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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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Hydroelectric power is considered to be an example of multiple transfers of energy because potential energy when the water is in the reservoir turns into kinetic energy making the water move to the dam and then the dam moves into the turbine turning it to electric energy.  
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4. Are the forces acting on the chandelier unbalanced?​
kakasveta [241]

Answer:

If it is not an object in motion, all forces are balanced.

3 0
3 years ago
I'll mark brainliest
irga5000 [103]

Answer:

A) 35 ft

B) 5 ft

C) Net displacement = distance covered by the dog to retrieve the stick - distance covered before the dog starts chewing the stick

Explanation:

A) Total distance covered by the dog = 20 + 15

                                  = 35 ft

B) Since the other distance covered by the dog before chewing the stick, after the retrieval, was in an opposite direction to the initial direction, then;

total displacement of the dog = 20 - 15

                                  = 5 ft

C) Net displacement = distance covered by the dog to retrieve the stick + distance covered before the dog starts chewing the stick

But, displacement involves a specified direction. The distance covered before the dog starts chewing the stick was in an opposite direction to the initial direction.

Thus,

Net displacement = distance covered by the dog to retrieve the stick - distance covered before the dog starts chewing the stick

7 0
3 years ago
Suppose the original segment of wire is stretched to 10 times its original length. How much charge must be added to the wire to
Debora [2.8K]

Here we want to study how the linear charge density changes as we change the measures of our body.

We will find that we need to add 9*Q of charge to keep the linear charge density unchanged.

<em>I will take two assumptions:</em>

The charge is homogeneous, so the density is constant all along the wire.

As we work with a linear charge density we work in one dimension, so the wire "has no radius"

Originally, the wire has a charge Q and a length L.

The linear charge density will be given by:

λ = Q/L

Now the length of the wire is stretched to 10 times the original length, so we have:

L' = 10*L

We want to find the value of Q' such that λ' (the <u>linear density of the stretched wire</u>) is still equal to λ.

Then we will have:

λ' = Q'/L' = Q'/(10*L) = λ = Q/L

Q'/(10*L) = Q/L

Q'/10 = Q

Q' = 10*Q

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If you want to learn more, you can read:

brainly.com/question/14514975

6 0
2 years ago
A turtle takes 3.5 minutes to walk 18 m toward the south along a deserted highway. A truck driver stops and picks up the turtle.
Akimi4 [234]

Answer:

3.59 m/s

Explanation:

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\frac{Total displacement}{Total time}

The turtle first walks 18m south, and then is taken 1,1Km (or 1100m) north. Thus, the total displacement  is 1082m north (1100m north - 18m south).

Now we have to calculate the total time, which will be equal to the sum of the time the turtle walked and the time it was taken by truck.

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To calculate the truck time we use the equation:

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Thus,

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Total time = 210s + 91.67s = 301.67s.

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Since the average velocity is a vector, it has a magnitude and a direction. In this case the magnitude is 3.59 m/s and the direction is north since the turtle's final displacement is north of where it started.

8 0
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