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ziro4ka [17]
3 years ago
12

(a) What is the angular speed ω about the polar axis of a point on Earth's surface at a latitude of 55° N? (Earth rotates about

that axis.) (b) What is the linear speed v of the point? What are (c) ω and (d) v for a point at the equator? (Note: Earth radius equals 6370 km and let one day be 24 hours)
Physics
1 answer:
Vaselesa [24]3 years ago
6 0

Answer

given,

radius of earth = 6370 km

earths latitude = 55° N

time  = 24 hour = 86400 s

angular speed = ω

a) \omega = \dfrac{2\pi}{T}

\omega = \dfrac{2\pi}{86400}

\omega = 7.27 \times 10^{-5}rad/s

b) v = r\omega cos\theta

v = 6370\times 7.27 \times 10^{-5}\times cos 55^0

v = 266 m/s

c) ω at equator

\omega = \dfrac{2\pi}{T}

\omega = \dfrac{2\pi}{86400}

\omega = 7.27 \times 10^{-5}rad/s

d) v = r\omega cos\theta

v = 6370\times 7.27 \times 10^{-5}\times cos 0^0

v = 463 m/s

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Un proyectil es lanzado horizontalmente desde una altura de 12 metro con una velocidad de 80 m/sg. a.Calcular el tiempo de vuelo
Naily [24]

Answer:

t= 1,56 s ,   x= 124,8 m , v = (80 i^ - 15,288 j ) m/s

Explanation:

Este es un ejercicio de lanzamiento proyectiles, comencemos por encontrar el tiempo que tarda en llegar al piso

        y = y₀ + v_{oy} t – ½ g t²

en este caso la altura inicial es y₀= 12 m y llega a y=0 , como es lanzado horizontalmente la velocidad vertical es cero (v_{oy}=0)

       0 = y₀ – ½ g t²

       t= √ (2 y₀/g)

calculemos

       t= √ ( 2 12 / 9,8)

       t= 1,56 s

El alcance del proyectil es la distancia horizontal recorrida  

        x = v₀ₓ t

        x = 80 1,56

        x= 124,8 m

La velocidad de impacto cuando toca el suelo

        vx = v₀ₓ = 80 ms

        v_{y} = v_{oy} – gt

        v_{y} = - 9,8 1,56

        v_{y} = - 15,288 m/s

la velocidad es

       v = (80 i^ - 15,288 j ) m/s

Traducttion  

This is a projectile launching exercise, let's start by finding the time it takes to reach the ground

        y = y₀ + v_{oy} t - ½ g t²

in this case the initial height is i = 12 m and it reaches y = 0, as it is thrown horizontally the vertical speed is zero (v_{oy} = 0)

       0 =y₀I - ½ g t²

       t = √ (2y₀ / g)

let's calculate

       t = √ (2 12 / 9.8)

       t = 1.56 s

Projectile range is the horizontal distance traveled

        x = v₀ₓ t

        x = 80 1.56

        x = 124.8 m

Impact speed when it hits the ground

        vₓ = v₀ₓ = 80 ms

        v_{y} = v_{oy} - gt

        v_{y} = - 9.8 1.56

        v_{y} = - 15,288 m / s

the speed is

       v = (80 i ^ - 15,288 j) m / s

7 0
3 years ago
If the pendulum took longer to complete one oscillation, how would the graph change?
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Answer:

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it takes 90 j of work to stretch a spring 0.2 m from its equilibrium position. How muc work is needed to stretch it an additiona
Vinvika [58]

Work needed: 720 J

Explanation:

The work needed to stretch a spring is equal to the elastic potential energy stored in the spring when it is stretched, which is given by

E=\frac{1}{2}kx^2

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k is the spring constant

x is the stretching of the spring from the equilibrium position

In this problem, we have

E = 90 J (work done to stretch the spring)

x = 0.2 m (stretching)

Therefore, the spring constant is

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Substituting,

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