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3 years ago

14

You throw a ball with a speed of 25.0 m/s at an angle of 40.0â—¦ above the horizontal directly toward a wall. The wall is 22.0 m

from the release point of the ball. How long does the ball take to reach the wall?
Show Options
A.11.15 s
B.11.5 s
C.1.115 s
D.1.15 s

2 answers:

Genrish500 [490]3 years ago

7 0

Answer:

D.1.15 s

Explanation:

When we decompose the velocity into two perpendicular components, the total vector and its components form a right triangle. Therefore, these components will be given by:

$cos\theta=\frac{v_x}{v}\\v_x=vcos\theta(1)\\sin\theta=\frac{v_y}{v}\\v_y=vsin\theta$

Using (1), we calculate the horizontal speed of the ball:

$v_x=25\frac{m}{s}cos(40^\circ)\\v_x=19.15\frac{m}{s}$

Recall that $v=\frac{x}{t}$ , solving for t:

$t=\frac{x}{v}\\t=\frac{22m}{19.15\frac{m}{s}}\\t=1.15s$

kirza4 [7]3 years ago

3 0

The horizontal speed is going to be the cosine of the given speed, therefore, the horizontal speed is 19.15 m/s. To find the time, divide the 22 m distance by the velocity. This results in 1.131 seconds, which is in between C and D.

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krok68 [10]

Answer:

5

Explanation:

Number of habitable planets = 1000

Fraction of planet with life = 1/10

Fraction of planet with life and civilization (before) = 1/4

Fraction of planet with life and civilization (now) =1/5

Therefore multiplying we have:

1000×1/10×1/4×1/5 = 5

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3 years ago

A particle's position is given by z(t) = −(6.50 m/s2)t2k for t ≥ 0. (Express your answer in vector form.) a. Find the particle's

blondinia [14]

Answer:

a) $z'(t) =v(t) = -13t$

Now we can replace the velocity for t=1.75 s

$v(1.75s) = -13*1.75 =-22.75 \frac{m}{s}$

For t = 3.0 s we have:

$v(3.0s) = -13*3.0 =-39 \frac{m}{s}$

b) $v_{avg}= \frac{z_f - z_i}{t_f -t_i}$

And we can find the positions for the two times required like this:

$z_f = z(3.0s) = -(6.5 \frac{m}{s^2}) (3.0s)^2=-58.5m$

$z_i = z(1.75s) = -(6.5 \frac{m}{s^2}) (1.75s)^2=-19.906m$

And now we can replace and we got:

$V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}$

Explanation:

The particle position is given by:

$z(t) = -(6.5 \frac{m}{s^2}) t^2, t\geq 0$

Part a

In order to find the velocity we need to take the first derivate for the position function like this:

$z'(t) =v(t) = -13t$

Now we can replace the velocity for t=1.75 s

$v(1.75s) = -13*1.75 =-22.75 \frac{m}{s}$

For t = 3.0 s we have:

$v(3.0s) = -13*3.0 =-39 \frac{m}{s}$

Part b

For this case we can find the average velocity with the following formula:

$v_{avg}= \frac{z_f - z_i}{t_f -t_i}$

And we can find the positions for the two times required like this:

$z_f = z(3.0s) = -(6.5 \frac{m}{s^2}) (3.0s)^2=-58.5m$

$z_i = z(1.75s) = -(6.5 \frac{m}{s^2}) (1.75s)^2=-19.906m$

And now we can replace and we got:

$V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}$

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3 years ago

An airplane flying at 115 m/s due east makes a gradual turn following a circular path to fly south. The turn takes 15 seconds to

bulgar [2K]

To solve this exercise it is necessary to apply the concepts related to Centripetal and Perimeter acceleration of a circle.

The perimeter of a circle is defined by

$P = 2\pi r$

Where,

r= radius

While centripetal acceleration is defined by

$a=\frac{v^2}{r}$

Where,

v= velocity

r= radius

PART A)

The distance of a body can be defined based on the speed and the time traveled, that is

x = v*t

For our values the distance is equal to

x = 15*115=1725m

The plane when going to make the turn from east to south makes a quarter of the circumference that is

$\frac{P}{4} = \frac{2\pi r}{4}$

The same route you take is the distance traveled, that is

$x = \frac{P}{4}$

$x = \frac{2\pi r}{4}$

$1725 = \frac{2\pi r}{4}$

$r = 1098.17m$

PART B)

With the radius is possible calculate he centripetal acceleration,

$a = \frac{v^2}{r}$

$a = \frac{115^2}{1098.17}$

$a = 12.04m/s^2$

Therefore the radius of the curva that the plane follows in making the turn is 1098.17m with a centripetal acceleration of $12.04m/s^2$

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