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3 years ago

What are the Characteristics of theta waves

1 answer:

4 0

When the delta brainwave frequencies increase into the frequency of theta brainwaves, active dreaming takes place and often becomes more experiential to the person. Typically, when this occurs there is rapid eye movement, which is characteristic of active dreaming. This is called REM, and is a well known phenomenon.. Hope that helps!

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Svetlanka [38]

Answer:

They are both extremely hot, they both produce a form of light, they both have/use fire(typically)

Explanation:

6 0

3 years ago

A mass m is attached to an ideal massless spring with spring constant k. In experiment 1 the mass oscillates with amplitude a, a

Trava [24]

Time period remains the same in both the experiment as change in amplitude does not affect time period.

What are the factors on which time period depends in SHM?

Time period is given by:

$T=2\pi \sqrt{\frac{m}{k} }$

where,

T = time period

m = mass

k = spring constant

In a straightforward harmonic motion, we see from the preceding formula that the time period depends only on the object's mass and spring constant (SHM). The time period will adjust to any variations in the object's mass or the spring constant.

What is Spring Constant?

A spring's "spring constant" is a property that quantifies the relationship between the force acting on the spring and the displacement it produces. In other words, it characterises a spring's stiffness and the extent of its range of motion.

Learn more about SHM here:

brainly.com/question/20885248

#SPJ4

6 0

1 year ago

The three stages of a train route took 1 hour ,2 hours ,and 4 hours . The first two stages were 80km and 200km of the train aver

luda_lava [24]

Answer:

the third stage was 480 km long

Explanation:

Stage 1:

Time = 1 hours

Speed = 80km

Stage 2:

Time = 2 hours

Speed = 200km

Stage 3:

Time = 4 hours

Let the Distance at the stage 3 be x

Average speed of the train route = 100 km/h

$\frac{ \text{speed at stage 1} + \text{speed at stage 2} + \text{speed at stage 3}}{3} = 0$

$\frac{ \text{speed at stage 1} + \text{speed at stage 2} + \text{speed at stage 3}}{3} = 100$

Lets find the speed at stage 1

Speed = $\frac{Distance }{Time}$

Speed = $\frac{80}{1}$

Speed 1= 80 km/hr

The speed at stage 2

Speed = $\frac{Distance }{Time}$

Speed = $\frac{200}{2}$

Speed 2 = 100 km/hr

The speed at stage 3

Speed = $\frac{Distance }{Time}$

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

Speed 3 = $\frac{x}{4}$

we kow that average is ,

$\frac{ \text{speed 1} + \text{speed 2} + \text{speed 3}}{3} = 100$

$\frac{ 80 + 100+ \frac{x}{4} }{3} = 100$

$\frac{ 180 + \frac{x}{4} }{3} = 100$

$\frac{ \frac{720 +x}{4} }{3} = 100$

$\frac{720 +x}{4} \times \frac{1}{3} = 100$

$\frac{720 +x}{12} = 100$

$720 +x = 100 \times 12$

$720 +x = 1200$

$x = 1200- 720$

x = 480

6 0

3 years ago

An advertisement claims that a particular automobile can "stop on a dime". What net force would actually be necessary to stop an

Luba_88 [7]

Answer:

F = 4399 KN

Explanation:

given,

mass of automobile = 890 kg

initial speed = 48 km/h

= 48 × 0.278 = 13.34 m/s

using equation of motion

v² = u² + 2 a × s

0 = 13.34² - 2 a ×0.018

$a = \dfrac{13.34^2}{0.036}$

a = 4943.21 m/s²

F = m × a

F = 890 × 4943.21

F = 4399456.9 N

F = 4399 KN

hence, the Net force is F = 4399 KN

4 0

3 years ago

A 0.0450 kg bullet is accelerated from rest to a speed of 425 m/s in a 2.25 kg rifle (which is inititally at rest). The pain of

Mrac [35]

Answer:

If the rifle is held loosely away from the shoulder, the recoil velocity will be of -8.5 m/s, and the kinetic energy the rifle gains will be 81.28 J.

Explanation:

By momentum conservation, and given the bullit and the recoil are in a straight line, the momentum analysis will be unidimentional. As the initial momentum is equal to zero (the masses are at rest), we have that the final momentum equals zero, so

$0=P_{f}=m_{b} *v_{b}+m_{r}*v_{r}$

now we clear $v_{r}$ and use the given data to get that

$v_{r}=-8.5\frac{m}{s}$

But we have to keep in mind that the bullit accelerate from rest to a speed of 425 m/s, then if the rifle were against the shoulder, the recoil velocity would be a fraction of the result obtained, but, as the gun is a few centimeters away from the shoulder, it is assumed that the bullit get to its final velocity, so the kick of the gun, gets to its final velocity $\bold{v_{r}}$ too.

Finally, using $v_{r}$ we calculate the kinetic energy as

$K=\frac{1}{2}m_{r}v_{r}^{2}=81.28J$

3 0

3 years ago