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3 years ago

What was the result of the Mexican army's victory over the Texan garrison at the Alamo?

2 answers:

3 0

The result of the Mexican victory was that fallen defenders became heroes to the cause of Texan independence. The Battle of the Alamo took place between February 23 and March 6, 1836 and became the central episode of the Texas Revolution . After this thirteen-day battle, the Mexican troops of General President Antonio Lopez de Santa Anna began an attack on San Antonio de Bexar, the current San Antonio in Texas. The Battle of the Alamo fought the army of Mexico against a group of Texan rebels, mostly American settlers. More than four thousand men from Santa Ana stood in front of the Alamo Fort , the last stronghold of the rebels, which barely reached 187. The Alamo was not a fortress prepared to withstand a siege. It is believed that all the rebels of the Alamo died in the siege, but Santa Anna came to lose up to about 900 men during the days that lasted the fight. However, the worst result for Santa Ana was precisely the resistance that the Texan rebels had in the Alamo, which fostered the fighting spirit of the Texans. A few days later, on March 14, 1836, Texas became independent from Mexico and a month later, Santa Ana was imprisoned.

KonstantinChe [14]3 years ago

3 0

The fallen defenders became heroes to the cause of Texan independence.

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A neutral atom contains 8 protons 9 neutrons, and 8 electrons the atom must be A. A oxygen isotope. B. A nitrogen isotope. C. A

Harrizon [31]

Answer:

A an Oxygen isotope

Explanation:

The number of protons is the Same as the atomic number

The atomic number is 9 is the atom has 9 protons

Thus, the atom is an oxygen isotope

6 0

4 years ago

Read 2 more answers

In a physics lab, light with a wavelength of 530 nm travels in air from a laser to a photocell in a time of 16.7 ns . When a sla

Harlamova29_29 [7]

Answer:

$\lambda'=78.086\ nm$

Explanation:

Given:

wavelength of light in the air, $\lambda=530\times 10^{-9}\ m$

time taken to travel from the source to the photocell via air, $t=16.7\ s$

time taken to reach the photocell via air and glass slab, $t'=21.3\times 10^{-9}\ s$

thickness of the glass slab, $x=0.87\ m$

Now we have the relation for time:

$\rm time=\frac{distance}{speed}$

hence,

$t=\frac{d}{c}$

c= speed of light in air

$16.7\times 10^{-9}=\frac{d}{3\times 10^8}$

$d=16.7\times 10^{-9}\times 3\times 10^8$

$d=5.01\ m$

For the case when glass slab is inserted between the path of light:

$\frac{(d-x)}{c} +\frac{x}{v} =t'$ (since light travel with the speed c only in the air)

here:

v = speed of light in the glass

$\frac{(5.01-0.87)}{3\times 10^8} +\frac{0.87}{v} =21.3\times 10^{-9}$

$v=4.42\times 10^7\ m.s^{-1}$

Using Snell's law:

$\frac{\lambda}{\lambda'} =\frac{c}{v}$

$\frac{530}{\lambda'} =\frac{3\times 10^8}{4.42\times 10^7}$

$\lambda'=78.086\ nm$

5 0

3 years ago

A student walks 4 block east, 7 blocks west, 1 blocks east then 2 blocks west in an hour. Her average speed is____ block/hour

Gemiola [76]

Average speed = (total distance) / (total time)

Average speed = (4+7+1+2 blox) / (1 hour)

Average speed = 14 blocks/hour

I'm gonna go out on a limb here and take a wild guess:

I'm guessing that there's another question glued onto the end of this one, and it asks you to find either her displacement or her average velocity. I'm so sure of this that I'm gonna give you the solution for that too. If there's no more question, then you won't need this, and you can just discard it. I won't mind.

Average velocity = (displacement) / (time for the displacement)

"Displacement" = distance and direction from the start point to the end point, regardless of how she got there.

Displacement = (4E + 7W + 1E + 2W)

Displacement = (5E + 9W)

Displacement = 4 blocks west

Average velocity = (4 blocks west) / (1 hour)

Average velocity = 4 blocks/hour West

4 0

3 years ago

For a projectile launched horizontally, which of the following best describes the downward component of a projectile's velocity?

Angelina_Jolie [31]

C. The downward component of the projectile's velocity continually increases

Explanation:

The motion of a projectile consists of two independent motions:

- A uniform motion (with constant velocity) along the horizontal direction

- A uniformly accelerated motion, with constant acceleration (equal to the acceleration of gravity) in the downward direction

Here we want to study the downward component of the projectile's velocity. Since the vertical motion is a uniformly accelerated motion, the vertical velocity is given by:

$v=u+at$