Answer:
5 m/s²
Explanation:
Acceleration = change in speed/time taken
20-5=15
15/3= 5 m/s²
Answer:
x=2.4t+4.9t^2
Explanation:
This equation is one of the kinematic equations to solve for distance. The original equation is as follows:
X=Xo+Vt+1/2at^2
We know that the ball starts at rest meaning that its initial velocity and position is zero.
X=0+Vt+1/2at^2
Since it is going down the ramp, you can use the acceleration of gravity constant. (9.81 m/s^2) and simplify that with the 1/2.
X=Vt+4.9t^2
Note: Since the positive direction in this problem is down, you are adding the 4.9t^2, but if a question says that the downward direction is negative, you would subtract those values.
Now, substitute in your velocity value.
X=2.4t+4.9t^2
Answer:
0.02325 m^3
Explanation:
P1 = 100000Pa , v1 = 0.0279m^3
P2= 120000Pa . V2 = ?
as mass and temperature are constant using ratio method for Boyle'slaw
P2V2= P1V1
or V2= P1V1/P2
V2 = 100000×0.0279/120000
V2= 0.02325 m^3
Answer:
height is 4 times
Explanation:
speed of ball hit by Luis is u
speed of ball hit by Ron is 2u
Let the maximum height traveled by the ball hit by Luis is h and by the ball hit by Ron is h'
At maximum height, the final velocity of ball is zero in both the cases.
Use third equation of motion, we get
For Luis :
![v^{2}=u^{2}-2gh](https://tex.z-dn.net/?f=v%5E%7B2%7D%3Du%5E%7B2%7D-2gh)
![0^{2}=u^{2}-2gh](https://tex.z-dn.net/?f=0%5E%7B2%7D%3Du%5E%7B2%7D-2gh)
.... (1)
For Ron :
![v^{2}=u^{2}-2gh](https://tex.z-dn.net/?f=v%5E%7B2%7D%3Du%5E%7B2%7D-2gh)
![0^{2}=4u^{2}-2gh'](https://tex.z-dn.net/?f=0%5E%7B2%7D%3D4u%5E%7B2%7D-2gh%27)
.... (2)
Divide equation (2) by equation (1), we get
![\frac{h'}{h}=4](https://tex.z-dn.net/?f=%5Cfrac%7Bh%27%7D%7Bh%7D%3D4)
h' = 4 h
Thus, the statement of third student is wrong.
Because the core is the hottest place at which it holds the most heat to bring it out to be towards africa.
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