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3 years ago

Napoleon's army captured the island of San Salvador, then moved on to occupy Louisiana and close the port of New Orleans.

1 answer:

3 0

A. true

Quote" Napoleon's army captured the island of San Salvador, then moved on to occupy Louisiana and close the port of New Orleans."

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Suppose that you are holding a cup of coffee in your hand identify all forces on the cup and reaction to each force

Stels [109]

Answer:

Newton's third law.

Explanation:

•your hand of holding UP/DOWN the cup has a reaction with force

•Coffee inside the cup is pushing outward to the edges of the cup

•The cup is pushing inward onto the coffee

"His third law states that for every action (force) in nature there is an equal and opposite reaction. In other words, if object A exerts a force on object B, then object B also exerts an equal and opposite force on object A."

3 0

3 years ago

Read 2 more answers

A rectangular piece of cardboard, whose area is 352 square centimeters, is made into an open box by cutting a 2-centimeter sq

Usimov [2.4K]

Answer:

Dimension of cardboard is 22 m by 16 m

Explanation:

Given that,

Area = 352 cm²

Side of each square cutting from corner = 2 cm

Volume of box = 432 cm³

Let the two sides are x and y.

The area of the rectangular piece is

$xy=352$

$y=\dfrac{352}{x}$ -------- (1)

The volume of the rectangular piece

$2(x-4)(\dfrac{352}{x}-4)=432$

$x^2-38x+352=0$

$(x-16)(x-22)=0$

x=16,22

Put the value of x in the equation (I)

For x = 16

$y=\dfrac{352}{16}=22$

For x = 22

$y=\dfrac{352}{22}=16$

Dimension of cardboard is 22 m by 16 m

8 0

3 years ago

In an Atwood's machine, one block has a mass of 602.0 g, and the other a mass of 717.0 g. The pulley, which is mounted in horizo

Wittaler [7]

Answer:

The acceleration of the both masses is 0.0244 m/s².

Explanation:

Given that,

Mass of one block = 602.0 g

Mass of other block = 717.0 g

Radius = 1.70 cm

Height = 60.6 cm

Time = 7.00 s

Suppose we find the magnitude of the acceleration of the 602.0-g block

We need to calculate the acceleration

Using equation of motion

$s=ut+\dfrac{1}{2}at^2$

Where, s = distance

t = time

a = acceleration

Put the value into the formula

$60.0\times10^{-2}=0+\dfrac{1}{2}\times a\times(7.00)^2$

$a=\dfrac{60.0\times10^{-2}\times2}{(7.00)^2}$

$a=0.0244\ m/s^2$

Hence, The acceleration of the both masses is 0.0244 m/s².

5 0

3 years ago

Use the definition of scalar product, a overscript right-arrow endscripts times b overscript right-arrow endscripts = ab cos θ,

makkiz [27]

Answer: $\theta=cos^{-1}0.991=7.69^o$

The following vectors have been given: $\vec{a}=3.0\widehat{i}+3.0\widehat{j}+3.0\widehat{k}\\ \vec{b}=5.0\widehat{i}+7.0\widehat{j}+6.0\widehat{k}$

The angle between these two vectors can be found by:

$cos\theta=\frac{\vec{a}.\vec{b}}{||\vec{a}|| ||\vec{b}||}\\
||\vec{a}=\sqrt{a_x^2+a_y^2+a_z^2}$

$\vec{a}.\vec{b}=a_xb_x+a_yb_y+a_zb_z\\ \vec{a}.\vec{b}=3\times5+3\times7+3\times6=15+21+18=54$

$||\vec{a}||=\sqrt{3^2+3^2+3^2}=\sqrt{27}\\ ||\vec{b}||=\sqrt{5^2+7^2+6^2}=\sqrt{110}$

$cos\theta=\frac{54}{\sqrt{27}\times\sqrt{110}}\\=0.991\\ \Rightarrow \theta=cos^{-1}0.991=7.69^o$

7 0

3 years ago

1. Any push or pull is called

timama [110]

Answer:

b ok

Explanation:

7 0

3 years ago