Answer:
1/3 the distance from the fulcrum
Explanation:
On a balanced seesaw, the torques around the fulcrum calculated on one side and on another side must be equal. This means that:

where
W1 is the weight of the boy
d1 is its distance from the fulcrum
W2 is the weight of his partner
d2 is the distance of the partner from the fulcrum
In this problem, we know that the boy is three times as heavy as his partner, so

If we substitute this into the equation, we find:

and by simplifying:

which means that the boy sits at 1/3 the distance from the fulcrum.
<h2>
Entire trip takes 1.22 seconds.</h2>
Explanation:
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 0 m/s
Acceleration, a = 9.81 m/s²
Time, t = 0.866 s
Substituting
s = ut + 0.5 at²
s = 0 x 0.866 + 0.5 x 9.81 x 0.866²
s = 3.68 m
Halfway is 3.68 m
Total height = 2 x 3.68 = 7.36 m
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 0 m/s
Acceleration, a = 9.81 m/s²
Time, t = ?
Displacement, s = 7.36 m
Substituting
s = ut + 0.5 at²
7.36 = 0 x t + 0.5 x 9.81 x t²
t = 1.22 s
Entire trip takes 1.22 seconds.
Answer:
The speed of the heavier fragment is 0.335c.
Explanation:
Given that,
Mass of the lighter fragment 
Mass of the heavier fragment 
Speed of lighter fragment = 0.893c
We need to calculate the speed of the heavier fragment
Let v is the speed of the second fragment after decay
Using conservation of relativistic momentum













Hence, The speed of the heavier fragment is 0.335c.
Answer:
35m/s[57o].
X = 35*Cos57 =
Y = 35*sin7 =Explanation:
learn man but there u go
Fresh snow is the surface that exhibits the highest albedo measurement.