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omeli [17]
3 years ago
6

A stone is thrown straight up from the edge of a roof, 925 feet above the ground, at a speed of 20 feet per second. Remembering

that the acceleration due to gravity is -32 feet per second squared, how high is the stone 6 seconds later?
Physics
1 answer:
Black_prince [1.1K]3 years ago
4 0
<h2>Answer: 469 feet</h2>

Explanation:

This problem is a good example of Vertical motion, where the main equation for this situation is:

y=y_{o}+V_{o}t-\frac{1}{2}gt^{2} (1)

Where:

y is the height of the stone at 6s (the value we want to find)

y_{o}=925ft is the initial height of the stone

V_{o}=20ft/s is the initial velocity of the stone

t=6s is the time  at which we need to find the height

g=32ft/s^{2} is the acceleration due to gravity

Having this clear, let's find y from (1):

y=925ft+(20ft/s)(6s)-\frac{1}{2}(32ft/s^{2})(6s)^{2} (2)

Finally:

y=469ft This is the height of the stone at t=6s

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<h3>Answer;</h3>

<em>B.)neither longitudinal nor transverse</em>

<h3><u>Explanation;</u></h3>
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3 years ago
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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.

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2 years ago
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3 years ago
a child riding a bicycle at 15 meters per second accelerates at -3,0 meters per second? for 4.0 seconds. What is the child's spe
Vika [28.1K]

Answer:

<h2> 27m/s</h2>

Explanation:

Given data

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substituting our data into the expression we have

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The velocity after 4 seconds is 27m/s

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