Vector A has a magnitude of 6.0 m and points 30° north of east. Vector B has a magnitude of 4.0 m and points 30° west of south. The resultant vector A+ B is given by

Answers

Answer 1

Answer:

Answer:

The resultant vector $\vec R = \vec A+\vec B$ is given by $\vec R = 3.196\,\hat{i}-0.464\,\hat{j}\,\,\,[m]$ .

Explanation:

Let $\vec A = 6\cdot (\cos 30^(\circ)\,\hat{i}+\sin 30^(\circ)\,\hat{j})$ and $\vec B = 4\cdot (-\sin 30^(\circ)\,\hat{i}-\cos 30^(\circ)\,\hat{j})$ , both measured in meters. The resultant vector $\vec R$ is calculated by sum of components. That is:

$\vec R = \vec A+\vec B$ (Eq. 1)

$\vec R = 6\cdot (\cos 30^(\circ)\,\hat{i}+\sin 30^(\circ)\,\hat{j})+4\cdot (-\sin 30^(\circ)\,\hat{i}-\cos 30^(\circ)\,\hat{j})$

$\vec R = (6\cdot \cos 30^(\circ)-4\cdot \sin 30^(\circ))\,\hat{i}+(6\cdot \sin 30^(\circ)-4\cdot \cos 30^(\circ))\,\hat{j}$

$\vec R = 3.196\,\hat{i}-0.464\,\hat{j}\,\,\,[m]$

The resultant vector $\vec R = \vec A+\vec B$ is given by $\vec R = 3.196\,\hat{i}-0.464\,\hat{j}\,\,\,[m]$ .

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In the circuit shown in the figure above, suppose that the value of R1 is 100 k ohms and the value of R2 is 470 k ohms. At which of the following locations in the circuit would you measure the highest voltage with your meter? A. Between points A and B B. Between points A and C C. Between points B and E D. Between points B and C

Three point mass particles are located in a plane: a. 3.77 kg located at the origin
b. 6.7106 kg at [(5.72 cm),(11.44 cm)]
c. 2.46181 kg at [(16.7024 cm),(0 cm)].

How far is the center of mass of the three particles from the origin? Answer in units of cm

Answers

The distance of the center of mass of the three particles from the origin is 6.1428 cm and 5.9316 cm.

Calculation of the distance:

Since

m1 = 3.77 kg (0, 0 )

m2 = 6.7106 kg ( 5.72 cm, 11.44 cm)

m3 = 2.46181 kg (16.7024 cm, 0 cm )

Now here we assume x and y be the coordinates with respect to the centre of mass.

So,

We know that

$x = (m_1x_1+m_2x_2+m_3x_3)/(m_1+m_2+m_3)\n\n = (3.77* 0 + 6.7106 * 5.72 + 2.46181 * 16.7024)/(3.77 + 6.7106 + 2.46181)$

= 6.1428 cm

Now

$y = (m_1y_1+m_2y_2+m_3y_3)/(m_1+m_2+m_3)\n\n = (3.77* 0 + 6.7106 * 11.44 + 2.46181 * 0)/(3.77 + 6.7106 + 2.46181)$

= 5.9316 cm

Learn more about mass here: brainly.com/question/16876455

Answer:

Explanation:

m1 = 3.77 kg (0, 0 )

m2 = 6.7106 kg ( 5.72 cm, 11.44 cm)

m3 = 2.46181 kg (16.7024 cm, 0 cm )

Let x and y be the coordinates of centre of mass.

$x = (m_(1)x_(1)+ m_(2)x_(2)+m_(3)x_(3))/(m_(1)+m_(2)+m_(3))$

$x = (3.77* 0+ 6.7106* 5.72 + 2.46181* 16.7024)/(3.77+6.7106+2.46181)$

x = 6.1428 cm

$y = (m_(1)y_(1)+ m_(2)y_(2)+m_(3)y_(3))/(m_(1)+m_(2)+m_(3))$

$y= (3.77* 0+ 6.7106* 11.44 + 2.46181* 0)/(3.77+6.7106+2.46181)$

y = 5.9316 cm

How does simple machines make work easier?

Answers

They make it so you would exert less force and make things easier to move

The simple machines can help make work easier by working quicker than the people and making other peoples jobs easier.

Explanation:

I hope this helped.

A tennis player hits a ball 2.0 m above the ground. The ball leaves his racquet with a speed of 20 m/s at an angle 5.0 ∘ above the horizontal. The horizontal distance to the net is 7.0 m, and the net is 1.0 m high. Does the ball clear the net?

Answers

Answer:

ball clears the net

Explanation:

$v_(o)$ = initial speed of launch of the ball = 20 ms^{-1}

$\theta$ = angle of launch = 5 deg

Consider the motion of the ball along the horizontal direction

$v_(ox)$ = initial velocity = $v_(o) Cos\theta = 20 Cos5 = 19.92 ms^(-1)$

$t$ = time of travel

$X$ = horizontal displacement of the ball to reach the net = 7 m

Since there is no acceleration along the horizontal direction, we have

$X = v_(ox) t \n7 = v_(ox) t\nt = (7)/(v_(ox))$ Eq-1

Consider the motion of the ball along the vertical direction

$v_(oy)$ = initial velocity = $v_(o) Sin\theta = 20 Sin5 = 1.74 ms^(-1)$

$t$ = time of travel

$Y_(o)$ = Initial position of the ball at the time of launch = 2 m

$Y$ = Final position of the ball at time "t"

$a_(y)$ = acceleration in down direction = - 9.8 ms⁻²

Along the vertical direction , position at any time is given as

$Y = Y_(o) + v_(oy) t + (0.5) a_(y) t^(2)\nY = 2 + (20 Sin5) ((7)/(20 Cos5)) + (0.5) (- 9.8) ((7)/(20 Cos5))^(2)\nY = 2.00758 m\n$

Since Y > 1 m

hence the ball clears the net

Lightning can sometimes occur on hot and humid summer evenings when there are no thunderstorms.(A) True
(B) False

Answers

Answer:

(B) False

Explanation:

No, it is not possible to have thunder without lightning. Thunder is a direct result of lightning.

B False because facts

If 3.00 ✕ 10−3 kg of gold is deposited on the negative electrode of an electrolytic cell in a period of 2.59 h, what is the current in the cell during that period? Assume the gold ions carry one elementary unit of positive charge.

Answers

Answer:

0.158 A.

Explanation:

Mass of gold deposited = 3 x 10^-3 kg

= 3 g

Molar mass = 196 g/mol

Number of moles = 3/196

= 0.0153 mol.

Faraday's constant,

1 coloumb = 96500 C/mol

Quantity of charge, Q = 96500 * 0.0153

= 1477.04 C.

Remember,

Q = I * t

t = 2.59 hr

= 2.59 * 3600 s

= 9324 s

Current, I = 1477.04/9324

= 0.158 A.

Answer:

0.158A

Explanation:

Using Faraday's first law of electrolysis which states that the mass(m) of a substance deposited or liberated at any electrode is directly proportional to the quantity of charge or electricity (Q) passed. i.e

m ∝ Q

m = Z Q

Where;

Z is the proportionality constant called electrochemical equivalent.

Faraday also observed that when 1 Faraday of electricity is equivalent to 96500C of charge.

Also,

Quantity of charge (Q), which is the product of current (I) passing through and the time taken (t) for the electrolysis, is given by;

Q = I x t; ----------------------(i)

With all of these in place, now let's go answer the question.

Since the gold ions carry one elementary unit of positive charge, now let's write the cathode-half reaction for gold (Au) as follows;

Au⁺ + e⁻ = Au ---------------------(ii)

From equation (ii) it can be deduced that when;

1 Faraday (96500C) of electricity is passed, 1 mole of Au forms ( = 197 grams of Au) [molar mass of Au = 197g]

Then, 3.00 x 10⁻³ kg (= 3 g of Au) will be formed by 3g x 96500C / 197g = 1469.5C

Therefore, the quantity of charge (Q) deposited is 1469.5C

Substitute this value (Q = 1469.5C) and time t = 2.59h (= 2.59 x 3600 s) into equation (i);

Q = I x t

1469.5 = I x 2.59 x 3600

1469.5 = I x 9324

Solve for I;

I = 1469.5 / 9324

I = 0.158A

Therefore, the current in the cell during that period is 0.158A

Note:

1 mole of gold atoms = 176g

i.e the molar mass of gold (Au) is 176g

How much heat is required to convert 0.3 kilogram of ice at 0Â°C to water at the same temperature? A. 334,584 J B. 167,292 J C. 100,375 J D. 450,759 J

Answers

Answer:

Option C is the correct answer.

Explanation:

Heat required to melt solid in to liquid is calculated using the formula

H = mL, where m is the mass and L is the latent heat of fusion.

Latent heat of fusion for water = 333.55 J/g

Mass of ice = 0.3 kg = 300 g

Heat required to convert 0.3 kilogram of ice at 0°C to water at the same temperature

H = mL = 300 x 333.55 = 100,375 J

Option C is the correct answer.

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Dave rows a boat across a river at 4.0 m/s. the river flows at 6.0 m/s and is 360 m across.a. in what direction, relative to the shore, does dave’s boat go? b. how long does it take dave to cross the river? c. how far downstream is dave’s landing point? d. how long would it take dave to cross the river if there were no current?