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3 years ago

A car traveling at 26 m/s starts to decelerate steadily. It comes to a complete stop in 5 seconds. What is its acceleration?

1 answer:

6 0

Acceleration or Deceleration = Velocity/Time
Deceleration = 26/5
Deceleration = 5.2 m/s²

Its supposed to be What is its "deceleration'' not acceleration.

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2 years ago

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2 years ago

most waves approach the shore at an angle. however, they bend to be nearly parallel to the shore as they approach it because bla

Alex

Most waves approach the shore at an angle. However, they bend to be nearly parallel to the shore as they approach it because when a wave reaches a beach or coastline, it releases a burst of energy that generates a current, which runs parallel to the shoreline.

Most waves approach shore at an angle. As each one arrives, it pushes water along the shore, creating what is known as a longshore current within the surf zone.
Waves approach the coast at an angle because of the direction of prevailing wind.
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1 year ago

Two circus performers rehearse a trick in which a ball and a dart collide. Horatio stands on a platform 9.8 m above the ground a

xenn [34]

Answer:

Time = 0.55 s

Height = 8.3 m

Explanation:

The ball is dropped and therefore has an initial velocity of 0. Its acceleration, g, is directed downward in the same direction as its displacement, $h_b$ .

The dart is thrown up in which case acceleration, g, acts downward in an opposite direction to its displacement, $h_d$ . Both collide after travelling for a time period, t. Let the height of the dart from the ground at collision be $h_d$ and the distance travelled by the ball measured from the top be $h_b$ .

It follows that $h_d+h_b=9.8$ .

Applying the equation of motion to each body (h = v_0t + 0.5at^2),

Ball:

$h_b=0\times t + 0.5\times 9.8t^2$ (since $v_{b0} =0$ .)

$h_b=4.9t^2$

Dart:

$h_d=17.8\times t - 0.5\times9.8t^2$ (the acceleration is opposite to the displacement, hence the negative sign)

$h_d=17.8\times t - 4.9t^2$

But

$h_b+h_d =9.8$

$17.8\times t - 4.9t^2+4.9t^2 =9.8$

$17.8\times t = 9.8$

$t = 0.55$

The height of the collision is the height of the dart above the ground, $h_d$ .

$h_d=17.8\times t - 4.9t^2$

$h_d=17.8\times 0.55 - 4.9\times(0.55)^2$

$h_d=9.79 - 1.48225$

$h_d=8.3$

8 0

3 years ago

A cart is driven by a large propeller or fan, which can accelerate or decelerate the cart. The cart starts out at the position x

mash [69]

Answer:

The acceleration of the cart is 1.0 m\s^2 in the negative direction.

Explanation:

Using the equation of motion:

Vf^2 = Vi^2 + 2*a*x

2*a*x = Vf^2 - Vi^2

a = (Vf^2 - Vi^2)/ 2*x

Where Vf is the final velocity of the cart, Vi is the initial velocity of the cart, a the acceleration of the cart and x the displacement of the cart.

Let x = Xf -Xi

Where Xf is the final position of the cart and Xi the initial position of the cart.

x = 12.5 - 0

x = 12.5

The cart comes to a stop before changing direction

Vf = 0 m/s

a = (0^2 - 5^2)/ 2*12.5

a = - 1 m/s^2

The cart is decelerating

Therefore the acceleration of the cart is 1.0 m\s^2 in the negative direction.

5 0

3 years ago