For the answer to the question above, first find out the gradient.
<span>m = rise/run </span>
<span>=(y2-y1)/(x2-x1) </span>
<span>the x's and y's are the points given: "After three hours, the velocity of the car is 53 km/h. After six hours, the velocity of the car is 62 km/h" </span>
<span>(x1,y1) = (3,53) </span>
<span>(x2,y2) = (6,62) </span>
<span>sub values back into the equation </span>
<span>m = (62-53)/(6-3) </span>
<span>m = 9/3 </span>
<span>m = 3 </span>
<span>now we use a point-slope form to find the the standard form </span>
<span>y-y1 = m(x-x1) </span>
<span>where x1 and y1 are any set of point given </span>
<span>y-53 = 3(x-3) </span>
<span>y-53 = 3x - 9 </span>
<span>y = 3x - 9 + 53 </span>
<span>y = 3x + 44 </span>
<span>y is the velocity of the car, x is the time.
</span>I hope this helps.
(This is from experience so sorry if it's wrong but) When wind instruments are played, sometimes the notes go flat or sharp depending on the speed the player blows air into the instrument as well as the warmth of the air. When playing a string instrument, the pitch can be changed in many ways. For example, when the player places their fingers on the string depending on which part of the tip of the finger you use, the tone of the sound and sometimes the pitch, changes. Looking at the question in a different way, you can change the pitch and the range of notes you can reach on the instrument (both wind and string) by changing the note you tune your instrument to. Hope this helps!!
Answer: Option (c) is the correct answer.
Explanation:
When the child is tossed up into the air then she gains kinetic energy as the child has moved from its initial position.
It is given that mass is 20 kg, velocity is 
, and height is 2 m.
Calculate the kinetic energy of child as follows.
kinetic energy = 
= 
= 
= 
Also, when child falls off the ground then she will have gravitational potential energy.
Calculate gravitational potential energy of child as follows.
Potential energy = m × g × h
= 
= 
Answer:
D. Calculate the area under the graph.
Explanation:
The distance made during a particular period of time is calculated as (distance in m) = (velocity in m/s) * (time in s)
You can think of such a calculation as determining the area of a rectangle whose sides are velocity and time period. If you make the time period very very small, the rectangle will become a narrow "bar" - a bar with height determined by the average velocity during that corresponding short period of time. The area is, again, the distance made during that time. Now, you can cover the entire area under the curve using such narrow bars. Their areas adds up, approximately, to the total distance made over the entire span of motion. From this you can already see why the answer D is the correct one.
Going even further, one can make the rectangular bars arbitrarily narrow and cover the area under the curve with more and more of these. In fact, in the limit, this is something called a Riemann sum and leads to the definition of the Riemann integral. Using calculus, the area under a curve (hence the distance in this case) can be calculated precisely, under certain existence criteria.
Dipper effect of an oncoming train get louder as it approaches and sound diminishes as it goes away sound traveling
