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2 years ago

Question 13

Physics

1 answer:

gulaghasi [49]2 years ago

7 0

Answer:

Unconditioned stimulus; unconditional response

Explanation:

Classical conditioning: this is a form of behavioural makeup that occurs when a neutral stimulus is paired with an associated stimulus. They have four principles namely:

- Unconditioned stimulus triggers a form of automatically response that is not controlled by your mind.

- Unconditioned response: this is the immediate reaction to an event or activitiy as a result of the unconditioned stimulus.

- Conditioned stimulus: this is works with a conscious trigger.

- Conditioned response: the paring of conditioned stimulus with unconditioned stimulus gives a conditioned response.

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what is a possible unit for the product VI, where V is the potential difference across a resistor and I is the current through t

liq [111]

Recall this equation for a device in a direct current circuit:
P = IV
P is the power dissipated by the device, I is the current through the device, and V is the voltage drop of the device.

If we choose to use the ampere as the unit of current and the volt as the unit of voltage, then the product of the current and the voltage will give the power with watts as the unit.

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2 years ago

Un electrón en un tubo de rayos catódicos acelera desde el reposo con una aceleración constante de 5.33x10¹²m/s² durante 0.150μs

Andre45 [30]

Can you translate that in English ? I'll try to help you out with that..

3 0

2 years ago

A golfer hits a golf ball at an angle of 25.0Â° to the ground. if the golf ball covers a horizontal distance of 301.5 m, what is

kvasek [131]

<u>Answer:</u>

Maximum height reached = 35.15 meter.

<u>Explanation:</u>

Projectile motion has two types of motion Horizontal and Vertical motion.

Vertical motion:

We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

Considering upward vertical motion of projectile.

In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g $m/s^2$ and final velocity = 0 m/s.

0 = u sin θ - gt

t = u sin θ/g

Total time for vertical motion is two times time taken for upward vertical motion of projectile.

So total travel time of projectile = 2u sin θ/g

Horizontal motion:

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 $m/s^2$ and time taken = 2u sin θ /g

So range of projectile, $R=ucos\theta*\frac{2u sin\theta}{g} = \frac{u^2sin2\theta}{g}$

Vertical motion (Maximum height reached, H) :

We have equation of motion, $v^2=u^2+2as$ , where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.

Initial velocity = vertical component of velocity = u sin θ, acceleration = -g, final velocity = 0 m/s at maximum height H

$0^2=(usin\theta) ^2-2gH\\ \\ H=\frac{u^2sin^2\theta}{2g}$

In the give problem we have R = 301.5 m, θ = 25° we need to find H.

So $\frac{u^2sin2\theta}{g}=301.5\\ \\ \frac{u^2sin(2*25)}{g}=301.5\\ \\ u^2=393.58g$

Now we have $H=\frac{u^2sin^2\theta}{2g}=\frac{393.58*g*sin^2 25}{2g}=35.15m$

So maximum height reached = 35.15 meter.

7 0

2 years ago

Consider a double Atwood machine constructed as follows: A mass 4m is suspended from a string that passes over a massless pulley

kenny6666 [7]

Answer:

Hello your question is incomplete attached below is the complete question

Answer : x ( acceleration of mass 4m ) = $\frac{g}{7}$

The top pulley rotates because it has to keep the center of mass of the system at equilibrium

Explanation:

Given data:

mass suspended = 4 meters

mass suspended at other end = 3 meters

first we have to express the kinetic and potential energy equations

The general kinetic energy of the system can be written as

T = $\frac{4m}{2} x^2 + \frac{3m}{2} (-x+y)^2 + \frac{m}{2} (-x-y)^2$

T = $4mx^2 + 2my^2 -2mxy$

also the general potential energy can be expressed as

U = $-4mgx-3mg(-x+y)-mg(-x-y)+constant=-2mgy +constant$

The Lagrangian of the problem can now be setup as

$L =4mx^2 +2my^2 -2mxy +2mgy + constant$

next we will take the Euler-Lagrange equation for the generalized equations :

Euler-Lagrange equation = $4x-y =0\\-2y+x +g = 0$

solving the equations simultaneously

x ( acceleration of mass 4m ) = $\frac{g}{7}$

The top pulley rotates because it has to keep the center of mass of the system at equilibrium

8 0

2 years ago

a block of mass m slides along a frictionless track with speed vm. It collides with a stationary block of mass M. Find an expres

shusha [124]

Answer:

Part a) When collision is perfectly inelastic

$v_m = \frac{m + M}{m} \sqrt{5Rg}$

Part b) When collision is perfectly elastic

$v_m = \frac{m + M}{2m}\sqrt{5Rg}$

Explanation:

Part a)

As we know that collision is perfectly inelastic

so here we will have

$mv_m = (m + M)v$

so we have

$v = \frac{mv_m}{m + M}$

now we know that in order to complete the circle we will have

$v = \sqrt{5Rg}$

$\frac{mv_m}{m + M} = \sqrt{5Rg}$

now we have

$v_m = \frac{m + M}{m} \sqrt{5Rg}$

Part b)

Now we know that collision is perfectly elastic

so we will have

$v = \frac{2mv_m}{m + M}$

now we have

$\sqrt{5Rg} = \frac{2mv_m}{m + M}$

$v_m = \frac{m + M}{2m}\sqrt{5Rg}$

6 0

2 years ago