Does a stick of dynamite contain force?

Answers

Answer 1
Answer:
No.  Force isn't something you can enclose in a box, an envelope,
or a bottle.

A stick of dynamite contains some amount of a chemical compound which,
when triggered by enough heat or shock, undergoes a chemical reaction
that proceeds very rapidly, and generates a large quantity of gases before
the gases can dissipate.  This creates very high pressure around the reaction,
and it's this pressure in a small volume that exerts great force for a very short
time.     The whole process is often described as an "explosion".
 

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In the qualifying round of the 50-yd freestyle in the sectional swimming championship, Dugan got an early lead by finishing the first 25.00 yd in 10.01 seconds. Dugan finished the return leg (25.00 yd distance) in 10.22 seconds. Determine Dugan's average velocity for the entire race. Number only, round to the nearest hundredth.
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In a generator, mechanical energy is converted into which of the following? A.an electromagnet

B.a moving rod

C.a magnetic field

D.electrical energy

Answers

In an electrical generator, you use steam, water, wind, or
bicycle pedals to turn the shaft, and electricity comes out.

Mechanical energy in  ===>  electrical energy out.

Answer:

eletrical ENERGY

Explanation:

D.

In the Bohr model of the hydrogen atom, an electron moves in a circular path around a proton. The speed of the electron is approximately 2.03 106 m/s.(a) Find the force acting on the electron as it revolves in a circular orbit of radius 5.35 10-11 m.
whats the magnitude in N ?


(b) Find the centripetal acceleration of the electron.
whats the magnitude in m/s2

Answers

In order to answer these questions, we need to know the charges on
the electron and proton, and then we need to know the electron's mass. 
I'm beginning to get the creepy feeling that, in return for the generous
5 points, you also want me to go and look these up so I can use them
in calculations ... go and collect my own straw to make the bricks with,
as it were. 

Ok, Rameses:

Elementary charge . . . . .  1.6 x 10⁻¹⁹  coulomb
                                        negative on the electron
                                        plussitive on the proton

Electron rest-mass . . . . .  9.11 x 10⁻³¹  kg


a).  The force between two charges is

      F  =  (9 x 10⁹) Q₁ Q₂ / R²

          =  (9 x 10⁹ m/farad) (-1.6 x 10⁻¹⁹C) (1.6 x 10⁻¹⁹C) / (5.35 x 10⁻¹¹m)²

          =     ( -2.304 x 10⁻²⁸) / (5.35 x 10⁻¹¹)²

          =          8.05 x 10⁻⁸  Newton .


b).  Centripetal acceleration  = 

                                               v² / r  .

                  A  =  (2.03 x 10⁶)² / (5.35 x 10⁻¹¹)

                     =      7.7 x 10²²  m/s² .

That's an enormous acceleration ... about  7.85 x 10²¹  G's !
More than enough to cause the poor electron to lose its lunch.

It would be so easy to check this work of mine ...
First I calculated the force, then I calculated the centripetal acceleration.
I didn't use either answer to find the other one, and I didn't use  "  F = MA "
either.

I could just take the ' F ' that I found, and the 'A' that I found, and the
electron mass that I looked up, and mash the numbers together to see
whether  F = M A .

I'm going to leave that step for you.   Good luck !

At height h above the surface of Earth, the gravitational acceleration is What is h? Note: The radius of Earth is 6380 km.

Answers


The acceleration of gravity is inversely proportional to
the square of the distance from Earth's center.

The acceleration of gravity is 9.8 m/s² on the Earth's surface ...
6380 km from the center.

If the acceleration of gravity at 'h' is 4.9 m/s² ... 1/2 of what it is
on the surface, then the distance from the center is

                 (6380 x √2) =  9,023 km  (rounded) ,

and 'h' is the distance above the surface

                     = (9,023 - 6,380) =  2,643 km  (rounded) .

The safe load, L, of a wooden beam supported at both ends varies jointly as the width, w, the square of the depth, d, and inversely as the length, l. A wooden beam 3in. wide, 6in. deep, and 11ft long holds up 1213lb. What load would a beam 6in. wide, 3in. deep and 12ft long of the same material support? (Round off your answer to the nearest pound.)

Answers

Answer:

L' = 555.95 lb

Explanation:

Analyzing the given conditions in the question, we get

The safe load, L is directly proportional to width (w) and square of depth (d²)

 also,

L is inversely proportional length (l) i.e L = k/l

combining the above conditions, we get an equation as:

 L = k(wd²/l)

 now, for the first case we have been given

w = 3 in

d = 6 in

l = 11 ft

L = 1213 lbs

 thus,

1213 lb = k ((3 × 6²)/11)

or

k = 123.54 lbs/(ft.in³)  

Now,

Using the calculated value of k to calculate the value of L in the second case  

in the second case, we have

w = 6 in

d =3 in

l = 12 ft

Final Safe load L' =  123.54 × (6 × 3²/12)

or

L' = 555.95 lb

Final answer:

The load that a beam of the given dimensions can support is derived using a joint variation formula. The constant of variaiton is found using the given beam's characteristics, and then applied on the second beam's measurements to find its load. The result is approximately 2025 pounds.

Explanation:

The question is related to joint variation and its principles in mathematics. First, it's necessary to define the equation of the joint variation:
L = k*(w*d^2)/l
where L is the load, w is the width, d is the depth, l is the length, and k is the constant of proportionality. From the information given in the question, we can derive that k equals to 1213 * 11 / (3 * 6^2), which is approximately 135. Then, we substitue the values of the second beam into the equation, w = 6, d = 3, l = 12 and k = 135, to get the load L for the second beam, which equates to about 2025 pounds.

Learn more about Joint Variation here:

brainly.com/question/33421771

#SPJ6

I need a equation for a problem in using a pulling force a mass a coefficient of static friction a coefficient of kinetic friction and solving for the frictional force

Answers

The minimum pulling force required to start it moving is
equal to the static frictional force.

F = (mass) x (gravity) x (coefficient of static friction).


Once it's moving, the force required to keep it moving at a steady speed,
without speeding up or slowing down, is the force of kinetic friction.

F = (mass) x (gravity) x (coefficient of kinetic friction).


In both equations, [ (mass) x (gravity)] is just the object's 'weight'.



How many people can die from an atomic bomb in 1 minute?

Answers

Answer:

no one says it that but

in hirosima their are 70000 people die instant