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VLD [36.1K]
3 years ago
5

Which statement is FALSE? *

Physics
2 answers:
Burka [1]3 years ago
7 0

Answer:

an example of a deflagration explosive is C4, is false.

Explanation:

Explosive materials may be categorized by the speed at which they expand. Materials that detonate (the front of the chemical reaction moves faster through the material than the speed of sound) are said to be "high explosives" and materials that deflagrate are said to be "low explosives".

C4 meaning:

C-4 or Composition C-4 is a common variety of the plastic explosive family known as Composition C, which uses RDX as its explosive agent. C-4 is composed of explosives, plastic binder, plasticizer to make it malleable, and usually a marker or odorizing taggant chemical.

dalvyx [7]3 years ago
3 0

Answer:

Some deflagration explosives have shock waves faster than the speed of sound.

Explanation:

A  example  of a deflagration explosives.

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What is the minimal mass of helium (density 0.18 kg/m3) needed to lift a balloon carrying two people in a basket, if the total m
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Answer:   
 M[min] = M[basket+people+ balloon, not gas] * ΔR/R[b] 
 Î”R is the difference in density between the gas inside and surrounding the balloon. 
 R[b] is the density of gas inside the baloon.   
 ==================================== 
 Let V be the volume of helium required. 
 Upthrust on helium = Weight of the volume of air displaced = Density of air * g * Volume of helium = 1.225 * g * V   
 U = 1.225gV newtons 
 ---- 
 Weight of Helium = Volume of Helium * Density of Helium * g 
 W[h] = 0.18gV N   
 Net Upward force produced by helium, F = Upthrust - Weight = (1.225-0.18) gV = 1.045gV N  -----

 
 Weight of 260kg = 2549.7 N 
 Then to lift the whole thing, F > 2549.7 
 So minimal F would be 2549.7 
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 =====   
 Let the density of the surroundings be R 
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3 0
3 years ago
You obtain a 100-W light bulb and a 50-W light bulb. Instead of connecting them in the normal way, you devise a circuit that pla
lesantik [10]

Answer:

When they are connected in series

     The  50 W bulb glow more than the 100 W bulb

Explanation:

From the question we are told that

     The power rating  of the first bulb is P_1  = 100 \ W

      The power rating of the second bulb is  P_2  =  50 \ W

     

Generally the power rating of the first bulb is mathematically represented as

      P_1  =  V^2 R

Where  V is the normal household voltage which is constant for both bulbs

  So  

        R_1  =  \frac{V^2}{P_1 }

substituting values

        R_1  =  \frac{V^2}{100}

Thus the resistance of the second bulb would be evaluated as

       R_2  =  \frac{V^2}{50}

From the above calculation we see that

        R_2  >  R_1

This power rating of the first bulb can also be represented mathematically as  

        P_  1  =  I^2_1  R_1

This power rating of the first bulb can also be represented mathematically as    

       P_  2  =  I^2_2 R_2

Now given that they are connected in series which implies that the same current flow through them so

       I_1^2 =  I_2^2

This means  that

       P \ \alpha  \  R

So  when they are connected in series

     P_2  >  P_1

This means that the 50 W bulb glows more than the 100 \ W bulb

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​Hooke's Law. The distance d when a spring is stretched by a hanging object varies directly as the weight w of the object. If th
nadezda [96]

Answer:

distance when the weight is 8 ​kg is 26.66 cm

Explanation:

given data

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distance when the weight is 8 ​kg

solution

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so d = 26.66

distance when the weight is 8 ​kg is 26.66 cm

7 0
3 years ago
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