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lana [24]
3 years ago
5

PLS HELP WORTH 100 POINTS!! Different things motivate different people. While you may be motivated to get good grades and get a

scholarship to an Ivy League college after high school, your classmate’s motivation might be to finish high school and go backpacking around the world. Do some research on the Internet and at the school library, and write about the four different types of external and internal motivations: extrinsic, intrinsic, identified, and introjected, and provide examples for each. Also, consider a situation in which you were highly motivated (it does not have to be academic), and describe your motivation and the outcome of the situation.
Physics
2 answers:
lord [1]3 years ago
5 0

Answer:

dang thats a lot of words

Explanation:

adelina 88 [10]3 years ago
4 0

Answer:

i had a different thing and i finished mine so i thought id try my best to help

Explanation:

So i had to write about something that motivated me

looking at myself in the mirror and hating how i looked. i was okay with that at first but when others around me started calling me over weight or "chubby" it go to me. and its motivated me to work out. ill fall behind and stop but then i just think about how that made me feel and i just work harder.

Thats what i had wrote

<h2>This is my idea of what your question is asking</h2>
  • an extrinsic motivation would be doing something and getting a reward, the reward of how my body would look therefor compliments i could receive. its like a like praise or reward
  • an intrinsic motivation would be doing something without getting a reward, it would be if i was just working out because i enjoyed it
  • identified motivation(regulation?) is if i worked out because i saw value in it
  • introjected motivation would be if i worked out to maintain self-esteem and my pride to avoid anxiety . Another example could be that im working out to please someone else and make them happy

I really hope this is helpful to someone

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The Back Saver Sit and Reach from the Fitnessgram measure this fitness component.
MAVERICK [17]
The correc answer is B.
7 0
3 years ago
When a falling meteoroid is at a distance above the Earth's surface of 3.40 times the Earth's radius, what is its acceleration d
Alchen [17]

Answer:

g = 0.85 ms^{-2}

Explanation:

g = \frac{GM}{h^{2} }

were; g is the acceleration due to Earth's gravity, G is Newton's gravitation constant (6.674 x 10^{-11} Nm^{2}kg^{-2}), M is the mass of the earth (5.972 x 10^{24} kg), and h is the distance of meteoroid to the earth.

h = 3.40 x R

  = 3.40 x 6371 km

h = 21661.4 km

  = 21661400 m

Thus,

g = \frac{6.674*10^{-11}*5.972*10^{24}  }{(21661400)^{2} }

  = \frac{3.9857 *10^{14} }{4.6922*10^{14} }

  = 0.84944

g = 0.85 ms^{-2}

The acceleration due to the Earth's gravitation is 0.85 ms^{-2}.

6 0
2 years ago
During the first 50 s a truck traveled at constant speed of 25 m/s. Find the distance that it is traveled.
Allushta [10]
Time=50s
speed=25m/s

Distance = speed×time
=25×50
=1250m

DISTANCE TRAVELLED IS =1250m

6 0
2 years ago
A 0.90-kg falcon is diving at 28.0 m/s at a downward angle of 35° . It catches a 0.325-kg pigeon from behind in midair. What is
pashok25 [27]

Answer:

22.11 m / s

Explanation:

The falcon catches the prey from behind means both are flying in the same direction ( suppose towards the left )

initial velocity of falcon = 28 cos 35 i - 28 sin 35 j  

( falcon was flying in south east direction making 35 degree from the east )

momentum = .9 ( 28 cos 35 i - 28 sin 35 j  )

= 20.64 i - 14.45 j

initial velocity of pigeon

= 7 i

initial momentum = .325 x 7i

= 2.275 i

If final velocity of composite mass of falcon and pigeon be V

Applying law of conservation of momentum

( .9 + .325) V = 20.64 i - 14.45 j +2.275 i

V = ( 22.915 i - 14.45 j ) / 1.225

= 18.70 i - 11.8 j

magnitude of V

= √ [  (18.7 )² + ( 11.8 )²]

= 22.11 m / s

6 0
3 years ago
Each driver has mass 79.0 kg. Including the masses of the drivers, the total masses of the vehicles are 800 kg for the car and 4
Mademuasel [1]

Answer:

Force exerted on the car driver by the seatbelt = 8139.4 N = 8.14 kN

Force exerted on the truck driver by the seatbelt = 1628.2 N = 1.63 kN

It is evident that the driver of the smaller vehicle has it worse. The car driver is in way more danger in this perfectly inelastic head-on collision with a bigger vehicle (the truck).

Explanation:

First of, we calculate the velocity of the vehicles after collision using the law of conservation of Momentum

Momentum before collision = Momentum after collision

Since the collision of the two vehicles was described as a head-on collision, for the sake of consistent convention, we will take the direction of the velocity of the bigger vehicle (the truck) as the positive direction and the direction of the car's velocity automatically is the negative direction.

Velocity of the truck before collision = 6.80 m/s

Velocity of the car before collision = -6.80 m/s

Let the velocity of the inelastic unit of vehicles after collision be v

Momentum before collision = (4000)(6.80) + (800)(-6.80) = 27200 - 5440 = 21,760 kgm/s

Momentum after collision = (4000 + 800)(v) = (4800v) kgm/s

Momentum before collision = Momentum after collision

21760 = 4800v

v = (21760/4800)

v = 4.533 m/s (in the direction of the big vehicle (the truck)

So, we then apply Newton's second law of motion which explains that the magnitude change in momentum is equal to the magnitude of impulse.

|Impulse| = |Change in momentum|

But Impulse = (Force exerted on each driver by the seatbelt) × (collision time) = (F×t)

Change in momentum = (Momentum after collision) - (Momentum before collision)

So, for the driver of the truck

Initial velocity = 6.80 m/s (the driver moves with the velocity of the truck)

Final velocity = 4.533 m/s

Change in momentum of the truck driver = (79)(6.80) - (79)(4.533) = 179.1 kgm/s

(F×t) = 179.1

F × 0.110 = 179.1

F = (179.1/0.11)

F = 1628.2 N = 1.63 kN

So, for the driver of the car

Initial velocity = -6.80 m/s (the driver moves with the velocity of the car)

Final velocity = 4.533 m/s

Change in momentum of the car driver = (79)(-6.80) - (79)(4.533) = -895.3 kgm/s

(F×t) = |-895.3|

F × 0.110 = 895.3

F = (895.3/0.11)

F = 8139.4 N = 8.14 kN

Hope this Helps!!!

3 0
3 years ago
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