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HACTEHA [7]
3 years ago
13

If vector A = 6i - 2j + 3k, determine

Physics
1 answer:
Mrac [35]3 years ago
5 0

Answer:

(a) 2\vec A=12\hat i-4\hat j+6\hat k

(b) \displaystyle \vec{U_A}=12/7\hat i-4/7\hat j+6/7\hat k

(c) -4\vec{U_A}=-48/7\hat i+16/7\hat j-24/7\hat k

Explanation:

<u>Vectors</u>

Given a vector

\vec A=6\hat i-2\hat j+3\hat k

We must determine the following:

a) A vector in the same direction as A with double magnitude 2A.

If the vector goes in the same direction but has a different magnitude, we only need to multiply each component by a common factor, in this case, by 2. Thus, the required vector is:

2\vec A=12\hat i-4\hat j+6\hat k

b) A unit vector in the same direction of A.

The unit vector needs to compute the magnitude of the vector:

\mid A\mid=\sqrt{6^2+2^2+3^2}

\mid A\mid=\sqrt{36+4+9}=\sqrt{49}=7

\mid A\mid=7

The unit vector is:

\displaystyle \vec{U_A}=\frac{\vec A}{\mid \vec A\mid}

\displaystyle \vec{U_A}=\frac{12\hat i-4\hat j+6\hat k}{7}

\displaystyle \vec{U_A}=12/7\hat i-4/7\hat j+6/7\hat k

c) A vector opposite to A with magnitude 4 m. We assume the original vector is also expressed in m.

The opposite vector to A is obtained simply by multiplying the unit vector by -1. To make its magnitude equal to 4, also multiply by 4. In all, we multiply the unit vector by -4:

-4\vec{U_A}=-4(12/7\hat i-4/7\hat j+6/7\hat k)

-4\vec{U_A}=-48/7\hat i+16/7\hat j-24/7\hat k

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What is the acceleration of a 349 kg object that moved with a force of 750 N?
zavuch27 [327]

Answer:

<h3>The answer is 2.15 m/s²</h3>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

where

f is the force

m is the mass

From the question we have

a =  \frac{750}{349}  \\  = 2.14899713...

We have the final answer as

<h3>2.15 m/s²</h3>

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4 0
3 years ago
A ball is dropped from a building taking 3sec to fall to the ground. Calculate:
GenaCL600 [577]

Answer:

Vf = 29.4 m/s

h = 44.1 m

Explanation:

Data:

  • Initial Velocity (Vo) = 0 m/s
  • Gravity (g) = 9.8 m/s²
  • Time (t) = 3 s
  • Final Velocity (Vf) = ?
  • Height (h) = ?

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

Final Velocity

Use formula:

  • Vf = g * t

Replace:

  • Vf = 9.8 m/s² * 3s

Multiply:

  • Vf = 29.4 m/s

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

Height

Use formula:

  • \boxed{h=\frac{g*(t)^{2}}{2}}

Replace:

  • \boxed{h=\frac{9.8\frac{m}{s^{2}}*(3s)^{2}}{2}}

Multiply time squared:

  • \boxed{h=\frac{9.8\frac{m}{s^{2}}*9s^{2}}{2}}

Simplify the s², and multiply in the numerator:

  • \boxed{h=\frac{88.2m}{2}}

It divides:

  • \boxed{h=44.1\ m}

What is the velocity when falling to the ground?

The final velocity is <u>29.4 meters per seconds.</u>

How high is the building?

The height of the building is <u>44.1 meters.</u>

3 0
3 years ago
4.2 mol of monatomic gas A interacts with 3.2 mol of monatomic gas B. Gas A initially has 9500 J of thermal energy, but in the p
Komok [63]

Answer:

14657.32 J

Explanation:

Given Parameters ;

Number of moles mono atomic gas A ,   n 1  =  4 .2 mol

Number of moles mono atomic gas B ,   n 2  =  3.2mol

Initial energy of gas A ,   K A  =  9500  J

Thermal energy given by gas A to gas B ,   Δ K  =  600 J

Gas constant   R  = 8.314  J / molK

Let  K B  be the initial energy of gas B.

Let T be the equilibrium temperature of the gas after mixing.

Then we can write the energy of gas A after mixing as

(3/2)n1RT = KA - ΔK

⟹ (3/2) x 4.2 x 8.314 x T = 9500 - 600

T = (8900 x 3 )/(2x4.2x8.314) = 382.32 K

Energy of the gas B after mixing can be written as

(3/2)n2RT = KB + ΔK

⟹ (3/2) x 3.2 x 8.314 x 382.32 = KB + 600

⟹ KB = [(3/2) x 3.2 x 8.314 x 382.32] - 600

⟹ KB = 14657.32 J

6 0
4 years ago
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