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3 years ago

What is the acceleration of an object that has a velocity of 18 m/s and is moving in a circle of radius 30m?

Physics

2 answers:

blsea [12.9K]3 years ago

5 0

Acceleration = V^2/r = 18^2/ 30 = 10.8 m/s^2

insens350 [35]3 years ago

4 0

Well, first of all, 18 m/s is the object's speed, but not its velocity.
Velocity consists of the object's speed AND direction, and this
particular object's direction is constantly changing, since it's
moving around a circle. So its velocity is constantly changing too.

Centripetal acceleration = (speed)² / (radius)

   = (18 m/s)² / (30 m)

   = (324 m²/s²) / (30 m)

   = 10.8 m/s²

   (about 1.1 G)

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If the wind or current is pushing your boat away from the dock as you prepare to dock, which line should you secure first?

Akimi4 [234]

Answer:

Bow Line

Explanation:

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1- We should cast off the bow and stern lines.

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2 years ago

What is the measure of a change in an object’s velocity?

tiny-mole [99]

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2 years ago

In traveling a distance of 2.5 km between points A and D, a car is driven at 99 km/h from A to B for t seconds and 48 km/h from

Andreyy89

Answer:

d = 1.954 Km

Explanation:

given,

total distance, D = 2.5 Km

in stretch A to B =

speed = 99 Km/h = 99 x 0.278 = 27.22 m/s time =t

in stretch B to C

time = 3.4 s

In stretch C to D

speed = 48 Km/h = 48 x 0.278 = 13.34 m/s time =t

we know,

distance = speed x time

distance of BC

using equation of motion

v = u + a t

27.22 = 13.34 - a x 3.4

a = 4.08 m/s²

uniform deceleration is equal to 4.08 m/s²

distance traveled in BC

$s = ut + \dfrac{1}{2}at^2$

$s = 13.34\times 3.4 + \dfrac{1}{2}\times 4.08 \times 3.4^2$

s = 68.94 m

$3000 = 99 \times \dfrac{1000\ t}{3600}+ 68.94 + 48\times \dfrac{1000\ t}{3600}$

3000 = 27.5 t + 68.94 + 13.33 t

40.83 t = 2931.06

t = 71.79 s

distance travel in AB

distance = s x t

d = 27.22 x 71.79

d = 1954 m

d = 1.954 Km

distance between A and B is equal to 1.954 Km.

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3 years ago

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