This is what I got:
Net force in the Y direction:
ΣFy = T1 - T2
F = ma
ma = T1 - T2
Isolate for T2
ma - T1 = -T2
Multiply by -1
T1 - ma = T2
100 - (3)(2) = T2
100 - 6 = T2
T2 = 94 N
Answer:
a
Generally from third equation of motion we have that
![v^2 = u^2 + 2a[s_i - s_f]](https://tex.z-dn.net/?f=v%5E2%20%3D%20%20u%5E2%20%2B%202a%5Bs_i%20-%20s_f%5D%20)
Here v is the final speed of the car
u is the initial speed of the car which is zero
is the initial position of the car which is certain height H
is the final position of the car which is zero meters (i.e the ground)
a is the acceleration due to gravity which is g
So
![v^2 = 0 + 2g[H - 0]](https://tex.z-dn.net/?f=v%5E2%20%3D%200%20%2B%202g%5BH%20-%200%5D%20)
=> 
b
Explanation:
Generally from third equation of motion we have that
Here v is the final speed of the car
u is the initial speed of the car which is zero
is the initial position of the car which is certain height H
is the final position of the car which is zero meters (i.e the ground)
a is the acceleration due to gravity which is g
So
=>
When 
we have that
=> 
=>
Answer:
B. 25 m/s/s
Explanation:
Centripetal acceleration is the square of the tangential velocity divided by the radius of curvature.
a = v² / r
Given v = 10 m/s and r = 4 m:
a = (10 m/s)² / 4 m
a = 25 m/s²
Answer:
d = 69 .57 meter
Explanation:
First case
Speed of car ( v ) = 20.5 mi/h = 9.164 M/S
distance ( d ) = 11.6 meter ( m = mass of the car )
Work done = 0.5 m v² = 0.5 * 9.164² * m J = 41.99 m J
Force = ( workdone /distance ) = ( 41.99 m / 11.6 ) = 3.619 m N
Second case
v = 50.2 mi/h = 22.44135 m/s
d = ?
Work done = 0.5 * 22.44² * m J = 251.7768 * m J
Since the braking force remains the same .
3.619 m = ( 251.7768 m / d )
d = 69 .57 meter
