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Leya [2.2K]
3 years ago
10

A commuter backs her car out of her garage with an acceleration of 1.40 m/s2.A) How long does it take her to reach a speed of 2.

00 m/s?B) If she then breaks to a stop in 0.800 s, what is her deceleration?
Physics
1 answer:
AURORKA [14]3 years ago
4 0

Answer:

(A) 1.43secs

(B) -2.50m/s^2

Explanation:

A commuter backs her car out of her garage with an acceleration of 1.40m/s^2

(A) When the speed is 2.00m/s then, the time can be calculated as follows

t= Vf-Vo/a

The values given are a= 1.40m/s^2 , Vf= 2.00m/s, Vo= 0

= 2.00-0/1.40

= 2.00/1.40

= 1.43secs

(B) The deceleration when the time is 0.800secs can be calculated as follows

a= Vf-Vo/t

= 0-2.00/0.800

= -2.00/0.800

= -2.50m/s^2

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Answer:

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a=\frac{h}{5(t_1)^2}

Explanation:

(The complete question is given in the attached picture. We need to find the acceleration in terms of h and t in this question)

We are given 3 stages of movement of elevator. We'll first model them each of the stage one by one to find the height covered in each stage. After that we'll find the total height covered by adding heights covered in each stage, and equate it to Total height h. From that we can find the formula for acceleration.

<h3></h3><h3>Stage 1</h3>

Constant acceleration, starts from rest.

Distance = y = \frac{1}{2}a(t_1)^2

Velocity = v_1=at_1

<h3>Stage 2</h3>

Constant velocity where

Velocity = v_o=v_1=at_1

Distance =

<h3>y_2=v_2(t_2)\\\text{Where~}t_2=4t_1 ~\text{and}~ v_2=v_1=at_1\\y_2=(at_1)(4t_1)\\y_2=4a(t_1)^2\\</h3><h3 /><h3>Stage 3</h3>

Constant deceleration where

Velocity = v_0=v_1=at_1

Distance =

y_3=v_1t_3-\frac{1}{2}a(t_3)^2\\\text{Where}~t_3=t_1\\y_3=v_1t_1-\frac{1}{2}a(t_1)^2\\\text{Where}~ v_1t_1=a(t_1)^2\\y_3=a(t_1)^2-\frac{1}{2}a(t_1)^2\\\text{Subtracting both terms:}\\y_3=\frac{1}{2}a(t_1)^2

<h3>Total Height</h3>

Total height = y₁ + y₂ + y₃

Total height = \frac{1}{2}a(t_1)^2+4a(t_1)^2+\frac{1}{2}a(t_1)^2 = 5a(t_1)^2

<h3 /><h3>Acceleration</h3>

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h = 5a(t₁)²

a=\frac{h}{5(t_1)^2}

6 0
3 years ago
When a mass of 0.350 kg is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The mass is now displ
pantera1 [17]

Answer:

The period is T =  0.700 \ s

Explanation:

From the question we are told that  

    The mass is m =  0.350  \ kg

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The spring constant for this is mathematically represented as

       k  = \frac{F}{x}

Where F is the force on the spring which is mathematically evaluated as

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    k  = \frac{3.43 }{ 0.12}

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melisa1 [442]

1.5m/s2

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Answer:

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Answer:

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