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Deffense [45]
4 years ago
15

Un carro tiene una velocidad angular de 95 rev/min. a) ¿Cuál es la velocidad tangencial del objeto si se colocan unos contrapeso

s a una distancia de 598 mm? b) Calcule la aceleración centrípeta.
Physics
1 answer:
Maru [420]4 years ago
6 0

Answer:

v = 5.949 m / s ,      a = 59.18 m / s²

Explanation:

Linear and angular quantities are related

           v = w r

we search the magnitudes to the SI system

           w = 95 rev / min (2π rad / 1rev) (1 min / 60 s) = 9.948 rad / s

           d = 598 mm = 0.598 m

a) let's calculate the linear velocity

           v = 9.948 0.598

           v = 5.949 m / s

b) Cenripetal acceleration

            a = v² / r

            a = 5.94 2 / 0.598

            a = 59.18 m / s²

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Answer:

speed is 81.03 mph

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Explanation:

given data

travel north = 70 mph

Stream current = 12 mph

direction = S 25° E

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to find out

speed and direction

solution

we will get component of resultant  that is

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so

( 12cos295 , 12 sin295 )    at ( 0, 70)

as that we can say

v sinθ + 12sin295 = 70      ....................1

v cosθ + 12 cos295  = 0     ......................2

so

vcosθ  = -5.0714

vsinθ = 80.8756

now by ratio

cosθ /sinθ  = -5.0714/ 80.8756

cot θ = -0.0627

θ = 93.58

so direction is N 3.58 W

and

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vcosθ  = - 12cos295

v = - 12cos295 / cos(93.58)

v = 81.03 mph

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3 years ago
You are driving home from school steadily at 97 km/h for 190 km . It then begins to rain and you slow to 60 km/h instantly. You
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4 years ago
While in a park, you walk west for 52 m, then you walk 34.6° north of west for 44 m, and finally you walk due north for 25 m. Fi
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Answer:

88m west and 50m north

Explanation:

1. Find the components of each vector in x and y:

(Please see the picture below)

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4 years ago
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