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3 years ago

14

_____ occurs when a subject opens there eyes and devotes their attention to an object and the eeg rhythm changes to fast, irregu

lar, shallow waves.

1 answer:

slega [8]3 years ago

3 0

Answer: B. Alpha block

Explanation:

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An object has a mass of 5.20g. The mass of the object in kg is

vampirchik [111]

Answer:

0.0520kg

Explanation:

divide by 1000

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3 years ago

a fast humvee drove from desert a to desert b. for the 12 hours, it travelled at an speed of 185 km/h. for the next 13 hours, it

ivanzaharov [21]

Answer:

I am not sure of the question?

Explanation:

List the question

8 0

3 years ago

A ball that has a mass of 0.25 kg spins in a circle at the end of a 1.6 m rope. the ball moves at a tangential speed of 12.2 m/s

NARA [144]

The centripetal force acting on the ball will be 23.26 N.The direction of the centripetal force is always in the path of the center of the course.

<h3>What is centripetal force?</h3>

The force needed to move a body in a curved way is understood as centripetal force. This is a force that can be sensed from both the fixed frame and the spinning body's frame of concern.

The given data in the problem is;

m is the mass of A ball = 0.25 kg

r is the radius of circle= 1.6 m rope

v is the tangential speed = 12.2 m/s

$\rm F_C$ is the centripetal force acting on the ball

The centripetal force is found as;

$\rm F_C = \frac{mv^2}{r} \\\\ F_C = \frac{0.25 \times (12.2)^2}{1.6} \\\\ F_C=23.26\ N$

Hence the centripetal force acting on the ball will be 23.26 N.

To learn more about the centripetal force refer to the link;

brainly.com/question/10596517

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2 years ago

What is the SI unit for intensity?

Allisa [31]

Answer:

The SI unit of intensity is the watt per square meter/metre (W/m^2.)

Explanation:

Intensity is equal to the power transferred per unit area. Since power is measured in watts (W) and 1 W = 1 J/s, then intensity can be viewed as how fast energy goes through a certain area.

In physics, intensity is often used when studying light, sound, or other phenomena that involve waves or energy transfer. (With waves, the power value is taken as the average power transfer over the wave's period.)

8 0

3 years ago

Bill throws a tennis ball to his dog. He throws the ball at a speed of 15 m/s at an angle of 30° to the horizontal. Assume he th

Sidana [21]

1a) Bill and the dog must have a speed of 13.0 m/s

1b) The speed of the dog must be 22.5 m/s

2a) The ball passes over the outfielder's head at 3.33 s

2b) The ball passes 1.2 m above the glove

2c) The player can jump after 2.10 s or 3.13 s after the ball has been hit

2d) One solution is when the player is jumping up, the other solution is when the player is falling down

Explanation:

1a)

The motion of the ball in this problem is a projectile motion, so it follows a parabolic path which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

In part a), we want to know at what speed Bill and the dog have to run in order to intercept the ball as it lands on the ground: this means that Bill and the dog must have the same velocity as the horizontal velocity of the ball.

The ball's initial speed is

u = 15 m/s

And the angle of projection is

$\theta=30^{\circ}$

So, the ball's horizontal velocity is

$v_x = u cos \theta = (15)(cos 30)=13.0 m/s$

And therefore, Bill and the dog must have this speed.

1b)

For this part, we have to consider the vertical motion of the ball first.

The vertical position of the ball at time t is given by

$y=u_yt+\frac{1}{2}at^2$

where

$u_y = u sin \theta = (15)(sin 30) = 7.5 m/s$ is the initial vertical velocity

$a=g=-9.8 m/s^2$ is the acceleration of gravity

The ball is at a position of y = 2 m above the ground when:

$2=7.5t + \frac{1}{2}(-9.8)t^2\\4.9t^2-7.5t+2=0$

Which has two solutions: t=0.34 s and t=1.19 s. We are told that the ball is falling to the ground, so we have to consider the second solution, t = 1.19 s.

The horizontal distance covered by the ball during this time is

$d=v_x t =(13.0)(1.19)=15.5 m$

The dog must be there 0.5 s before, so at a time

t' = t - 0.5 = 0.69 s

So, the speed of the dog must be

$v_x' = \frac{d}{t'}=\frac{15.5}{0.69}=22.5 m/s$

2a)

Here we just need to consider the horizontal motion of the ball.

The horizontal distance covered is

$d=98 m$

while the horizontal velocity of the ball is

$v_x = u cos \theta = (34)(cos 30)=29.4 m/s$

where u = 34 m/s is the initial speed.

So, the time taken for the ball to cover this distance is

$t=\frac{d}{v_x}=\frac{98}{29.4}=3.33 s$

2b)

Here we need to calculate the vertical position of the ball at t = 3.33 s.

The vertical position is given by

$y= h + u_y t + \frac{1}{2}at^2$

where

h = 1.2 m is the initial height

$u_y = u sin \theta = (34)(sin 30)=17.0 m/s$ is the initial vertical velocity

$a=g=-9.8 m/s^2$ is the acceleration of gravity

Substituting t = 3.33 s,

$y=1.2+(17)(3.33)+\frac{1}{2}(-9.8)(3.33)^2=3.5 m$

And sinc the glove is at a height of y' = 2.3 m, the difference in height is

y - y' = 3.5 - 2.3 = 1.2 m

2c)

In order to intercept the ball, he jumps upward at a vertical speed of

$u_y' = 7 m/s$

So its position of the glove at time t' is

$y'= h' + u_y' t' + \frac{1}{2}at'^2$

where h' = 2.3 m is the initial height of the glove, and t' is the time from the moment when he jumps. To catch the ball, the height must be

y' = y = 3.5 m (the height of the ball)

Substituting and solving for t', we find

$3.5 = 2.3 + 7t' -4.9t'^2\\4.9t'^2-7t'+12 = 0$

Which has two solutions: t' = 0.20 s, t' = 1.23 s. But this is the time t' that the player takes to reach the same height of the ball: so the corresponding time after the ball has been hit is

$t'' = t -t'$

So we have two solutions:

$t'' = 3.33 s - 0.20 s = 3.13 s\\t'' = 3.33 s - 1.23 s = 2.10 s$

So, the player can jump after 2.10 s or after 3.13 s.

2d)

The reason for the two solutions is the following: the motion of the player is a free fall motion, so initially he jump upwards, then because of gravity he is accelerated downward, and therefore eventually he reaches a maximum height and then he falls down.

Therefore, the two solutions corresponds to the two different part of the motion.

The first solution, t'' = 2.10 s, is the time at which the player catches the ball while he is in motion upward.

On the other hand, the second solution t'' = 3.13 s, is the time at which the player catches the ball while falling down.

Learn more about projectile motion:

brainly.com/question/8751410

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7 0

3 years ago