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Allushta [10]
3 years ago
10

33. A zebra is at rest 60 m away from a charging lion

Physics
1 answer:
Ann [662]3 years ago
3 0

Answer:

the lion will miss the zebra by 3 meters

Explanation

The zebra has gone 23.75m after 5 seconds

and the lion can only go 20 m/s

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2. Determine the units of the quantity described by each of the following
notka56 [123]

Explanation:

a.) kg(m/s)(1/s) = force (N)

b.) (kg/s)(m/s) = force (N)

c.) (kg/s)(m/s) = force (N)

d.) (kg/s)(m/s) = force (N)

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What is true about all scientists
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A 60 kg student is standing in the train station next to her 10 kg suitcase when her train is called. (A) Estimate how much work
ASHA 777 [7]

Answer

given,

mass of student = 60 Kg

mass of suitcase = 10 Kg

a) Work done by picking of Suitcase is equal to zero

b) acceleration = 0.1 m/s²

   distance = 2 m

Using second law conservation

F = m a

F = 10 x 0.1 = 1 N

c) Work done

  W = F x s

  W = 1 x 2 = 2 J

d) When the are moving with constant speed acceleration is equal to zero

F = m a

F = 10 x 0 = 0 N

W = F x s = 0 x s = 0 J

e)   work done = change in kinetic energy

        K.E = 2 J

5 0
3 years ago
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

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