Answer:
greater
Explanation:
Let the initial velocity is u .
Acceleration due to gravity on earth, ge = g
Acceleration due to gravity on moon, gm = g / 6
Let the maximum height reached on earth is h and the maximum height reached on the moon is h'.
Use third equation of motion
![v^{2}=u^{2}- 2gh](https://tex.z-dn.net/?f=v%5E%7B2%7D%3Du%5E%7B2%7D-%202gh)
At maximum height, the final velocity = 0.
For earth
0 = u^2 - 2 gh
h = u^2 / 2g ..... (1)
For moon
0 = u^2 - 2 g/6 x h'
h' = 6u^2 / 2g = 6 h (From (1)
So, the maximum height reached at moon is 6 times the maximum height reached at earth.
Force = mass*acceleration so
3.6*2.5 =9 Newtons
We can approach this in another way.
We know that sin(∅) = height / hypotenuse.
Thus, for x, height is 1 and hypotenuse is 3. Using Pythagoras theorem,
3² = 1² + b²
b = √8
cos(x) = b/hypotenuse
cos(x) = √8 / 3
Now, lets consider y:
sec(y) = 1 / cos(y) = 1 / base / hypotenuse = hypotenuse / base
The hypotenuse is 25 and the base is 24. We again apply Pythagoras theorem to find the third side, which works out to be:
height = 7
sin(y) = height / hypotenuse
sin(y) = 7/25
Now, sin(x + y) =
sin(x)cos(y) + sin(y)cos(x)
= (1/3)(24/25) + (√8 / 3)(7/25)
= 8/25 + 7√8/75
= (24 + 14√2) / 75
To simplify his insane equation, do M*G*H (Mass*Gravity*Height)
so like this.
2 (Mass)*9.8 (Gravity)*40 (height)
then when you multiply, you get
784 kg m^2/s
Answer:
990 J
Explanation:
Kinetic energy is:
KE = ½ mv²
Given m = 55 kg and v = 6 m/s:
KE = ½ (55 kg) (6 m/s)²
KE = 990 J