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3 years ago

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

Physics

1 answer:

Nezavi [6.7K]3 years ago

6 0

Answer:

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

Explanation:

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Given vectors D (3.00 m, 315 degrees wrt x-axis) and E (4.50 m, 53.0 degrees wrt x-axis), find the resultant R= D + E. (a) Write

Eva8 [605]

Answer:

R = ( 4.831 m , 1.469 m )

Magnitude of R = 5.049 m

Direction of R relative to the x axis= 16°54'33'

Explanation:

Knowing the magnitude and directions relative to the x axis, we can find the Cartesian representation of the vectors using the formula

$\vec{A}= | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )$

where $| \vec{A} |$ its the magnitude and θ.

So, for our vectors, we will have:

$\vec{D}= 3.00 m \ ( \ cos(315) \ , \ sin (315) \ )$

$\vec{D}= ( 2.121 m , -2.121 m )$

and

$\vec{E}= 4.50 m \ ( \ cos(53.0) \ , \ sin (53.0) \ )$

$\vec{E}= ( 2.71 m , 3.59 m )$

Now, we can take the sum of the vectors

$\vec{R} = \vec{D} + \vec{E}$

$\vec{R} = ( 2.121 \ m , -2.121 \ m ) + ( 2.71 \ m , 3.59 \ m )$

$\vec{R} = ( 2.121 \ m + 2.71 \ m , -2.121 \ m + 3.59 \ m )$

$\vec{R} = ( 4.831 \ m , 1.469 \ m )$

This is R in Cartesian representation, now, to find the magnitude we can use the Pythagorean theorem

$|\vec{R}| = \sqrt{R_x^2 + R_y^2}$

$|\vec{R}| = \sqrt{(4.831 m)^2 + (1.469 m)^2}$

$|\vec{R}| = \sqrt{23.338 m^2 + 2.158 m^2}$

$|\vec{R}| = \sqrt{25.496 m^2}$

$|\vec{R}| = 5.049 m$

To find the direction, we can use

$\theta = arctan(\frac{R_y}{R_x})$

$\theta = arctan(\frac{1.469 \ m}{4.831 \ m})$

$\theta = arctan(0.304)$

$\theta = 16\°54'33''$

As we are in the first quadrant, this is relative to the x axis.

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3 years ago

Consider the following distinct forces:______________________________.

Kamila [148]

Answer:

1 and 2

Explanation:

It is given that, force exerted by air is negligible in any way.

Also, it is given that channel is in the shape of a segment of a circle with its center at O.

When it is within the frictionless channel at position 'Q',

A gravity exerts force on the ball in the downward direction. On the other hand, the channel pointing from Q to O also exerts a force on the ball.

However, there is no any force in the direction of motion. On the other hand, he channel pointing from O to Q does not exert a force on the ball.

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What is the Underground water

Semmy [17]

Answer:

water found underground is called undergroundwater

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In high air pressure the molecules are

Anvisha [2.4K]

Aaaaa is the answer

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If 600 J of work is required to move 50 coulombs of charge

Usimov [2.4K]

Answer:

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3 years ago

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