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malfutka [58]
2 years ago
3

Jorge camina en línea recta por una plaza, cruzándola de forma diagonal, recorriendo 60 m en 3 min. A partir de esta situación,

responde: a) ¿Cuál es la distancia que recorre? b) ¿Cuál es el módulo del desplazamiento? c) ¿Qué valor tendrá su rapidez y el módulo de su velocidad?
Physics
1 answer:
Virty [35]2 years ago
8 0

Answer:

a)    d = 60m (distance)

b)    D = 60m (displacement)

c)    v = 20 m/min

d)   |v| = 20 m/min

Explanation:

a) The distance traveled by Jorge is 60m

d = 60m

b) The module of the displacement D, is equal to the values of the distance d, because Jorge walked in a straight line.

D = d = 60m

c) The speed of Jorge is given by the following formula:

v=\frac{d}{t}

d: distance = 60m

t: time of the walk = 3min

v=\frac{60m}{3min}=20\frac{m}{min}

The speed is 20 m/min

The module of the Jorge's velocity is:

|v|=\frac{D}{t}

D: displacement = d = 60m

t: time = 3 min

|v|=\frac{60m}{3min}=20\frac{m}{min}

The module of Jorge's velocity is 20 m/min

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Answer:

λ = 2042 nm

Explanation:

given data

screen distance d = 11 m

spot s = 4.5 cm = 4.5 ×10^{-2} m

separation L = 0.5 mm = 0.5 ×10^{-3} m

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what is λ

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we will find first angle between first max and central bright

that is tan θ = s/d

tan θ = 4.5 ×10^{-2}  / 11

θ = 0.234

and we know diffraction grating for max

L sinθ  = mλ

here we know m = 1  so put all value and find λ

L sinθ  = mλ

0.5 ×10^{-3}  sin(0.234)  = 1 λ

λ = 2042.02 ×10^{-9}  m

λ = 2042 nm

3 0
2 years ago
A shuttle on Earth has a mass of 4.5 E 5 kg. Compare its weight on Earth to its weight while in orbit at a height of 6.3 E 5 met
faltersainse [42]

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83%

Explanation:

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In orbit, the weight is:

w = GMm / (R+h)²

where h is the height of the shuttle above the surface of the Earth.

The ratio is:

w/W = R² / (R+h)²

w/W = (R / (R+h))²

Given that R = 6.4×10⁶ m and h = 6.3×10⁵ m:

w/W = (6.4×10⁶ / 7.03×10⁶)²

w/W = 0.83

The shuttle in orbit retains 83% of its weight on Earth.

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2 years ago
The transmutation of a radioactive uranium isotope, 234/92 U, into a radon isotope, 222/86 Rn, involves a series of three nuclea
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Explanation:

3 0
2 years ago
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A steel aircraft carrier is 370 m long when moving through the icy North Atlantic at a temperature of 2.0 °C. By how much does t
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Answer:

The carrier lengthen is 0.08436 m.

Explanation:

Given that,

Length = 370 m

Initial temperature = 2.0°C

Final temperature = 21°C

We need to calculate the change temperature

Using formula of change of temperature

\Delta T=T_{f}-T_{i}

\Delta T=21-2.0

\Delta T=19^{\circ}C

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Using formula of length

\Delta L=\alpha_{steel}\times L_{0}\times\Delta T

Put the value into the formula

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8 0
2 years ago
1. A 9.4×1021 kg moon orbits a distant planet in a circular orbit of radius 1.5×108 m. It experiences a 1.1×1019 N gravitational
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26 days

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m = 9.4×1021 kg

r= 1.5×108 m

F = 1.1×10^ 19 N

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==> 1.1 × 10^{19} = v^{2} × 6.26×10^{13}

==> v^{2} =  1.1 × 10^{19} ÷ 6.26×10^{13}

==> v^{2} = 0.17571885 × 10^{6}

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T= 2πr ÷ v

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