What is the resultant of a and B if a= -2i - 5j and B = I - 4j

Answers

Answer 1

Answer: a= -2i - 5j
B = I - 4j

a + B = -2i -5j + 1i - 4j

a + B = -1i - 9j

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In 1780, in what is now referred to as "Brady's Leap," Captain Sam Brady of the U.S. Continental Army escaped certain death from his enemies by running over the edge of the cliff above Ohio's Cuyahoga River in (Figure 1) , which is confined at that spot to a gorge. He landed safely on the far side of the river. It was reported that he leapt 22 ft (≈ 6.7 m) across while falling 20 ft (≈ 6.1 m).What is the minimum speed with which he’d need to run off the edge of the cliff to make it safely to the far side of the river?

Express your answer to two significant figures and include the appropriate units.

Answers

The minimum speed with which the captain Sam Brady of the US continental army had to run off the edge of the cliff to make it safely to the far side of the river is $\boxed{19.667\text{ ft/s}}$ or $\boxed{5.998\text{ m/s}}$ or $\boxed{6\text{ m/s}}$ or $\boxed{599.8\text{ cm/s}}$ .

Further explanation:

As Captain Sam Brady jumps from the cliff, he moves in two dimension under the action of gravity.

Given:

The height of free fall of the captain Brady is $20\text{ ft}$ or $6.1\text{ m}$ .

The horizontal distance moved by the captain Brady is $22\text{ ft}$ or $6.7\text{ m}$ .

Concept:

The time required to free fall of a body can be calculated by using the expression given below.

$\left( { - s}\right)=ut-(1)/(2)g{t^2}$ ……. (1)

The displacement is considered negative because the captain is moving in vertically downward direction.

Here, $s$ is the distance covered by the body in free fall, $u$ is the initial velocity of the object, $g$ is the acceleration due to gravity and $t$ is the time taken in free fall of a body.

As the Caption jumps off the cliff, he has his velocity in the horizontal direction. The velocity of the captain in vertical direction is zero.

Substitute $0$ for $u$ in the equation (1) .

$s=(1)/(2)g{t^2}$

Rearrange the above expression for $t$ .

$\boxed{t=\sqrt {\frac{{2s}}{g}}}$ …… (2)

Converting acceleration due to gravity in $\text{ft}/\text{s}^2$ .

$\begin{aligned}g&=\left( {9.81\,{\text{m/}}{{\text{s}}^{\text{2}}}} \right)\left( {\frac{{1.0\,{\text{ft/}}{{\text{s}}^{\text{2}}}}}{{0.305\,{\text{m/}}{{\text{s}}^{\text{2}}}}}} \right) \n&=32.16\,{\text{ft/}}{{\text{s}}^{\text{2}}} \n \end{aligned}$

Substitute $20\text{ ft}$ for $s$ and $32.16\,{\text{ft/}}{{\text{s}}^{\text{2}}}$ for $g$ in equation (2) .

$\begin{aligned}t&=\sqrt {\frac{{2\left( {20\,{\text{ft}}} \right)}}{{\left( {32.16\,{\text{ft/}}{{\text{s}}^{\text{2}}}} \right)}}} \n&=1.116\,{\text{s}} \n \end{aligned}$

Therefore, the time taken by captain to free fall a height $20\text{ ft}$ is $1.116\text{ s}$ .

In the same time interval captain has to move $22\text{ ft}$ in horizontal direction. The acceleration is zero in horizontal direction. So, the velocity will be constant throughout the motion in the horizontal direction.

The distance travelled by captain in the horizontal direction is given by,

$x=v\cdot t$

Rearrange the above expression for $v$ .

$\boxed{v=(x)/(t)}$ …… (3)

Here, $x$ is the distance travelled in horizontal direction, $v$ is the velocity of the captain and $t$ is the time.

Substitute $22\text{ ft}$ for $x$ and $1.116\text{ s}$ for $t$ in equation (3) .

$\begin{aligned}v&=\frac{{22\,{\text{ft}}}}{{1.116\,{\text{s}}}} \n&=19.71\,{\text{ft/s}} \n \end{aligned}$

Thus, the minimum speed with which the captain Sam Brady of the US continental army had to run off the edge of the cliff to make it safely to the far side of the river is $\boxed{19.667\text{ ft/s}}$ or $\boxed{5.998\text{ m/s}}$ or $\boxed{6\text{ m/s}}$ or $\boxed{599.8\text{ cm/s}}$ .

Learn more:

1. Energy density stored in capacitor brainly.com/question/9617400

2. Kinetic energy of the electrons brainly.com/question/9059731

3. Force applied by the car on truck brainly.com/question/2235246

Keywords:

Free fall, projectile, gravity, 1780, Brady’s, leap, Captain, Sam Brady, US, continental army, enemies, Ohio’s, Cuyahoga river, 22 ft, 6.7 m, 20 ft, 6.1 m, minimum speed, run off, edge, cliff, safely, far side, river, 19.71 ft/s, 6 m/s, 6 meter/s, 5.99 m/s, 599.8 cm/s.

Final answer:

Using the principles of projectile motion from Physics, Captain Sam Brady would need to run with an initial horizontal speed of approximately 19.64 ft/s to reach the far side of the river.

Explanation:

This problem can be solved using basic Physics, specifically projectile motion. Here, Captain Sam Brady had to run off the edge of the cliff to make it safely to the far side of the river which is 22 ft away while falling 20 ft down. We assume that he jumps horizontally (i.e., his initial vertical velocity is 0).

Firstly, we calculate the time for the vertical fall. Using the equation t = sqrt (2h/g) where h is height and g is the acceleration due to gravity (32.2 ft/s²), we get time t ≈ 1.12s (rounded to two significant figures).

Next, we can use this time to figure out his initial horizontal velocity needed. The equation v = d/t where v is velocity, d is distance, and t is time gives us v ≈ 19.64 ft/s (rounded to two significant figures).

So, Captain Sam Brady would need to run with an initial horizontal speed of approximately 19.64 ft/s to make it safely across the river.

Learn more about Projectile Motion here:

brainly.com/question/20627626

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A large sphere has a mass of 175 kg and is suspended by a chain from the ceiling. The mass of the chain is 12.0 kg. What is the tension in the chain?1830, N1610, N1720, N1940 N.

Answers

Each point in the chain supports the weight of all the mass below it.

At the bottom end of the chain, the weight is (175 x 9.8) = 1,715 N .
At the top of the chain, the weight is (175 + 12) x (9.8) = 1,833 N .

The tension in the chain varies linearly from 1,715N at the bottom
to 1,833N at the top.

Why do prokaryotic cells usually reproduce more quickly than eukaryotic cells? Choose 1 answer:

(Choice A) A Eukaryotic cells are more structurally complex than prokaryotic cells.

(Choice B) B Eukaryotic cells have less DNA than prokaryotic cells.

(Choice C) C Eukaryotic cells have more cell walls than prokaryotic cells.

(Choice D) D Eukaryotic cells are smaller than prokaryotic cells.

Answers

Prokaryotic cells reproduce more quickly than eukaryotic cells because the latter are more structurally complex. The correct option is A.

Structural complexity of cells

Eukaryotic cells have nuclei and membrane-bound organelles such as mitochondrion. This is unlike prokaryotic cells whose DNA lies freely in the cytoplasm and lacks membrane-bound organelles.

In order words, eukaryotic cells are structurally more complex than prokaryotic cells. Thus, reproductive times in prokaryotes are usually quicker than in eukaryotes.

More on the structural complexity of cells can be found here: brainly.com/question/3502696

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I dont get why its c i dont under the graph for the vertical component in option c

Answers

The vertical component grows at an increasing rate because
gravity is ACCELERATING the object vertically. So its speed
keeps growing. Speed is the slope of the displacement graph.

If you had a planet to choose from to live in what would it be?

Answers

I personally would live on Mars cuz that is red cuz

I would personally like to live on mars because it is like our temperature, but a little bit colder.

Sudhi is studying the motion of a toy car. she places the car in the middle of a ramp and releases it. which best describes the reference point in this scenario?

Answers

From the motion of the toy car, the reference point of this scenario is the starting position of the car.

What is reference point?

The reference point of an object is defined as the position through the measurement of an object is taken.

In studying the motion of a toy car, if she places the car in the middle of a ramp and releases it, we can conclude that the reference point of this scenario is the starting position of the car.

The starting position of the car will serve as point of reference because, the distance, speed, and time of motion of the car can be obtained with respect to the starting point.

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Thank you for posting your question here. Below are the choices that can be found elsewhere:

a. the top of the ramp
b. the bottom of the ramp
c. the ending position of the car
d. the starting position of the car

I think the answer is D which is the starting position of the car. I hope it helps.

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