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Scorpion4ik [409]

3 years ago

5

The association area is responsible for perceiving and attending to stimuli, and the association area is r

esponsible for identifying them.

1 answer:

AlladinOne [14]3 years ago

4 0

Answer:

Parietal on the first blank and temporal on the second

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Consider a thin circular disk that has been heated up to 400 °C and then left inside a chamber to cool down. The chamber surface

kogti [31]

Answer:

hello your question lacks the required diagram attached below is the complete question with the required diagram

answer : Qtotal = 807.4 Mw

Explanation:

Given Data :

disk properties :

∈ = 0.65

D = 200 mm

Ts = 400⁰c

attached below is the detailed solution

The total rate of Heat transferred from the disk

Qtotal = 807.4 Mw

7 0

3 years ago

When heating a liquid in a test tube, which of the following safety precautions should you take?. . 1.Wear latex gloves.. 2. Hea

marusya05 [52]

4 for sure because you dont want anything spilling on you or others that is harmful.

8 0

3 years ago

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What term was used to describe the final digit

xxMikexx [17]

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Significant digits (also called significant figures or “sig figs” for short) indicate the precision of a measurement. A number with more significant digits is more precise. For example, 8.00 cm is more precise than 8.0 cm.

7 0

3 years ago

A wave hits a wall as shown. As the wave interacts with a wall, which kind of wave interaction is shown? absorption diffraction

dangina [55]

The answer is reflection.

The drawing is simple but illustrates the concept beautifully.

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A loaded ore car has a mass of 950 kg. and rolls on rails ofnegligible friction. It starts from rest ans is pulled up a mineshaf

stiks02 [169]

(a) 10241 W

In this situation, the car is moving at constant speed: this means that its acceleration along the direction parallel to the slope is zero, and so the net force along this direction is also zero.

The equation of the forces along the parallel direction is:

$F - mg sin \theta = 0$

where

F is the force applied to pull the car

m = 950 kg is the mass of the car

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

$\theta=30.0^{\circ}$ is the angle of the incline

Solving for F,

$F=mg sin \theta = (950)(9.8)(sin 30.0^{\circ})=4655 N$

Now we know that the car is moving at constant velocity of

v = 2.20 m/s

So we can find the power done by the motor during the constant speed phase as

$P=Fv = (4655)(2.20)=10241 W$

(b) 10624 W

The maximum power is provided during the phase of acceleration, because during this phase the force applied is maximum. The acceleration of the car can be found with the equation

$v=u+at$

where

v = 2.20 m/s is the final velocity

a is the acceleration

u = 0 is the initial velocity

t = 12.0 s is the time

Solving for a,

$a=\frac{v-u}{t}=\frac{2.20-0}{12.0}=0.183 m/s^2$

So now the equation of the forces along the direction parallel to the incline is

$F - mg sin \theta = ma$

And solving for F, we find the maximum force applied by the motor:

$F=ma+mgsin \theta =(950)(0.183)+(950)(9.8)(sin 30^{\circ})=4829 N$

The maximum power will be applied when the velocity is maximum, v = 2.20 m/s, and so it is:

$P=Fv=(4829)(2.20)=10624 W$

(c) $5.82\cdot 10^6 J$

Due to the law of conservation of energy, the total energy transferred out of the motor by work must be equal to the gravitational potential energy gained by the car.

The change in potential energy of the car is:

$\Delta U = mg \Delta h$

where

m = 950 kg is the mass

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

$\Delta h$ is the change in height, which is

$\Delta h = L sin 30^{\circ}$

where L = 1250 m is the total distance covered.

Substituting, we find the energy transferred:

$\Delta U = mg L sin \theta = (950)(9.8)(1250)(sin 30^{\circ})=5.82\cdot 10^6 J$

8 0

3 years ago