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3 years ago

10 PTS! HELP!

1 answer:

6 0

Shortsightedness is corrected<span> using a concave (curved inwards) lens which is placed in front of a myopic eye, moving the image back to the retina and making it clearer. Longsightedness is </span>corrected <span>using a </span>convex <span>Hyperopia is known to you probably as farsightedness(</span>outward facing) lens.

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A boy 12.0 m above the ground in a tree throws a ball for his dog, who is standing right below the tree and starts running the i

miss Akunina [59]

Answer:

Explanation:

Height covered = 12m

time to fall by 12 m

s = 1/2 gt²

12 = 1/2 g t²

t = 1.565 s

Horizontal distance of throw

= 8.5 x 1.565

= 13.3 m

This distance is to be covered by dog during the time ball falls ie 1.565 s

Speed of dog required = 13.3 / 1.565

= 8.5 m /s

b ) dog will catch the ball at a distance of 13.3 m .

4 0

3 years ago

If f(x) = -x2+x-1 then value of f(f(2)) is

qaws [65]

Answer:

Explanation:

<u>Function</u>

f(x) = -x² + x - 1

<u>Solving</u>

Substitute 2 in place of x
f(2) = -(2)² + 2 - 1
f(2) = -4 + 2 - 1
f(2) = -3
f(f(2)) = f(-3)
f(-3) = -(-3)² - 3 - 1
f(-3) = -9 - 3 - 1
f(f(2)) = <u>-13</u>

6 0

2 years ago

The hydronium ion concentration in an aqueous solution is [H+] = 5.3 x10-4 M. What is the hydroxide ion concentration? Is this s

VladimirAG [237]

Answer:a

Explanation:

Given

$[H^+]=5.3\times 10^{-4}\ M$

and $pH+pOH=14$

Also $pH=-\log [H^+]$

therefore $pH=-\log (5.3\times 10^{-3})$

$pH=-(-4-\log (5.3))$

$pH=3.275$

Thus $pOH=14-3.275$

$pOH=10.275$

and $pOH=-\log [OH^{-}]$

$[OH]^{-1}=10^{-10.25}$

$[OH]^{-1}=1.88\times 10^{-11}\ M$

As the pH is less than 7 therefore solution is acidic

6 0

3 years ago

How far can a bullet at 450 m/s travel for 4.7 s?

marta [7]

The bullet can travel about distance of (2115 m)

Velocity can be regarded as the ratio of distance to time

Velocity= (distance / time)

Given: velocity= 450 m/s

Time= 4.7s

Distance= ( velocity × time)

<em>If we substitute the values</em>

Distance = ( 450 × 4.7)

= 2115 m

Therefore, the bullet can travel about 2115 m

Learn more at : brainly.com/question/13639113?referrer=searchResults

5 0

3 years ago

A bullet is shot horizontally from shoulder height (1.5 m) with an initial speed 200 m/s. (a) How much time elapses before the b

guapka [62]

<h2>Answer: (a)t=0.553s, (b)x=110.656m</h2>

Explanation:

This situation is a good example of the projectile motion or parabolic motion, in which the travel of the bullet has two components: x-component and y-component. Being their main equations as follows:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$V_{o}=200m/s$ is the bullet's initial speed

$\theta=0$ because we are told the bullet is shot horizontally

$t$ is the time since the bullet is shot until it hits the ground

y-component:

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

Where:

$y_{o}=1.5m$ is the initial height of the bullet

$y=0$ is the final height of the bullet (when it finally hits the ground)

$g=9.8m/s^{2}$ is the acceleration due gravity

Now, for the first part of this problem, the time the bullet elapsed traveling, we will use equation (2) with the conditions given above:

$0=1.5m+200m/s.sin(0) t-\frac{9.8m/s^{2}.t^{2}}{2}$ (3)

$0=1.5m-\frac{9.8m/s^{2}.t^{2}}{2}$ (4)

Finding $t$ :

$t=\sqrt{\frac{1.5m(2)}{9.8m/s^{2}}}$ (5)

Then we have the time elapsed before the bullet hits the ground:

$t=0.553s$ (6)

For the second part of this problem, we are asked to find how far does the bullet traveled horizontally. This means we have to use the equation (1) related to the x-component:

$x=V_{o}cos\theta t$ (1)

Substituting the knonw values and the value of $t$ found in (6):

$x=200m/s.cos(0)(0.553s)$ (7)

$x=200m/s(0.553s)$ (8)

Finally:

$x=110.656m$

4 0

3 years ago