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aev [14]
2 years ago
8

What is the role of the midbrain, and which part of the midbrain is important for arousing the forebrain?

Physics
1 answer:
Rainbow [258]2 years ago
6 0

Answer:

The fore-brain consists of the cerebrum, thalamus, and hypothalamus. Mid-brain relays information between fore-brain and hind-brain. Mid-brain plays important role in developing motor functions like eye movement, auditory functions and visual processing.  The crus cerebri connects the cerebral hemispheres to the cerebellum, contains a portion having a network of neurons that is responsible for arousal and alertness.

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How is air pressure related to elevation?
exis [7]
Air pressure decreases with altitude

hope that helps you!
6 0
3 years ago
The weight of a body of certain mass becomes zero in space.why?write with reasons​
Anvisha [2.4K]

Answer:

Weight is what you get when a certain amount of gravity is acting on that mass, and something, like the surface of a planet, is resisting that action. In space, when falling freely, there's nothing resisting the pull of gravity so weight disappears. Mass however stays.

hope this helps u

Explanation:

7 0
3 years ago
Which of the following states of matter is always a good conductor of electricity? solid, liquid, gas, or plasma **:/ Thank you!
pogonyaev

Plasma...I believe is always a good conductor of electricity. I was tempted to say a solid, but not all solids are the same in composition and that goes for liquid and gas as well.

Hopefully this helped and good luck.

8 0
3 years ago
A train is accelerating at a rate of 2 km/hr/s.  If its initial velocity is 20 km/hr, what is its velocity after 30 seconds?
Mademuasel [1]
Here it is. *WARNING* VERY LONG ANSWER

________________________________________... 
<span>11) If Galileo had dropped a 5.0 kg cannon ball to the ground from a height of 12 m, the change in PE of the cannon ball would have been product of mass(m),acceleration(g)and height(h) </span>

<span>The change in PE =mgh=5*9.8*12=588 J </span>
<span>______________________________________... </span>
<span>12.) The 2000 Belmont Stakes winner, Commendable, ran the horse race at an average speed = v = 15.98 m/s. </span>

<span>Commendable and jockey Pat Day had a combined mass =M= 550.0 kg, </span>

<span>Their KE as they crossed the line=(1/2)Mv^2 </span>

<span>Their KE as they crossed the line=0.5*550*(15.98)^2 </span>

<span>Their KE as they crossed the line is 70224.11 J </span>

<span>______________________________________... </span>
<span>13)Brittany is changing the tire of her car on a steep hill of height =H= 20.0 m </span>

<span>She trips and drops the spare tire of mass = m = 10.0 kg, </span>

<span>The tire rolls down the hill with an intial speed = u = 2.00 m/s. </span>

<span>The height of top of the next hill = h = 5.00 m </span>

<span>Initial total mechanical energy =PE+KE=mgH+(1/2)mu^2 </span>

<span>Initial total mechanical energy =mgH+(1/2)mu^2 </span>

<span>Suppose the final speed at the top of second hill is v </span>

<span>Final total mechanical energy =PE+KE=mgh+(1/2)mv^2 </span>

<span>As mechanical energy is conserved, </span>

<span>Final total mechanical energy =Initial total mechanical energy </span>

<span>mgh+(1/2)mv^2=mgH+(1/2)mu^2 </span>

<span>v = sq rt [u^2+2g(H-h)] </span>

<span>v = sq rt [4+2*9.8(20-5)] </span>

<span>v = sq rt 298 </span>

<span>v =17.2627 m/s </span>

<span>The speed of the tire at the top of the next hill is 17.2627 m/s </span>
<span>______________________________________... </span>
<span>14.) A Mexican jumping bean jumps with the aid of a small worm that lives inside the bean. </span>

<span>a.)The mass of bean = m = 2.0 g </span>

<span>Height up to which the been jumps = h = 1.0 cm from hand </span>

<span>Potential energy gained in reaching its highest point= mgh=1.96*10^-4 J or 1960 erg </span>

<span>b.) The speed as the bean lands back in the palm of your hand =v=sq rt2gh =sqrt 0.196 =0.4427 m/s or 44.27 cm/s </span>
<span>_____________________________ </span>
<span>15.) A 500.-kg horse is standing at the top of a muddy hill on a rainy day. The hill is 100.0 m long with a vertical drop of 30.0 m. The pig slips and begins to slide down the hill. </span>

<span>The pig's speed a the bottom of the hill = sq rt 2gh = sq rt 2*9.8*30 =sq rt 588 =24.249 m/s </span>
<span>__________________________________ </span>
<span>16.) While on the moon, the Apollo astronauts Neil Armstrong jumped up with an intitial speed 'u'of 1.51 m/s to a height 'h' of 0.700 m, </span>

<span>The gravitational acceleration he experienced = u^2/2h = 2.2801 /(2*0.7) = 1.629 m/s^2 </span>
<span>______________________________________... </span>

<span>EDIT </span>
<span>1.) A train is accelerating at a rate = a = 2.0 km/hr/s. </span>

<span>Acceleration </span>

<span>Initial velocity = u = 20 km/hr, </span>

<span>Velocity after 30 seconds = v = u + at </span>

<span>Velocity after 30 seconds = v = 20 km/hr + 2 (km/hr/s)*30s = </span>

<span>Velocity after 30 seconds = v = 20 km/hr + 60 km/hr = 80 km/ hr </span>

<span>Velocity after 30 seconds = v = 80 km/hr=22.22 m/s </span>
<span>_______________________________- </span>
<span>2.) A runner achieves a velocity of 11.1 m/s 9 s after he begins. </span>

<span>His acceleration = a =11.1/9=1.233 m/s^2 </span>

<span>Distance he covered = s = (1/2)at^2=49.95 m</span>
7 0
3 years ago
Three people pull simultaneously on a stubborn donkey. Jack pulls directly ahead of the donkey with a force of 77.3 N, Jill pull
DiKsa [7]

Answer:

F_{net} = 232.8 N

towards right so it is -15 degree

Explanation:

Net force in forward direction due to all three is given as

F_x = F_1 + F_2cos45 + F_3cos45

here we know that

F_1 = 77.3 N

F_2 = 61.7 N

F_3 = 147 N

F_x = 77.3 + 61.7 cos45 + 147 cos45

F_x = 224.9 N

Similarly in Y direction we will have

F_y = F_3 sin45 - F_2 sin45

F_y = (147 - 61.7)sin45

F_y = 60.3 N

Now the net force on the donkey is given as

F_{net} = \sqrt{F_x^2 + F_y^2}

F_{net} = \sqrt{224.9^2 + 60.3^2}

F_{net} = 232.8 N

Now direction of force is given as

tan\theta = \frac{F_y}{F_x}

tan\theta = \frac{60.3}{224.9}

\theta = 15^o towards right so it is -15 degree

3 0
2 years ago
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