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frosja888 [35]
3 years ago
10

Multiple-Concept Example 9 reviews the concepts that are important in this problem. A drag racer, starting from rest, speeds up

for 391 m with an acceleration of +18.9 m/s2. A parachute then opens, slowing the car down with an acceleration of -9.92 m/s2. How fast is the racer moving 332 m after the parachute opens?​
Physics
1 answer:
Mademuasel [1]3 years ago
6 0

Answer:

V = 90.51 m/s

Explanation:

From the given information:

Initial speed (u) = 0

Distance (S) = 391 m

Acceleration (a) = 18.9 m/s²

Using the relation for the equation of motion:

v² - u² = 2as

v² - 0² = 2as

v² = 2as

v = \sqrt{2as}

v = \sqrt{2*18.9*391}

v = 121.57 m/s

After the parachute opens:

The initial velocity = 121.57 m/ss

Distance S' = 332 m

Acceleration = -9.92 m/s²

How fast is the racer can be determined by using the relation:

V=  \sqrt{v^2 + 2aS'}

V = \sqrt{121.57^2+ 2 (-9.92)(332)}

V = 90.51 m/s

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s2008m [1.1K]

To understand this question, you need to understand the concept of acceleration first. Have you ever been in a car and noticed that it was getting faster and faster? That "speeding up" of the car is known as acceleration! Acceleration is essentially the rate at which you speed up.

Okay, so we now know what acceleration is. What are its units? The unit of acceleration is the change in velocity over a period of time: \frac{∆v}{t}

If you haven't learned about velocity yet, just think about it as speed for now. The funny-looking triangle, ∆, is a symbol for "the change of." For example, if I started walking at 3 \frac{feet}{second} then sped up to 5 \frac{feet}{second}, then the change in my speed would be 2 \frac{feet}{second}, because I started walking 2 \frac{feet}{second} faster!

Okay, enough with all the explanations. Hopefully, you understand the units now. Let's take a look at the question. A car accelerates from 4 \frac{meters}{second} to 16 \frac{meters}{second}  in 4 seconds. What would the acceleration be? Let's set up an equation:

a = \frac{∆v}{t}

a is the acceleration, ∆v is the change in velocity, and t is the time elapsed.

Now, let's plug in our values! ∆v is the change in velocity, and to find that we simply have to subtract 16 \frac{meters}{second} by 4 \frac{meters}{second}. That makes sense, right? Back to the equation.

a = \frac{∆v}{t}
a = \frac{16-4}{4}

(16 - 4 is the change in velocity, and 4 is the number of seconds the car was accelerating)

a = \frac{12}{4}

a = 3 (\frac{meters}{second^{2}})

We have our answer! The car's acceleration is 3 meters per second^{2}.

(You might be thinking: Wait. Meters per second squared? The reason for that is because acceleration is the rate at which the speed increases! That makes the unit \frac{\frac{meters}{second}}{second}, which can be simplified down to \frac{meters}{second^{2} })

Let me know if you need clarification on anything I explained here!
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Answer:

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How do gravitational and electic force compare
velikii [3]

Electric Forces. ... Just like objects that have mass exert gravitational forces on each other, objects that are charged will also exert electric forces on each other. The electric force is directly proportional to the charge of the two objects and inversely proportional to the distance between them squared.

3 0
3 years ago
The launch velocity of a toy car launcher is determined to be 5 m/s. If the car is to be launched from a height of 0.5 m, where
lora16 [44]

Answer:

1.6 m

Explanation:

Given that the launch velocity of a toy car launcher is determined to be 5 m/s. If the car is to be launched from a height of 0.5 m.

The time for landing should be calculated by using the second equation of motion formula

h = Ut + 1/2gt^2

Let U = 0

0.5 = 1/2 × 9.8 × t^2

0.5 = 4.9t^2

t^2 = 0.5 / 4.9

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t = 0.32 s

The target should be placed so that the toy car lands on it at:

Distance = 5 × 0.32

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Therefore, the target should be placed so that the toy car lands on it 1.6 metres away.

7 0
3 years ago
A man pulled a 13.0 kg object 11.8 cm vertically with his teeth. (a) How much work (in J) was done on the object by the man in t
jonny [76]

Answer:

(a)The work done by the man is -15.03J.

(b)The force exerted on the object is 127.4N.

Explanation:

Mass of the object pulled by the man is -13kg

Object is lifted 11.8 cm vertical with his teeth it means (displacement = +11.8cm = +0.118m)

Acceleration due to gravity is 9.8 \mathrm{m} / \mathrm{s}^{2}

(a) <u>Calculating the work done</u>:

Work done = mgh

Where "m" is mass of an object, "g" is acceleration due to gravity and "h" is the displacement.

\text { Work }=-13 \times 9.8 \times(+0.118 \mathrm{m})

\text { Work }=-15.03 \mathrm{J}

The work done by the man is -15.03J.

(b) <u>Calculating the force</u>:

Probably the man and the object are close to the exterior of the earth. If the rigidity required to maintained the object of consistent velocity interior the gravitational field of the earth is \mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2}

Thus the weight of the object is balanced by the force of the man's teeth on the object. That is

F = mg

\mathrm{F}=13 \times 9.8

F = 127.4N

The force exerted on the object is 127.4N.

4 0
2 years ago
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