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Andrew [12]
2 years ago
5

The emission of electrons from metals that have absorbed photons is called the

Physics
1 answer:
valina [46]2 years ago
4 0

Answer:

The answer to your question is the Photoelectric effect.

Explanation:

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Dr. Clairmont will have his psychology students training dogs from the local animal shelter to sit using various types of reinfo
olga2289 [7]

Answer:Habituation is a simple learned behavior in which an animal gradually stops responding to a repeated stimulus.

Imprinting is a specialized form of learning that occurs during a brief period in young animals—e.g., ducks imprinting on their mother.

In classical conditioning, a new stimulus is associated with a pre-existing response through repeated pairing of new and previously known stimuli.

In operant conditioning, an animal learns to perform a behavior more or less frequently through a reward or punishment that follows the behavior.

Some animals, especially primates, are capable of more complex forms of learning, such as problem-solving and the construction of mental maps.

Introduction

If you own a dog—or have a friend who owns a dog—you probably know that dogs can be trained to do things like sit, beg, roll over, and play dead. These are examples of learned behaviors, and dogs can be capable of significant learning. By some estimates, a very clever dog has cognitive abilities on par with a two-and-a-half-year-old human!

Explanation:

6 0
3 years ago
A compunds whose empirical formula is xf3 consists of 65% f by mass. what is the atomic mass of x
Reika [66]

Empirical formula of compound is XF3

Compound consist of 65% F

In 100g of compound there is 65 g of F

= 65 / 19 moles of Fluorine = 3.421 moles

So moles of X = 3.421 / 3 = 1.140 moles

And in 100 g  X consist of 35 g

So the molar mass of X = 35 / 1.140 = 30.71 g = 31 approximately

And it is the mass of phosphorus

So the empirical formula for the compound is PX3

4 0
3 years ago
Read 2 more answers
What is the function of the phytoplankton in ocean ecosystems?
vesna_86 [32]
Ok ok! It's A. Producer
5 0
3 years ago
A 150kg motorcycle starts from rest and accelerates at a constant rate along a distance of 350m. The applied force is 250N and t
notsponge [240]

A) The net force on the motorbike is 205.9 N

B) The acceleration of the motorbike is 1.37 m/s^2

C) The final speed is 5.2 m/s

D) The elapsed time is 3.80 s

Explanation:

A)

There are two forces acting on the motorbike:

- The applied force, F = 250 N, forward

- The frictional force, F_f, backward

The frictional force can be written as

F_f = \mu mg

where

\mu=0.03 is the coefficient of kinetic friction

m=150 kg is the mass of the motorbike

g=9.8 m/s^2 is the acceleration of gravity

Therefore the net force is given by

\sum F = F - F_f = F - \mu mg

And substituting, we find

\sum F=250 - (0.03)(150)(9.8)=205.9 N

2)

The acceleration of the motorbike can be found by using Newton's second law, which states that the net force is equal to the product between mass and acceleration:

\sum F = ma

where

m is the mass

a is the acceleration

In this problem, we have

\sum F = 205.9 N is the net force

m = 150 kg is the mass

Solving for a, we find the acceleration:

a=\frac{\sum F}{m}=\frac{205.9}{150}=1.37 m/s^2

C)

Since the motion of the motorbike is a uniformly accelerated motion, we can use the following suvat equation:

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

For this motorbike, we have:

u = 0 (it starts from rest)

a=1.37 m/s^2

s = 350 m

Solving for v,

v=\sqrt{u^2+2as}=\sqrt{0+2(1.37)(9.8)}=5.2 m/s

4)

For this part of the problem, we can use the following suvat equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the elapsed time

Here we have:

v = 5.2 m/s

u = 0

a=1.37 m/s^2

Solving for t, we find

t=\frac{v-u}{a}=\frac{5.2-0}{1.37}=3.80 s

Learn more about Newton's laws of motion and accelerated motion:

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brainly.com/question/2562700

#LearnwithBrainly

6 0
3 years ago
what is haopening to the speed and acceleration as a car travels from rest to 72 km/hr in 4.0 seconds?
Deffense [45]

From the given information, we don't know. All we can tell is the net effect of all the changes during those 4 seconds ... the "averages".

The average change in speed is an increase of 72 km/hr (20 m/s) during that time.

The average acceleration is a constant 18 km/hr/sec (5 m/s^2) during the same time.

But both the speed and the acceleration may have gone up or down many times during the 4 seconds.

6 0
3 years ago
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