Waves sometimes travel by squeezing together and then do what

Give an example of something that is cyclic and non - cyclic

LiRa [457]

a clock .. and i guess a non functioning clock ?

7 0

3 years ago

In order to stop a heavier (or more massive) object, which of the following is true?

goblinko [34]

Answer:the force that object be lower

Explanation:bc I got it right

5 0

3 years ago

A parallel plate capacitor is constructed with a dielectric slab with κ = 1.5 inserted between the plates. The area of each plat

maksim [4K]

Answer:

The value is $V = 4.533 \ V$

Explanation:

From the question we are told that

The dielectric constant is k = 1.5

The area of each plate is $A = 5 \ cm^2 = 0.0005 m^2$

The distance between the plates is $d= 1 \ mm = 0.001 \ m$

The charge on the capacitor is $Q = 6*10^{-11} \ C$

Generally the electric field in a vacuum is mathematically represented as

$E_o = \frac{V_o}{d}$

Generally $V_o$ is the voltage of the capacitor which is mathematically represented as

$V_o = \frac{Q}{C_o}$

Here $C_o$ is the capacitance of the capacitor in a vacuum which is mathematically represented as

$C_o = \frac{\epsilon_o * A}{d}$

Here $epsilon_o$ is a constant with value $epsilon_o= 8.85*10^{-12} C^2 \cdot N^{-1} \cdot m^{-2}$

=> $C_o = \frac{8.85*10^{-12} * 0.0005 }{0.001}$

=> $C_o = 4.425 *10^{-12} \ F$

So

$V_o = \frac{6*10^{-11} }{ 4.425 *10^{-12}}$

$V_o = 13.6 \ V$

So

$E_o = \frac{ 13.6}{0.001}$

=> $E_o = 13600 \ V/m$

The electric field when the dielectric slab is inserted is mathematically represented as

$E = \frac{E_o}{k}$

=> $E = \frac{13600}{1.5}$

=> $E =9067 \ V/m$

Generally the electric field between the plates is mathematically evaluated as

$E_{N} = E_o - E$

=> $E_{N} = 13600- 9067$

=> $E_{N} = 4533 \ V/m$

Generally the potential difference between the plates is

$V = 4533 * 0.001$

=> $V = 4.533 \ V$

6 0

4 years ago

A ball is traveling uphill with an initial velocity of 5.0 m/s and an acceleration of -2.0 m/s^2. A) How fast is the ball travel

Rzqust [24]

Answer:

A) The ball is traveling at 5.0 m/s (magnitude) when the ball returns to its release point.

B) The maximum uphill position is at 6.25 m from the release point.

C) On the way up, the velocity of the ball at x = 6.0 m is 1 m/s and on the way down it is - 1m/s.

Explanation:

Hi there!

The position and velocity of the ball can be calculated using the following equations:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position of th ball at time t.

x0 = initial position.

v0 = initial velocity.

a = acceleration.

t = time.

v = velocity at time t.

A) Let´s place the origin of the frame of reference at the point at which the ball has a velocity of 5.0 m/s. Then, x0 = 0.

When the ball returns to the initial point, its position will be 0. Then using the equation of position we can calculate at which time the ball is at x = 0:

x = x0 + v0 · t + 1/2 · a · t²

0 m = 5.0 m/s · t - 1/2 · 2.0 m/s² · t²

0 m = 5.0 m/s · t - 1.0 m/s² · t²

0 m = t (5.0 m/s - 1.0 m/s² · t)

t = 0 (this is logic becuase the ball starts at x = 0)

and

5.0 m/s - 1.0 m/s² · t = 0

t = -5.0 m/s / -1.0 m/s²

t = 5.0 s

With this time, we can calculate the velocity of the ball:

v = v0 + a · t

v = 5.0 m/s - 2.0 m/s² · 5.0 s

v = -5.0 m/s

The ball is traveling at 5.0 m/s when the ball returns to its release point.

B) Let´s use the equation of velocity to obtain the time at which the ball is at its maximum uphill position:

v = v0 + a · t

0 = 5.0 m/s - 2.0 m/s² · t

-5.0 m/s/ -2.0 m/s² = t

t = 2.5 s

Now, using the equation of position, let´s find the position of the ball at t = 2.5 s. This position will be the maximum uphill position because at that time the velocity is 0:

x = x0 + v0 · t + 1/2 · a · t²

x = 5.0 m/s · 2.5 s - 1/2 · 2.0 m/s² · (2.5 s)²

x = 6.25 m

The maximum uphill position is at 6.25 m from the release point.

C) First, let´s find the time at which the ball is 6.0 meters uphill from the releasing point:

x = x0 + v0 · t + 1/2 · a · t²

6.0 m = 5.0 m/s · t - 1/2 · 2 m/s² · t²

0 = -1 m/s² · t² + 5.0 m/s · t - 6.0 m

Solving the quadratic equation using the quadratic formula:

a = -1

b = 5

c = -6

t = [-b ± √(b² - 4ac)]/2a

t₁ = 2 s (on its way up)

t₂ = 3 s (on its way down)

Now, let´s calculate the velocity of the ball at those times:

v = v0 + a · t

v = 5.0 m/s - 2 m/s² · 2 s = 1 m/s

v = 5.0 m/s - 2 m/s² · 3 s = -1 m/s

On the way up, the velocity of the ball at x = 6.0 m is 1 m/s and on the way down it is - 1m/s.

7 0

3 years ago

Consider water flowing through a cylindrical pipe with a variable cross-section. The velocity is v at a point where the pipe dia

Harrizon [31]

Answer:

one ninth

Explanation:

d = 1 cm , v = v

D = 3d, V = ?

By the equation of continuity,

A V = a v

3.14 x D^2 / 4 x V = 3.14 x d^2 / 4 x v

9d^2 x V = d^2 x v

V = v / 9

Thus, the velocity becomes one ninth the initial velocity

5 0

3 years ago