The vector V is at a 72° angle and has a magnitude of 12.
Describe this vector using vector notation (Use the & and symbols.) (6 points)
Question 9
What is the angle of the vector described below? (3 points).
Please answer all questions will mark brainliest and give 5 stars
V=2x+3y

The Vector V Is At A 72 Angle And Has A Magnitude Of 12.Describe This Vector Using Vector Notation (Use

Answers

Answer 1

Answer:

Part 1

The vector can be expressed as follows;

r = 12, θ = 72°, ·v = (12, ∠72°), ·v ≈ 3.71·[tex]\mathbf{\hat i}[/tex] + 11.41·[tex]\mathbf{\hat j}[/tex] or

[tex]\mathbf{\cdot v}=\begin{bmatrix}3.71\\ 11.41\end{bmatrix} = \begin{pmatrix}3.71\\ 11.41 \end{pmatrix}[/tex]

Part 2

The angle of the vector is 56.31°

Explanation:

Part 1

The given parameters are;

The direction of the vector = 72°

The vector magnitude = 12

Therefore, the vector can be described in the following forms;

Polar form

1) Direct notation;

r = 12, θ = 72°

2) Ordered set

·v = (12, ∠72°)

Rectangular vector notation;

1) Unit vector notation

·v = vₓ[tex]\mathbf{\hat i}[/tex] + [tex]v_y[/tex][tex]\mathbf{\hat j}[/tex]

Where;

vₓ = 12 × cos(72) ≈ 3.71

[tex]v_y[/tex] = 12 × sin(72) ≈ 11.41

Therefore;

·v ≈ 3.71·[tex]\mathbf{\hat i}[/tex] + 11.41·[tex]\mathbf{\hat j}[/tex]

2) Matrix notation

[tex]\mathbf{\cdot v}=\begin{bmatrix}3.71\\ 11.41\end{bmatrix} = \begin{pmatrix}3.71\\ 11.41 \end{pmatrix}[/tex]

Part 2

The given vector is V = 2·x + 3·y

Therefore, the angle of the vector, θ = tan⁻¹(y/x) =  tan⁻¹(3/2) ≈ 56.31°

The angle of the vector = 56.31°.


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What force is needed to move a barrel 45-m if 3600 J of work are accomplished?​

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The answer is 80 N

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[tex]f = \frac{3600}{45} \\ [/tex]

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A high school bus travels 240 km in 6.0 h. What is its average speed for the trip? (in km/h).

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40 km/h

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Look at the formula speed is equal to the distance over time or s = d/t.

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Use the formula: 240/6.0

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A student is given a 5.4 g mineral sample. She determines that the volume of her sample is 2 cm3. What is the density of her mineral

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If a bicyclist travels at 15 km/h, how long will it take her to travel 30 km?​

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If two cars are 5 m apart, and one car has a mass of 2,565 kg and the other
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Answers

Gravitational force between two cars are 2.92 × 10 ⁻¹¹ Nm²/kg.

The universal force of attraction happened between two objects is called as gravitational force.

The mass of the two cars, m₁ = 2565 kg

                                            m₂ = 4264 kg

                                             r = 5m

                                            G = 6.67 ₓ 10 ⁻11 Nm²/kg

How gravitational force can be calculated?

Definition of gravitational force: The force of attraction between any two bodies is directly proportional to the product of the masses and inversely proportional to the square of the distance between them.

                            F ∝ m₁ m₂ ÷ r²

where the force exerted between the two masses m₁ and m₂,

                            F = G (m₁ m₂) ÷ r²

where as,  F - Gravitational force between two bodies,

                 G - Gravitational constant,

                 m₁ - mass of the first body,

                m₂ - mass of the second body,

                r² - square of the distance between two bodies.

                        F= 6.67 × 10⁻¹¹ Nm²/kg ( 2565 kg × 4264 kg) ÷ ( 5²)

         Hence,

                         F = 2.92 × 10⁻⁵ N

Option D is correct answer.

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Explain and reason why the the moon rotates around the earth every lunar cycle
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The level of toluene (a flammable hydrocarbon) in a storage tank may fluctuate between 10 and 400 cm from the top of the tank. since it is impossible to see inside the tank, an open-end manometer with water or mercury as the manometer fluid is to be used to determine the toluene level. one leg of the manometer is attached to the tank 500 cm from the top. a nitrogen blanket at atmospheric pressure is maintained over the tank contents. felder, richard m.; rousseau, ronald w.; bullard, lisa g.. elementary principles of chemical processes, 4th edition (page 81). wiley. kindle edition.

Answers

Complete Question  

The complete question is shown on the first and second uploaded image  

Answer:

When water is used the reading is  [tex]    R =  2281.6 \  cm  [/tex]

When mercury is used the reading is [tex]   R =  23.83 \ cm [/tex]  

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Explanation:

From the question we are told that  

    The length of the leg of the manometer to the top of the tank is  d =  500cm

   The toluene level where in the tank where the height of the manometer fluid level in the open arm is equal to the height where the manometer is connected to the tank is  h =150 cm  

   The manometer reading is  R  

Generally at the point where the height of the open arm is equal to the height of the of the point connected to the tank ,  

   The pressure at the height of the both arms of the manometer corresponding to the base of the tank are equal  

     i.e   [tex]P_1 = P_2[/tex]

Here [tex]P_1[/tex] is the pressure of the manometer at the point corresponding to the base of the tank and this is mathematically represented as  

       [tex] P_{atm} + P_1 =  P_{atm} + P_t[/tex]

Here [tex]P_t[/tex] is the pressure due to the toluene level in the tank and in the arm of the manometer connected to the tank and this is mathematically represented as  

         [tex]P_t  =  \rho_t  * g  * h_i[/tex]

Here  

      [tex]\rho_t [/tex] is the density of toluene with value  [tex]\rho_t =  867 kg/m^3 [/tex]

       

 [tex]h_i[/tex] is the height of the connected arm above the point equivalent to the base of the tank , this mathematically represented as

          [tex]h_i =  d - h + R[/tex]

  and  [tex] P_2 [/tex] is the the pressure at the open arm of the manometer at the point equivalent to the base of the base of the tank and this is mathematically represented as

         [tex] P_2 =  \rho_f * g *  h_f [/tex]

Here  

       [tex]\rho_f[/tex] is the density of the fluid in use , if it is water the density is  

       [tex]\rho_w =  1000 \  kg /m^3 [/tex]

and  if it is  mercury the density is  

        [tex]\rho_m =  13600 \  kg /m^3 [/tex]

[tex]h_f[/tex] is the height of the  fluid in the open arm of the manometer from the point equivalent to the base of the tank which is equivalent the manometer reading R

So when the fluid is water we have

      [tex] P_{atm} +  \rho_t* g *(d - h + R) =  P_{atm} + \rho_f * g *  h_f[/tex]

=>   [tex]   \rho_t* (d - h + R) =   \rho_w *  h_f[/tex]        

=>   [tex]    867  (500 - 150 + R) =    1000 *  R [/tex]

=>    [tex]    R =  2281.6 \  cm  [/tex]

So when the fluid is mercury we have      

  [tex]   \rho_t* (d - h + R) =   \rho_m *  h_f[/tex]          

=>   [tex]    867  (500 - 150 + R) =   13600  *  R [/tex]  

=>   [tex]   R =  23.83 \ cm [/tex]  

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So the best fluid to use is mercury because for water a slight change in toluene level will cause a  large change in height .

       

A ball is thrown horizontally from the top of
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2
.
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Answer in units of s.
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Answer in units of m/s.

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sorry dear I did not understand

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Answer:

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Explanation:

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Answers

Answer:

Explanation:

average speed = 90 km / hr

Let the total distance be 2 d .

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time taken   to cover distance d at 50 km /h = d / 50

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He should locate the chloroform stored in a transparent container in chemical storage and pour directly into his beaker from that location.

He should locate the chloroform stored in a transparent container in chemical storage and should take it to the fume hood to pour.

He should locate the chloroform stored in a dark container in chemical storage and should take it to the fume hood to pour.

He should locate the chloroform stored in a dark container in chemical storage and pour directly into his beaker from that location.

Answers

Answer:

The correct option is the third option

Explanation:

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From the above explanation, it can be deduced that Malik should locate the chloroform stored in a dark container in chemical storage and should take it to the fume hood to pour.

Answer:

He should locate the chloroform stored in a dark container in chemical storage and should take it to the fume hood to pour.

Explanation:

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Answers

Answer: Approximately 302 m/s^2

================================================

Work Shown:

s = starting velocity = 0

f = final velocity  = 46

d = distance = 3.5

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[tex]f^2 = s^2 + 2a*d \ \ \text{ ..... one of the kinematics equations}\\\\46^2 = 0^2 + 2a*3.5\\\\2116 = 7a\\\\7a = 2116\\\\a = \frac{2116}{7}\\\\a \approx 302.28571\\\\a \approx 302[/tex]

The acceleration to three sig figs is roughly 302 m/s^2

The acceleration is so large because the ball's final velocity is incredibly fast in such a short amount of time.

1. If a ball is rolling horizontally on the floor with a velocity of 2.5 m/s, what are the x and y
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Vx =
Vy =

Answers

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Answers

One way:

The ball returned to the thrower's hand after 5 seconds.  It spent 2.5 seconds going up, and 2.5 seconds coming back down.

Use the formula Distance = (1/2) (acceleration) (time squared)

Distance = (1/2) (gravity) (2.5 sec)

Distance = (1/2) (9.8 m/s ²) (6.25 sec² )

Distance = (4.9 m/s²) (6.25 sec²)

Distance = 30.63 meters

Another way:

The thrower threw the ball straight up at 25 m/s.

It sailed upward for 2.5 seconds.

At the top of its trip, before it started falling, its speed was zero.

Its average speed was (1/2) (25+0) = 12.5 m/s

At an average speed of 12.5 m/s for 2.5s, the ball rose (12.5x2.5) = 31.25m.

(Why are the two answers different ?

Because if the ball was thrown straight up at 25 m/s, it would actually return to the thrower in 5.1 seconds, not 5.0 seconds.

So if you use the acceleration of gravity to solve the problem, you get one answer, but if you use 5.0sec to solve it, you get a slightly different one.)

A runner is jogging in a straight line at a steady vr= 7.3 km/hr. When the runner is L= 2.1 km from the finish line, a bird begins flying straight from the runner to the finish line at vb= 29.2 km/hr (4 times as fast as the runner). When the bird reaches the finish line, it turns around and flies directly back to the runner After this first encounter, the bird then turns around and flies from the runner back to the finish line, turns around again and flies back to the runner. The bird repeats the back and forth trips until the runner reaches the finish line. How far does the bird travel from the beginning (including the distance traveled to the first encounter

Answers

Answer:

Explanation:

Time taken by jogger to travel the distance to finishing line = 2.1 / 7.3

= .28767 hr

Bird will keep flying for this time period

distance covered by bird = speed x time

= 29.2 x .28767 km

= 8.4 km .

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Answers

Answer:

UR ANSWER IS WAVELENGTH

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PLEASE ANSWER QUICKLY!!!

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Answer:

50%

Explanation:

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Answer:

thank u for the happiest year of my life

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B. Inclined plane
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O D. Lever

Answers

Answer:

the answer is A.) Wedge

A knife to cut a piece of bread is a A. Wedge

What is a simple machine?

A simple machine, any of several devices with few or no moving parts that are used to modify motion and the magnitude of a force in order to perform work.

since , a wedge is a simple machine  with two inclined planes which when put together forms a sharped edge, which forms a triangular shaped tool which can be used to separate portion of  two objects .

hence , a knife to cut a piece of bread is a A. Wedge

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Usually, the relationship between mass and weight on Earth is highly proportional; objects that are a hundred times more massive than a one-liter bottle of soda almost always weigh a hundred times more—approximately 1,000 newtons, which is the weight one would expect on Earth from an object with a mass slightly greater ...



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Answer:

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