The potential energy would be zero. Only kinetic energy is present in this case. To find out what the answer is we do the equation: mv^2/2 soo...
KE =mv^2/2
KE= 1(2^2)/2 which the answer will come up by 2 Joules.
You will use the height of the bridge from the ground.
Solution:
Formula to be used is y=Viy(t)+g(t^2)/2
Where:
Vi=initial velocity which is 0 m/s
y=10 m
Gravitational acceleration or g =9.8m/s^2
T= time you need
Substitute all the given to the formula
10m=(0m/s)(t)+(9.8m/s^2)(t^2)/2
10mx2=9.8m/s^2(t^2)
Now isolate the variable you want to find which is T or time
10mx2/9.8m/s^2=t^2
20m/9.8m/s^2=t^2
Square root of 2.04= square root of t^2
T=1.43 secs
The answer is 1.43 seconds
Answer:
answer c: a mass and volume
Explanation:
no matter what elements are chosen, they all have a mass and volume measurement I believe
The magnitude of the vector C is 96.32m
<h3>How to solve for the magnitude of vector c</h3>
Ax = AcosθA
= 40 cOS 20
= 37.59
Ay = AsinθA
-40sin20
= -13.68
Bx = B cos θ B
= 75Cos50
= 48.21
By = BsinθB
= 75sin50
= 57.45
Cx = AX + Bx
= 37.59 + 48.21
= 85.8
Cy = Ay + By
= -13.65 + 57.45
= 43.77
The magnitude is solved by
|c| = 
= √85.8² + 43.77²
= 96.32m
The magnitude of the vector c is 96.32m
Read more on the magnitude of a vector here:
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Answer:0.58 m
Explanation:
The initial velocity of the ball is u = 2.0 m/s
The height of the table is, h = 1.0 m
The ball falls in vertical direction under acceleration due to gravity.
Time taken for ball to hit the floor:
h= ut + 0.5gt² ( from the equation of motion)
1.0 m=2.0 m/s × t+0.5 × 9.8 m/s²× t²
Solving this for t,
t = 0.29 s ( we have neglected the negative value of t)
In the same time, the ball would cover a horizontal distance of :
s = u t
⇒s = 2.0 m/s×0.29 s = 0.58 m
Thus, the landing spot is 0.58 m away from the table.
