A 0.05-kg car starts from rest at a height of 0.95 m. Assuming no friction, what is the kinetic energy of the car when it reaches the bottom of the hill? (Assume g = 9.81 m/s2.)

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Answer 1

Answer:

Answer: What is the answer?

Explanation:

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Hope that helped =)

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Answer:

Explanation:

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Hi, the answer for this question is element!

Hope this helped!

Plz mark brainliest!

~Nate

1. calculate agni's average speed during the race.2. the average speed of agni within the first 20 km ( number 1 in the graph ) can be calculated using the slope of the graph :
distance : time = 20 km : 1/2 hour = 40 km/hour

B. 1. what quantities can be identified in the graph?
    2. explain the meaning of the horizontal lines in the graph.
   3. calculate agni's velocity during the race.
   4. calculate the speed reached by agni at the end of the race.

on independence day, there was a bicycle race organized in a neighborhood. an observer took notes on the progress of one participant, named agni. who finally won the race... the collected data is shown in " graph of agni's bicycle track " as shown below

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The formula to find the speed/velocity is :
v = Δx/Δt = ( $( x_(final) - x_(initial) )/( t_(travel) )$

B.
1. From the graph given, we can get an information such as the distance travelled, and the time given.
2. If the graph are horizontal, it means that the Agni's speed is zero because Agni doesn't move.
3. If the question ask the velocity during all time, we can say that the velocity is :
. $v = (x)/(t) = (50 km)/(2.5 hour) = 20 (km)/(h)$
4. So, the final velocity can be found by looking the third part of the graph. So,..
$v = (x)/(t) = ((50-35) km )/(0.5 hour) = 30 (km)/(h)$

Hope this will help you :)

If a car can go from 0 to 60 mi/hr in 8.0 seconds, what would be its final speed after 5.0 seconds if its starting speed were 50 mi/hr?

Answers

$a= (V)/(t) \ \ \ \Rightarrow\ \ a= (60\ [mi/hr])/(8\ [s]) \n \nthe\ final\ speed\ after\ 5.0\ seconds\n \nV=50\ [mi/hr]+5\ [s]\ \cdot (60\ [mi/hr])/(8\ [s])=50\ [mi/hr] + (5)/(8) \cdot 60\ [mi/hr]=\n \n=50\ [mi/hr]+37.5\ [mi/hr]=87.5\ [mi/hr]$

$v_1=60\ (mi)/(h)\n\nt_1=8.0\ s=(8)/(3600)\ h=(1)/(450)\ h\n--------------\na=(v)/(t)\n\na=(60)/((1)/(450))\ (mi)/(h^2)=27000\ (mi)/(h^2)\n--------------\nv_o=50\ (mi)/(h)\n\nt=5.0\ s=(5)/(3600)\ h=(1)/(720)\ h\n--------------$

$v=v_o+at\n\n\nv=50\ (mi)/(h)+27000\ (mi)/(h^2)\cdot(1)/(720)\ h=50\ (mi)/(h)+37.5\ (mi)/(h)=87.5\ (mi)/(h)\leftarrow Answer$

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