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Elenna [48]
4 years ago
10

Suppose that each component of a certain vector is doubled. (a) by what multiplicative factor does the magnitude of the vector c

hange? (b) by what multiplicative factor does the direction angle of the vector change?
Physics
1 answer:
gladu [14]4 years ago
7 0
<span>a. The magnitude of the vector is doubled as well. Let's say we have a 2-dimensional vector with components x and y. It's magnitude lâ‚ is given by: lâ‚ = âš(x² + y²) If we double the components x and y, the new magnitude lâ‚‚ is: lâ‚‚ = âš((2x)² + (2y²)) With a bit of algebra... lâ‚‚ = âš(4x² + 4y²) lâ‚‚ = âš4(x² + y²) lâ‚‚ = 2âš(x² + y²) We can write the new magnitude lâ‚‚ in terms of the old magnitude lâ‚. lâ‚‚ = 2lâ‚ Therefore, the new magnitude is double the old one. It should be clear that this relationship applies to 3D (and 1D) vectors as well. b. The direction angle is unchanged. The direction angle θ₠for a 2-dimensional vector is given by: θ₠= arctan(y / x) If we double both components, we get: θ₂ = arctan(2y / 2x) θ₂ = arctan(y / x) θ₂ = θ₠The new direction angle is the same as the old one.</span>
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The hour and minute hands of a tower clock like Big Ben in London are 2.79 m and 4.44 m long and have masses of 58.2 kg and 90 k
LuckyWell [14K]

Answer: 895.85 x 10^-6 J or 8.96 x 10^-4 J

Explanation:

Angular kinetic energy E in Joules

E = ½Iw^2

W is angular velocity in radians/sec

1 radian/sec = 9.55 rev/min

I is moment of inertia in kgm^2

I = cMR^2

M is mass (kg), R is radius (meters)

c = 1/3 for a rod around its end, R = length

For minute hand

I = (1/3)(90)(4.44)^2 = 0.33 x 90 x 19.7136 = 585.49

w= 1 rev/hour = 1 rev/3600sec = 2pi/ 3600 = pi/1800 rad/s

KE = 1/2(1/3)(90)(4.44)^2(pi/1800)^2 = 0.5 x 0.33 x 90 x 19.7136 x 3.05 x 10^-6

KE = 0.00089 J

For hour hand

I = (1/3)(58.2)(2.79)^2 = 0.33 x 58.2 x 2.79^2 = 149.5

w = 1 rev/12hour = 1 rev/(12x3600sec) = 2pi/ 12x3600 = pi/21600 rad/s

KE = 1/2(1/3)(90)(4.44)^2(pi/21600)^2 = 0.5 x 0.33 x 90 x 19.7136 x 2.12 x 10^-8

KE = 5.85 x 10^-6 J

Therefore total kinetic energy = 895.85 x 10^-6 J

4 0
4 years ago
Two hockey pucks, labeled A and B, are initially at rest on asmooth ice surface and are separated by a distance of18.0 m. Simult
zlopas [31]

Answer:

8.505 m

Explanation:

Let V1 and V2 be velocities of puck A and B respectively

Since A and B move in the same direction, so the relative velocity will be V1+V2=3.5+3.9=7.4m/s

Or

Vr=7.4 m/s

Distance=S= 18 m

Time =t=?

S=Vr×t

==> t=S/Vr

==> t= 18/7.4=2.43 sec

At this time both will strike together

<em><u>Distance by puck A</u></em>

<em>V1=3.5 m/s</em>

Time=t= 2.43 sec

Distance covered=d=?

d=V1×t=3.5×2.43=8.505 m

So, puck A will cover 8.505 meters before collision

6 0
3 years ago
A sound wave has a speed of 340 m/s and a frequency of 1000 hz. The period of the sound is closest to?
harina [27]
<h2>Relationship Between Frequency and Period</h2>

The frequency and the period are inversely proportional.

f=\dfrac{1}{T} where T is the period

<h2>Solving the Question</h2>

We're given:

  • <em>v</em> = 340 m/s
  • <em>f</em> = 1000 Hz

Because the frequency and the period are reciprocals of each other, we can find the period of the sound by finding the reciprocal of the frequency:

T=\dfrac{1}{1000}

<h2>Answer</h2>

T=\dfrac{1}{1000}

6 0
2 years ago
If the electron just misses the upper plate as it emerges from the field, find the speed of the electron as it emerges from the
Scilla [17]

The speed of the electron as it emerges from the field is; 388.587 m/s

<h3>What is the speed of the electron?</h3>

Initial speed; v₀x = 1.1 * 10⁶ m/s

Acceleration in horizontal direction = 0 m/s²

distance; s_x = 2 cm = 0.02 m

Thus, formula to find time here is;

t = s_x/v₀x

t = 0.02/(1.1 * 10⁶)

t = 1.82 * 10⁻⁶ s

Now for the vertical distance; v,y_o = 0 m/s

Thus, the equation of motion becomes;

s_y = ¹/₂at²

0.005 = ¹/₂a(1.82 * 10⁻⁶)²

Solving for a gives;

a = 3.02 * 10¹³ m/s²

Thus the speed of the electron as it emerges from the field is;

v² = u² + 2as

v = √(0² + 2(3.02 * 10¹³ * 0.005))

v = 388.587 m/s

Read more about Electron speed at; brainly.com/question/15094100

#SPJ1

Complete Question is;

An electron is projected with an initial speed v0 = 1.1 * 10⁶ m/s into the uniform field between the parallel plates. The distance between the plates is 1 cm and the length of the plates is 2 cm. Assume that the field between the plates is uniform and directed vertically downward, and that the field outside the plates is zero. The electron enters the field at a point midway between the plates. E = N/C

find the speed of the electron as it emerges from the field?.

6 0
1 year ago
A converging lens brings rays of light together at a focal point. The bending of light rays is the result ofA. A combination of
Tcecarenko [31]

Answer:

.C. Refraction of light passing through the lens.

Explanation:

The convergence or divergence of the rays of light is only due to refraction of light.  When a ray of light is made to pass through a lens, it bend's from it's original path. This bending of light from its original path is called Refraction. Whereas in Reflection the ray of light is reflected back into the same medium.

Hence the correct option is C. Refraction of light passing through the lens.

5 0
3 years ago
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