A table has a height of 36.0 inches. How many centimeters is this? (Note: 1 in

An information signal consists of a 25 Hz and a 75 Hz sine waves summed together. It is sampled at a frequency of 500 Hz, the hi

denpristay [2]

Answer:

Explanation:

An information contains

25Hz and 75Hz sine wave

Sample frequency is 500Hz

The analogy signal are generally

y(t) = Asin(2πx/λ - wt), w=2πf

y1(t)=Asin(2πx/λ - wt)

y1(t)=Asin(2πx/λ - 2π•25t)

y1(t)=Asin(2πx/λ - 50πt)

Similarly

y2(t)=Asin(2πx/λ - 150πt)

Using Nyquist theorem

Nyquist Theorem states that in order to adequately reproduce a signal it should be periodically sampled at a rate that is 2 times the highest frequency you wish to record.

From sampling

f(nyquist)=f(sample)/2

f(nyquist)=500/2

f(nyquist)=250Hz

From signal

The highest frequency is 150Hz

F(nyquist) = 2×F(highest)

f(nyquist)= 2×150

f(nyquist)= 300Hz

Sample per frequency Ns is given as

Ns=F(sample)/F(highest signal)

Ns=500/150

Ns=3.33sample/period

This is above nyquist rate of 2sample/period

So signal below 300Hz reproduced without aliasing.

The highest resulting frequency is 300Hz

6 0

3 years ago

Gamma rays x rays visible light and radio waves are all types of

bazaltina [42]

Answer:

Electromagnetic waves

Explanation:

Electromagnetic waves are waves that consist of oscillating electric and magnetic fields, that oscillate perpendicularly to each other and perpendicularly to the direction of propagation of the wave (for such a reason, these waves are also called transverse waves).

Electromagnetic waves always travel in a vacuum at the same speed, called speed of light:

$c=3.0\cdot 10^8 m/s$

and they are classified into 7 different types, according to their frequency. From lowest to highest frequency, we have:

Radio waves

Microwaves

Infrared

Visible light

Ultraviolet

X-rays

Gamma rays

Therefore, gamma rays, x-rays, visible light and radio waves are all types of electromagnetic waves with different frequencies.

6 0

2 years ago

What is the correct equation kinetic enery

Ivahew [28]

Kinetic energy is directly proportional to the mass of the object and to the square of its velocity: K.E. = 1/2 m v2.

4 0

3 years ago

Conduction is the transfer of heat in: 1 gases 2 liquids 3 solids

kykrilka [37]

Answer:

Conduction heat transfer is the transfer of heat by means of molecular excitement within a material without bulk motion of the matter.

Explanation:

Conduction heat transfer in gases and liquids is due to the collisions and diffusion of the molecules during heir random motion.

4 0

2 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

#LearnwithBrainly

7 0

3 years ago