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3 years ago

An example of a high explosive is what?

2 answers:

5 0

The answer is Dynamite.
Explosive, any substance or device that can be made to produce a volume of rapidly expanding gas in an extremely brief period. Chemical explosives are of two types; detonating, or high explosives and deflagrating, or low, explosives. Detonating explosives, such as TNT and dynamite, are characterized by extremely rapid decomposition and development of high pressure, whereas deflagrating explosives, such as black and smokeless powders, involve merely fast burning and produce relatively low pressures.

frosja888 [35]3 years ago

3 0

Unit nine Quiz

1. False

2. A Sniffer

3. True

4. True

5. True

6. Primary explosives

7. False

8. 7000 miles per hour

9. A microscope

10. Dynamite

11 . Black powder

12. True

13. False

14. False

15. Color-coded chips

~GENERAL OUT~

You might be interested in

Calcium chloride is used to make pickles. The calcium atom transfers electrons to chlorine atoms, leading to the formation of ch

GarryVolchara [31]

Calcium chloride contains ionic bonds.

Pennies contain metallic bonds.

Hydrochloric acid contains covalent bonds.

You're welcome.

3 0

3 years ago

Read 2 more answers

If a marathon runner averages 9.51 mi/hr, how long does it take him to run a 26.220-mile marathon. Express your answers in hours

ASHA 777 [7]

Explanation:

Speed of the marathon runner, v = 9.51 mi/hr

Distance covered by the runner, d = 26.220 mile

Let t is the time taken by the marathon runner. We know that the speed of the runner is given by total distance divided by total time taken. Mathematically, it is given by :

$v=\dfrac{d}{t}$

$t=\dfrac{d}{v}$

$t=\dfrac{26.220\ mi}{9.51\ mi/hr}$

t = 2.75 hours

Since, 1 hour = 60 minutes

t = 165 minutes

Since, 1 minute = 60 seconds

t = 9900 seconds

Hence, this is the required solution.

7 0

3 years ago

How could you weaken the force of gravity between cars and the Earth?<br>**I WILL MARK BRAINLEST**

scoundrel [369]

Answer:

The answer you have selected is correct.

Explanation:

Increase radius, force of gravity decreases

6 0

3 years ago

an aircraft landing on an air craft carrier is brought to a complete stop from an inital velocity of 215km/hr in 2.7 seconds. wh

worty [1.4K]

u= 215 km/hr = 215 * 1000/ 3600 = aprx 60m/s
v=0
t=2.7sec
v= u - at
u= at
60/2.7 = 22.23 m/s^2

Hope it helps

8 0

2 years ago

Planet 1 orbits Star 1 and Planet 2 orbits Star 2 in circular orbits of the same radius. However, the orbital period of Planet 1

hichkok12 [17]

Answer:

The mass of Star 2 is Greater than the mass of Start 1. (This, if we suppose the masses of the planets are much smaller than the masses of the stars)

Explanation:

First of all, let's draw a free body diagram of a planet orbiting a star. (See attached picture).

From the free body diagram we can build an equation with the sum of forces between the start and the planet.

$\sum F=ma$

We know that the force between two bodies due to gravity is given by the following equation:

$F_{g} = G\frac{m_{1}m_{2}}{r^{2}}$

in this case we will call:

M= mass of the star

m= mass of the planet

r = distance between the star and the planet

G= constant of gravitation.

so:

$F_{g} =G\frac{Mm}{r^{2}}$

Also, if the planet describes a circular orbit, the centripetal force is given by the following equation:

$F_{c}=ma_{c}$

where the centripetal acceleration is given by:

$a_{c}=\omega ^{2}r$

where

$\omega = \frac{2\pi}{T}$

Where T is the period, and $\omega$ is the angular speed of the planet, so:

$a_{c} = ( \frac{2\pi}{T})^{2}r$

or:

$a_{c}=\frac{4\pi^{2}r}{T^{2}}$

so:

$F_{c}=m(\frac{4\pi^{2}r}{T^{2}})$

so now we can do the sum of forces:

$\sum F=ma$

$F_{g}=ma_{c}$

$G\frac{Mm}{r^{2}}=m(\frac{4\pi^{2}r}{T^{2}})$

in this case we can get rid of the mass of the planet, so we get:

$G\frac{M}{r^{2}}=(\frac{4\pi^{2}r}{T^{2}})$

we can now solve this for $T^{2}$ so we get:

$T^{2} = \frac{4\pi ^{2}r^{3}}{GM}$

We could take the square root to both sides of the equation but that would not be necessary. Now, the problem tells us that the period of planet 1 is longer than the period of planet 2, so we can build the following inequality:

$T_{1}^{2}>T_{2}^{2}$

So let's see what's going on there, we'll call:

$M_{1}$ = mass of Star 1

$M_{2}$ = mass of Star 2

So:

$\frac{4\pi^{2}r^{3}}{GM_{1}}>\frac{4\pi^{2}r^{3}}{GM_{2}}$

we can get rid of all the constants so we end up with:

$\frac{1}{M_{1}}>\frac{1}{M_{2}}$

and let's flip the inequality, so we get:

$M_{2}>M_{1}$

This means that for the period of planet 1 to be longer than the period of planet 2, we need the mass of star 2 to be greater than the mass of star 1. This makes sense because the greater the mass of the star is, the greater the force it applies on the planet is. The greater the force, the faster the planet should go so it stays in orbit. The faster the planet moves, the smaller the period is. In this case, planet 2 is moving faster, therefore it's period is shorter.

6 0

3 years ago