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2 years ago

Best friends Kate and Diana are at the park, watching their children play together. Kate reminds Diana about

1 answer:

3 0

Kate's recollection of these different events along her life best exemplifies the use of her episodic memory.

<h3>What is episodic memory?</h3>

The term episodic memory makes reference to conscious personal background experiences that were collected along life.

Episodic memory is also defined as the collection of all life-day experiences collected by a person.

The episodic memory may be, for example, the first day when a person drove his/her car or a bike.

Learn more about episodic memory here:

brainly.com/question/25040884

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What do you mean by peltier coefficient?

devlian [24]

Answer:

The Peltier coefficient is a measure of the amount of heat carried by electrons or holes

Explanation:

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3 years ago

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A small dog is trained to jump straight up a distance of 1.2 m. How much kinetic energy does the 7.2-kg dog need to jump this hi

LUCKY_DIMON [66]

Answer:84.672 joules.

Explanation:

1) Data:

m = 7.2 kg
h = 1.2 m
g = 9.8 m / s²

2) Physical principle

Using the law of mechanical energy conservation principle, you have that the kinetic energy of the dog, when it jumps, must be equal to the final gravitational potential energy.

3) Calculations:

The gravitational potential energy, PE, is equal to m × g × h

So, PE = m × g × h = 7.2 kg × 9.8 m/s² × 1.2 m = 84.672 joules.

And that is the kinetic energy that the dog needs.

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3 years ago

A rock is thrown horizontally from a bridge with a speed of 29.0 m/s. if the rock is 23.7 meters above the river at the moment o

mash [69]

It would be 1.5 meters im sure form that distance to me is that nswe

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2 years ago

Ten identical steel wires have equal lengths L and equal "spring constants" k. The Young's modulus of each wire is Y. The wires

svlad2 [7]

Answer:

option (B)

Explanation:

Young's modulus is defined as the ratio of longitudinal stress to the longitudinal strain.

Its unit is N/m².

The formula for the Young's modulus is given by

$Y=\frac{F \times L}{A\times \delta L}$

where, F is the force applied on a rod, L is the initial length of the rod, ΔL is the change in length of the rod as the force is applied, A is the area of crossection of the rod.

It is the property of material of solid. So, when the 10 wires are co joined together to form a new wire of length 10 L, the material remains same so the young' modulus remains same.

8 0

2 years ago

Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.

SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

(b) m1 = 915kg;

m2 = 605kg;

m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

$F_{pull}=m_{T}.a$

$m_{T}=\frac{F_{pull}}{a}$

$m_{T}=\frac{3615}{1.516}$

$m_{T}=2384.5$

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

For $m_{1}$ :

The only force acting On the $m_{1}$ box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

$m_{1} = \frac{F_{12}}{a}$

$m_{1} = \frac{1387}{1.516}$

$m_{1}$ = 915kg

For $m_{2}$ :

There are two forces acting on $m_{2}$ : tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

$m_{2} = \frac{F_{23}-F_{12}}{a}$

$m_{2} = \frac{2304-1387}{1.516}$

$m_{2}$ = 605kg

For $m_{3}$ :

$m_{3} = m_{T} - (m_{1}+m_{2})$

$m_{3} = 2384.5-1520.0$

$m_{3}$ = 864.5kg

8 0

3 years ago