Ask question

Ask question

All categories

Gelneren [198K]

4 years ago

11

The difference between an innate behavior and a learned behavior is that a learned behavior is __________ and an innate behavior

is not.
A.

based on inborn responses

B.

instinctive

C.

based on reflexes

D.

a result of experience

2 answers:

ehidna [41]4 years ago

8 0

Answer:

A result of experience.

Explanation:

edg 2020

lukranit [14]4 years ago

4 0

ANSWER:

D. A result of experience.

The difference between an innate behaviour and learned behaviour is that a learned behaviour is a result of experience and an innate behaviour is not.

You might be interested in

In a series lrc circuit, the frequency at which the circuit is at resonance is f0. If you double the resistance, the inductance,

taurus [48]

When you double capacitance and inductance, the new resonance frequency becomes f/2.

Resonance frequency:

The resonance frequency of RLC series circuit, is the frequency at which the capacity reactance is equal to inductive reactance.

It can also be defined as the natural frequency of an object where it tends to vibrate at a higher amplitude.

Xc = Xl

which gives the value for resonance frequency:

$f = \frac{1}{2\pi \sqrt{LC} }$

where;

f is the resonance frequency

L is the inductance

C is the capacitance

When you double capacitance and inductance, the new resonance frequency becomes;

$f' = \frac{1}{2\pi \sqrt{2L2C} }$

$f' = \frac{1}{2\pi \sqrt{4LC} }$

$f' = \frac{1}{\pi \sqrt{LC} }\frac{1}{2}$

$f' = \frac{1}{2} f$

Thus from above,

When you double capacitance and inductance, the new resonance frequency becomes f/2.

Learn more about resonance frequency here:

<u>brainly.com/question/13040523</u>

#SPJ4

6 0

2 years ago

If a certain silver wire has a resistance of 3.00 Ï at 11.0Â°c, what resistance will it have at 25.0Â°c?

Romashka [77]

11.30 times the nuber of toys

7 0

3 years ago

A car, starting from the origin, travels

sweet [91]

Answer:

The net displacement of the car is 3 km West

Explanation:

Please see the attached drawing to understand the car's trajectory: First in the East direction for 4 km (indicated by the green arrow that starts at the origin (zero), and stops at position 4 on the right (East).

Then from that position, it moves back towards the West going over its initial path, it goes through the origin and continues for 3 more km completing a moving to the West a total of 7 km. This is indicated in the drawing with an orange trace that end in position 3 to the left (West) of zero.

So, its NET displacement considered from the point of departure (origin at zero) to the final point where the trip ended, is 3 km to the west.

8 0

3 years ago

Read 2 more answers

A 2000 kg truck traveling north at 34 km/h turns east and accelerates to 58 km/h. (a) What is the change in the truck's kinetic

barxatty [35]

Explanation:

It is given that,

Mass of the truck, m = 2000 kg

Initial velocity of the truck, u = 34 km/h = 9.44 m/s

Final velocity of the truck, v = 58 km/h = 16.11 m/s

(a) Change in truck's kinetic energy, $\Delta E=\dfrac{1}{2}m(v^2-u^2)$

$\Delta E=\dfrac{1}{2}\times 2000\ kg\times (16.11^2-9.44^2)$

$\Delta E=170418.5\ J$

$\Delta E=1.7\times 10^5\ J$

(b) Change in momentum of the truck, $\Delta p=m(v-u)$

$\Delta p=2000\ kg\times (16.11-9.44)$

$\Delta p=13340\ kg-m/s$

Hence, this is the required solution.

6 0

4 years ago

A basketball player of height 2.40 m is standing 2.60 m in front of a convex spherical mirror of radius of curvature 4.00 m. Det

IrinaVladis [17]

Answer:

The size of the image is 1.04 m.

Explanation:

Given that,

Height of object = 2.40 m

Distance of object = 2.60 m

Radius of curvature =4.00 m

Focal length $f=\dfrac{R}{2}=\dfrac{4.00}{2}=2.00$

We need to calculate the image distance

Using mirror formula

$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$

$\dfrac{1}{v}=\dfrac{1}{2.00}+\dfrac{1}{2.60}$

$\dfrac{1}{v}= \dfrac{23}{26}$

$v=1.13\ cm$

We need to calculate the height of the image

Using formula of magnification

$m=\dfrac{h'}{h}=-\dfrac{v}{u}$

Put the value into the formula

$\dfrac{h'}{2.40}=-\dfrac{1.13}{-2.60}$

$h'=\dfrac{1.13}{2.60}\times2.40$

$h'=1.04\ m$

Hence, The size of the image is 1.04 m

8 0

3 years ago