Answer:
a)The parachutist in the air for 12.63 seconds.
b)The parachutist falls from a height of 293 meter.
Explanation:
Vertical motion of parachutist:
Initial speed, u = 0m/s
Acceleration, a = 9.81 m/s²
Displacement, s = 50 m
We have equation of motion, v² = u² + 2as
Substituting
v² = 0² + 2 x 9.81 x 50
v = 31.32 m/s
Time taken for this
31.32 = 0 + 9.81 x t
t = 3.19 s
After 50m we have
Initial speed, u = 31.32m/s
Acceleration, a = -2 m/s²
Final speed , v = 3 m/s
We have equation of motion, v² = u² + 2as
Substituting
3² = 31.32² - 2 x 2 x s
s = 243 m
Time taken for this
3 = 31.32 - 2 x t
t = 9.44 s
a) Total time = 3.19 + 9.44 = 12.63 s
The parachutist in the air for 12.63 seconds.
b) Total height = 50 + 243 = 293 m
The parachutist falls from a height of 293 meter.
Answer:
<h2>Because water can lose and accept H + ions .......</h2>
You can just use basic
trigonometry to solve for the x & y components.
<span>vector a = 10cos(30) i +
10sin(30) j = <5sqrt(3), 5></span>
vector b is only slightly harder because the angle is relative
to vector a, and not the positive x-axis. Anyway, this just makes vector b with
an angle of 135deg to the positive x-axis.
<span>vector b = 10cos(135) i +
10sin(135) j = <-5sqrt(2), 5sqrt(2)></span>
So
now we can do the questions:
r = a + b
r = <5sqrt(3)-5sqrt(2), 5+5sqrt(2)>
(a)
5sqrt(3)-5sqrt(2)
(b)
5+5sqrt(2)
(c)
|r|
= sqrt( (5sqrt(3)-5sqrt(2))2 + (5+5sqrt(2))2 )
=
12.175
(d)
θ = tan-1 (
(5+5sqrt(2)) / (5sqrt(3)-5sqrt(2)) )
θ
= 82.5deg
<span> </span>
Answer:
t = 0.2845Nm (rounded to 4 decimal places)
Explanation:
The disk rotates at a distance of an arc length of 28cm
Arc length = radius × central angle × π/180
28cm = 10cm × central angle × π/180
Central angle = 
× 180/π ≈ 160.4°
Torque (t) = rFsin(central angle) , where F is the applied force
Radius in meters = 10/100 = 0.1m
t = 0.1m × 16N × sin160.4°
t = 0.2845Nm (rounded to 4 decimal places)
Answer:
<u>-8</u>
Explanation:
if he starts at ten and takes 10 steps left he'll be at -10... then if he takes 2 steps to the right , he's at -8 on the number-line
