac{q_2}{R_2} ]

The objective is to obtain the the magnitude of electric for different cases

And the potential difference for other cases

Considering a

                      r  = 4.00 m

           E = \frac{((2+1)*10^{-6})*8.99*10^9}{16}

                = 1.685*10^3 N/C

Considering b

           r = 0.700 m \ , R_2 > r > R_1

This implies that the electric field would be

            E = \frac{1}{4\pi \epsilon_0}\frac{q_1}{r^2}

             This because it the electric filed of the charge which is below it in distance that it would feel

            E = 8*99*10^9  \frac{2*10^{-6}}{0.4900}

               = 36.69*10^3 N/C

   Considering c

                      r  = 0.200 m

=>   r

 The electric field = 0

     This is because the both charge are above it in terms of distance so it wont feel the effect of their electric field

       Considering d

                  r  = 4.00 m

=> r > R_1 >r>R_2

Now the potential difference is

                  V =\frac{1}{4\pi \epsilon_0} \frac{q_1 + \ q_2}{r} = 8.99*10^9 * \frac{3*10^{-6}}{4} = 6.7*10^3 V

This so because the distance between the charge we are considering is further than the two charges given  

          Considering e

                       r = 1.00 m R_2 = r > R_1

                V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2}  ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{1.00} \frac{1.00*10^{-6}}{1.00} ] = 26.79 *10^3 V

          Considering f

              r = 0.700 m \ , R_2 > r > R_1

                      V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2}  ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.700} \frac{1.0*10^{-6}}{1.00} ] = 34.67 *10^3 V

          Considering g

             r =0.500\m , R_1 >r =R_1

   V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2}  ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V

          Considering h

                r =0.200\m , R_1 >R_1>r

  V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2}  ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V

           Considering i    

   r =0\ m \ , R_1 >R_1>r

  V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2}  ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V

8 0
3 years ago
Explain in your own words how the Doppler Effect is also applicable in our study of light.
arlik [135]

well in my own words, i'd saw the the doppler effect is similar to light because sound has a speed, and light does too.

so my theory is if you go fast enough everything would just become black, or maybe white? idk its hard to explain

but what my point is, is taht the doppler effect works in the same way, like if a car is moving towards you the sound is being emitted from the car and being pushed by the speed of the car making it have a much higher pitch, when the car is going away however it drops to a lower pitch due the the sound waves being DRAGGED by the car.

there hoped this helped I guess

8 0
2 years ago
Calculate the molecular weight of Aluminium hydroxide​
Vedmedyk [2.9K]

Al(OH)3 = 26.98 + [(16×3) + (1.01×3)] = 26.98 + 51.03 = 78.01 and the unit will be g/mol

<h3><em>Al(OH)3 = 78.01 g/mol</em></h3>
5 0
3 years ago
The amount of gas that a helicopter uses is directly proportional to the number of hours spent flying. the helicopter flies for
igomit [66]

Answer:

The helicopter uses 35 gallons to fly for 5 hours.

Explanation:

The amount of gas that a helicopter uses for flying varies directly proportional to the number of hours spent flying.

g ∝ T

where g represents amount of gas and T time of flight.

Then,

\therefore \frac{g_1}{g_2}=\frac{T_1}{T_2}

The helicopter files 4 hours and uses 28 gallons of fuel.

Here, g₁= 28 gallons, T₁=4 hours

g₂=?, T₂=5 hours.

\therefore \frac{g_1}{g_2}=\frac{T_1}{T_2}

\Rightarrow \frac{28}{g_2}=\frac{4}{5}

⇒28×5= g₂×4

⇒ g₂×4=28×5

\Rightarrow g_2=\frac{28\times 5}{4}

\Rightarrow g_2=35 gallons

The helicopter uses 35 gallons to fly for 5 hours.

5 0
3 years ago
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