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3 years ago

11

After school, several students skateboard at the nearby park. Your best friend, Erwin, is a good skateboarder and happens to fal

l when attempting a fancy trick. After falling, Erwin gets back up and starts repeating the trick over and over again. How would a cognitive psychologist interpret Erwin getting back up and repeating the trick many times?

1 answer:

Alchen [17]3 years ago

3 0

Answer:

A person who never gives up.

Explanation:

due to his passion for skateboarding he try's his never gives up until he will finally learns the trick.

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In the Olympic shot-put event, an athlete throws the shot with an initial speed of 12.0m/s at a 40.0? angle from the horizontal.

HACTEHA [7]

A) Horizontal range: 16.34 m

B) Horizontal range: 16.38 m

C) Horizontal range: 16.34 m

D) Horizontal range: 16.07 m

E) The angle that gives the maximum range is $41.9^{\circ}$

Explanation:

A)

The motion of the shot is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.

The vertical motion is a uniformly accelerated motion, so we can use the following suvat equation to find the time of flight:

$s=u_y t + \frac{1}{2}at^2$ (1)

where

s = -1.80 m is the vertical displacement of the shot to reach the ground (negative = downward)

$u_y = u sin \theta$ is the initial vertical velocity, where

u = 12.0 m/s is the initial speed

$\theta=40.0^{\circ}$ is the angle of projection

So

$u_y=(12.0)(sin 40.0^{\circ})=7.7 m/s$

$a=g=-9.8 m/s^2$ is the acceleration due to gravity (downward)

Substituting the numbers, we get

$-1.80 = 7.7t -4.9t^2\\4.9t^2-7.7t-1.80=0$

which has two solutions:

t = -0.21 s (negative, we ignore it)

t = 1.778 s (this is the time of flight)

The horizontal motion is instead uniform, so the horizontal range is given by

$d=u_x t$

where

$u_x = u cos \theta=(12.0)(cos 40^{\circ})=9.19 m/s$ is the horizontal velocity

t = 1.778 s is the time of flight

Solving, we find

$d=(9.19)(1.778)=16.34 m$

B)

In this second case,

$\theta=42.5^{\circ}$

So the vertical velocity is

$u_y = u sin \theta = (12.0)(sin 42.5^{\circ})=8.1 m/s$

So the equation for the vertical motion becomes

$4.9t^2-8.1t-1.80=0$

Solving for t, we find that the time of flight is

t = 1.851 s

The horizontal velocity is

$u_x = u cos \theta = (12.0)(cos 42.5^{\circ})=8.85 m/s$

So, the range of the shot is

$d=u_x t = (8.85)(1.851)=16.38 m$

C)

In this third case,

$\theta=45^{\circ}$

So the vertical velocity is

$u_y = u sin \theta = (12.0)(sin 45^{\circ})=8.5 m/s$

So the equation for the vertical motion becomes

$4.9t^2-8.5t-1.80=0$

Solving for t, we find that the time of flight is

t = 1.925 s

The horizontal velocity is

$u_x = u cos \theta = (12.0)(cos 45^{\circ})=8.49 m/s$

So, the range of the shot is

$d=u_x t = (8.49)(1.925)=16.34$ m

D)

In this 4th case,

$\theta=47.5^{\circ}$

So the vertical velocity is

$u_y = u sin \theta = (12.0)(sin 47.5^{\circ})=8.8 m/s$

So the equation for the vertical motion becomes

$4.9t^2-8.8t-1.80=0$

Solving for t, we find that the time of flight is

t = 1.981 s

The horizontal velocity is

$u_x = u cos \theta = (12.0)(cos 47.5^{\circ})=8.11 m/s$

So, the range of the shot is

$d=u_x t = (8.11)(1.981)=16.07 m$

E)

From the previous parts, we see that the maximum range is obtained when the angle of releases is $\theta=42.5^{\circ}$ .

The actual angle of release which corresponds to the maximum range can be obtained as follows:

The equation for the vertical motion can be rewritten as

$s-u sin \theta t + \frac{1}{2}gt^2=0$

The solutions of this quadratic equation are

$t=\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g}$

This is the time of flight: so, the horizontal range is

$d=u_x t = u cos \theta (\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g})=\\=\frac{u^2}{-2g}(1+\sqrt{1+\frac{2gs}{u^2 sin^2 \theta}})sin 2\theta$

It can be found that the maximum of this function is obtained when the angle is

$\theta=cos^{-1}(\sqrt{\frac{2gs+u^2}{2gs+2u^2}})$

Therefore in this problem, the angle which leads to the maximum range is

$\theta=cos^{-1}(\sqrt{\frac{2(-9.8)(-1.80)+(12.0)^2}{2(-9.8)(-1.80)+2(12.0)^2}})=41.9^{\circ}$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

8 0

3 years ago

A flat coil of wire consisting of 15 turns, each with an area of 40 cm 2, is positioned perpendicularly to a uniform magnetic fi

zheka24 [161]

Answer:

0.54 A

Explanation:

Parameters given:

Number of turns, N = 15

Area of coil, A = 40 cm² = 0.004 m²

Change in magnetic field, ΔB = 5.1 - 1.5 = 3.6 T

Time interval, Δt = 2 secs

Resistance of the coil, R = 0.2 ohms

To get the magnitude of the current, we have to first find the magnitude of the EMF induced in the coil:

|V| = |(-N * ΔB * A) /Δt)

|V| = | (-15 * 3.6 * 0.004) / 2 |

|V| = 0.108 V

According to Ohm's law:

|V| = |I| * R

|I| = |V| / R

|I| = 0.108 / 0.2

|I| = 0.54 A

The magnitude of the current in the coil of wire is 0.54 A

6 0

3 years ago

When dogs see police dogs do they think "oh no its the cops"?

ira [324]

No they say "Watch out it's the fuzz"

4 0

2 years ago

Read 2 more answers

Consider a venturi with a small hole drilled in the side of the throat. This hole is connected via a tube to a closed reservoir.

Marina CMI [18]

Answer:

( $P_1-P_2$ )=1913.31 N/m^2

Explanation:

given:

$\frac{A_t}{A_1}$ =0.85

$V_1$ =90 m/s

γ∞=1.23 kg/m^3

solution:

since outside pressure is atm pressure vaccum can be defined by ( $P_1-P_2$ )

$V_1$ =√2( $P_1-P_2$ )/γ∞[ $\frac{A_t}{A_1}^2$ -1]

( $P_1-P_2$ )=1913.31 N/m^2

6 0

2 years ago

What is know as law of inertia?

schepotkina [342]

Answer:

<em><u>mark</u></em><em><u> </u></em><em><u>me</u></em><em><u> </u></em><em><u>brianliest</u></em><em><u> </u></em><em><u>plz</u></em>

Explanation:

Law of inertia, also called Newton's first law, postulate in physics that, if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force.
Law of Inertia states that a body in a state of rest or uniform motion remains in the same state until and unless an external force acts on it.
A body continues to be in its state of rest or in uniform motion along a straight line unless an external force is applied on it. This law is also called law of inertia.

5 0

2 years ago