The linear search is adequate for searching through ________ arrays, but not through _______

Help?? Pictures for each state

Over [174]

Explanation:

Kentucky: image of blue grass state the atlantic

4 0

3 years ago

Find the x-component of this

valentina_108 [34]

The x-component of this vector inclined from the horizontal axis is 15m.

<h3>What is vector?</h3>

A vector is the representation of a physical quantity in magnitude and direction.

A vector incline at an angle in has two components. On in x direction and other is in y direction.

For the given length of vector = 45.6 m and the angle of inclination from the +x axis θ = 70.8°

The x component of the given vector is

= 45.6cos 70.8°

= 14.99 m

Thus, the x- component of this vector is 15m.

Learn more about vector.

brainly.com/question/13322477

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4 0

2 years ago

A ball of mass m is thrown straight upward from ground level at speed v0. At the same instant, at a distance D above the ground,

n200080 [17]

Answer:

a. $t = \frac{v_{0} +/- \sqrt{v_{0} ^{2} - gD} }{g}$ b. D = v₀²/2g

Explanation:

Here is the complete question

A ball is thrown straight up from the ground with speed v₀ . At the same instant, a second ball is dropped from rest from a height D , directly above the point where the first ball was thrown upward. There is no air resistance

Find the time at which the two balls collide.

Express your answer in terms of the variables D ,v₀ , and appropriate constants..

t = ?!

Part B

Find the value of D in terms of v₀ and g so that at the instant when the balls collide, the first ball is at the highest point of its motion.

Express your answer in terms of the variables v₀ and g .

D =?!

Solution

The distance moved by the ball dropped from distance,D with velocity v₀, H₁ = D - (v₀t - gt²/2) = D + v₀t + gt²/2.

The distance moved by the ball thrown straight upward with velocity v₀ is H₂ = v₀t - gt²/2.

The two balls collide when their vertical distances are equal. That is H₁ = H₂

So, D - v₀t + gt²/2 = v₀t - gt²/2

Collecting like terms

D + gt²/2 + gt²/2 = v₀t + v₀t

D +gt² = 2v₀t

gt² - 2v₀t + D = 0.

Using the quadratic formula,

$t = \frac{-(-2v_{0} ) +/- \sqrt{(-2v_{0} )^{2} - 4 X g XD} }{2g} = \frac{2v_{0} +/- \sqrt{4v_{0} ^{2} - 4gD} }{2g} = \frac{v_{0} +/- \sqrt{v_{0} ^{2} - gD} }{g}$

B. At its highest point, the velocity of the first ball, v = 0. Using v² = u² - 2gs where s = highest point of first ball when they collide and u = v₀.

0 = v₀² - 2gs

s = v₀²/2g.

Also, the time it takes the first ball to reach its highest point is gotten from v = u - gt. At highest point, v = 0 and u = v₀. So,

0 = v₀ - gt₀

t₀ = v₀/g

Also H = s₁ + s where s₁ = distance moved by second ball in time t₀ for collision = v₀t₀ - gt₀²/2.

So, H = v₀t₀ - gt₀²/2 + v₀²/2g = v₀(v₀/g) - g(v₀/g)²/2 + v₀²/2g = v₀²/2g - v₀²/2g + v₀²/2g = v₀²/2g

6 0

3 years ago

A projectile is launched at ground level with an initial speed of 54.5 m/s at an angle of 35.0° above the horizontal. It strikes

Alchen [17]

<h2>Answer: x=125m, y=48.308m</h2>

Explanation:

This situation is a good example of the projectile motion or parabolic motion, in which we have two components: x-component and y-component. Being their main equations to find the position as follows:

x-component:

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

Where:

$V_{o}=54.5m/s$ is the projectile's initial speed

$\theta=35\°$ is the angle

$t=2.80s$ is the time since the projectile is launched until it strikes the target

$x$ is the final horizontal position of the projectile (the value we want to find)

y-component:

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

Where:

$y_{o}=0$ is the initial height of the projectile (we are told it was launched at ground level)

$y$ is the final height of the projectile (the value we want to find)

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

Having this clear, let's begin with x (1):

$x=(54.5m/s)cos(35\°)(2.8s)$ (3)

$x=125m$ (4) This is the horizontal final position of the projectile

For y (2):

$y=0+(54.5m/s)sin(35\°)(2.8s)-\frac{(9.8m/s^{2})(2.8s)^{2}}{2}$ (5)

$y=48.308m$ (6) This is the vertical final position of the projectile

4 0

4 years ago

A proton travels with a speed of 5.02×10⁶ m/s in a direction that makes an angle of 60.0° with the direction of a magnetic ficld

lesantik [10]

Answer: The magnitude of the proton's acceleration is 0.748 ×10^14 m/s²

Explanation:
the velocity ,v, of ththe proton = 5.02×10^6 m/s
Magnitude , B , of the magnetic field = 0.180 T

First , we need to find the magnitude of the Force on the proton. This is given by the relation :
F = q(v x B) = qvBsinθ

where 'q' is the charge of proton , q= 1.6×10^-19 C
θ is the angle the proton makes with the direction of the magnetic field

Putting the respective values of v, B ,θ in the above equation, we get:

F = (1.6×10^-19 C)(5.02×10^6 m/s)(0.180T) sin60°
∴ F = 1.25 ×10^-13 N

Now , from Newton's second law we know that ,
F=m×a

∴ a = F/m

Mass of a proton = 1.67×10^27 kg
a= 1.25 × 10^-13 N / 1.67 × 10^27 kg

a= 0.748 × 10^14 m/s² =acceleration of the proton

(To know more about Magnetic Fields : brainly.com/question/9095546)

5 0

2 years ago

The linear search is adequate for searching through ________ arrays, but not through ________ ones

The linear search is adequate for searching through arrays, but not through ones