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2 years ago

A 1500 kg car

1 answer:

5 0

<h2>We determine the final velocity, </h2><h2>v</h2><h2>f</h2><h2>, of the car. We do this by applying the kinematic equation,</h2><h2 /><h2>v</h2><h2>2</h2><h2>f</h2><h2>=</h2><h2>v</h2><h2>2</h2><h2>i</h2><h2>+</h2><h2>2</h2><h2>a</h2><h2>d</h2><h2 /><h2>where </h2><h2>v</h2><h2>i</h2><h2> is the initial velocity, a is the acceleration, and d is the distance covered across the movement. We can express the acceleration using the equation,</h2><h2 /><h2>a</h2><h2>=</h2><h2>F</h2><h2>m</h2><h2 /><h2>where F is the applied force and m is the mass of the car. We use the following values for the variables:</h2><h2 /><h2>v</h2><h2>i</h2><h2>=</h2><h2>5.25</h2><h2> </h2><h2>m</h2><h2>s</h2><h2>−</h2><h2>1</h2><h2>F</h2><h2>=</h2><h2>1250</h2><h2> </h2><h2>N</h2><h2>m</h2><h2>=</h2><h2>1500</h2><h2> </h2><h2>k</h2><h2>g</h2><h2>d</h2><h2>=</h2><h2>42.8</h2><h2> </h2><h2>m</h2><h2>We proceed with the solution.</h2><h2 /><h2>v</h2><h2>2</h2><h2>f</h2><h2>=</h2><h2>v</h2><h2>2</h2><h2>i</h2><h2>+</h2><h2>2</h2><h2>a</h2><h2>d</h2><h2>v</h2><h2>2</h2><h2>f</h2><h2>=</h2><h2>v</h2><h2>2</h2><h2>i</h2><h2>+</h2><h2>2</h2><h2>F</h2><h2>d</h2><h2>m</h2><h2>Take the positive root.</h2><h2>v</h2><h2>f</h2><h2>=</h2><h2>√</h2><h2>v</h2><h2>2</h2><h2>i</h2><h2>+</h2><h2>2</h2><h2>F</h2><h2>d</h2><h2>m</h2><h2>=</h2><h2>√</h2><h2>(</h2><h2>5.25</h2><h2> </h2><h2>m</h2><h2>s</h2><h2>−</h2><h2>1</h2><h2>)</h2><h2>2</h2><h2>+</h2><h2>2</h2><h2>(</h2><h2>1250</h2><h2> </h2><h2>N</h2><h2>)</h2><h2>(</h2><h2>42.8</h2><h2> </h2><h2>m</h2><h2>)</h2><h2>1500</h2><h2> </h2><h2>k</h2><h2>g</h2><h2>≈</h2><h2>9.9</h2><h2> </h2><h2>m</h2><h2>s</h2><h2>−</h2><h2>1</h2><h2>Help improve Study.com. Report an Error</h2>

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What is your operational definition for a fast reaction time?

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Answer:

The ability to react to a certain stimulus with a speedy and effective manner

Explanation:

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3 years ago

Suppose you are drinking root beer from a conical paper cup. The cup has a diameter of 10 centimeters and a depth of 13 centimet

sveta [45]

Answer:

The level of the root beer is dropping at a rate of 0.08603 cm/s.

Explanation:

The volume of the cone is :

$V=\frac {1}{3}\times \pi\times r^2\times h$

Where, V is the volume of the cone

r is the radius of the cone

h is the height of the cone

The ratio of the radius and the height remains constant in overall the cone.

Thus, given that, r = d / 2 = 10 / 2 cm = 5 cm

h = 13 cm

r / h = 5 / 13

r = {5 / 13} h

$V=\frac {1}{3}\times \frac {22}{7}\times ({{{\frac {5}{13}\times h}}})^2\times h$

$V=\frac {550}{3549}\times h^3$

Also differentiating the expression of volume w.r.t. time as:

$\frac {dV}{dt}=\frac {550}{3549}\times 3\times h^2\times \frac {dh}{dt}$

Given: $\frac {dV}{dt}$ = -4 cm³/sec (negative sign to show leaving)

h = 10 cm

So,

$-4=\frac{550}{3549}\times 3\times {10}^2\times \frac {dh}{dt}$

$\frac{55000}{1183}\times \frac {dh}{dt}=-4$

$\frac {dh}{dt}=-0.08603\ cm/s$

<u>The level of the root beer is dropping at a rate of 0.08603 cm/s.</u>

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3 years ago

J.J. Thomson discovered the electron by noticing that: A. molecules with the same atoms exhibited the same chemical properties.

GenaCL600 [577]

J.J. Thomson discovered the electron by noticing that a beam of particles could be influenced by an electric or magnetic force.. That is option B.

<h3>What is an electron?</h3>

An electron can be defined as the part of an atom that is negatively charged and is found revolving round the nucleus of an atom.

J.J. Thomson was the scientist that discovered electrons through subjecting two oppositely-charged electric plates around the cathode ray.

He noticed that the particles where deflected by both the magnetic and electric fields.

Learn more about cathode rays here:

brainly.com/question/4441361

#SPJ1

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2 years ago

A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the lens strength (a.k.a, lens p

Kipish [7]

Answer:

20.0 cm

Explanation:

Here is the complete question

The normal power for distant vision is 50.0 D. A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the power of her eyes. What is the closest object she can see clearly?

Solution

Now, the power of a lens, P = 1/f = 1/u + 1/v where f = focal length of lens, u = object distance from eye lens and v = image distance from eye lens.

Given that we require a 10 % increase in the power of the lens to accommodate the image she sees clearly, the new power P' = 50.0 D + 10/100 × 50 = 50.0 D + 5 D = 55.0 D.

Also, since the object is seen clearly, the distance from the eye lens to the retina equals the distance between the image and the eye lens. So, v = 2.00 cm = 0.02 m

Now, P' = 1/u + 1/v

1/u = P'- 1/v

1/u = 55.0 D - 1/0.02 m

1/u = 55.0 m⁻¹ - 1/0.02 m

1/u = 55.0 m⁻¹ - 50.0 m⁻¹

1/u = 5.0 m⁻¹

u = 1/5.0 m⁻¹

u = 0.2 m

u = 20 cm

So, at 55.0 dioptres, the closet object she can see is 20 cm from her eye.

8 0

2 years ago

Assume that the physics instructor would like to have normal visual acuity from 21 cm out to infinity and that his bifocals rest

shutvik [7]

This is note the complete question, the complete question is:

One of the lousy things about getting old (prepare yourself!) is that you can be both near-sighted and farsighted at once. Some original defect in the lens of your eye may cause you to only be able to focus on some objects a limited distance away (near-sighted). At the same time, as you age, the lens of your eye becomes more rigid and less able to change its shape. This will stop you from being able to focus on objects that are too close to your eye (far-sighted). Correcting both of these problems at once can be done by using bi-focals, or by placing two lenses in the same set of frames. An old physicist instructor can only focus on objects that lie at distance between 0.47 meters and 5.4 meters.

Assume that the physics instructor would like to have normal visual acuity from 21 cm out to infinity and that his bifocals rest 2.0 cm from his eye. What is the refractive power of the portion of the lense that will correct the instructors nearsightedness?

Answer: 3.04 D

Explanation:

when an object is held 21 cm away from the instructor's eyes, the spectacle lens must produce 0.47m ( the near point) away.

An image of 0.47m from the eye will be ( 47 - 2 )

i.e 45 cm from the spectacle lens since the spectacle lens is 2cm away from the eye.

Also, the image distance will become negative

gap between lense and eye = 2cm

Therefore;

image distance d₁ = - 45cm = - 0.45m

object distance d₀ = 21 - 2 = 19cm = 0.19m

P = 1/f = 1/ d = 1/d₀ + 1/d₁ = 1/0.19 + (-1/0.45)

P = 1/f = 5.26315789 - 2.22222222

P = 1/f = 3.04093567 ≈ 3.04 D

5 0

3 years ago