<span>A. </span>Let’s
say the horizontal component of the velocity is vx and the vertical is vy. <span>
Initially at t=0 (as the mug leaves the counter) the
components are v0x and v0y.
<span>v0y = 0 since the customer slides it horizontally so applied
force is in the x component only.
<span>The equations for horizontal and vertical projectile motion
are:
x = x0 + v0x t
y = y0 + v0y t - 1/2 g t^2 = y0 - 1/2 g t^2 </span></span></span>
Setting the origin to be the end corner of the
counter so that x0=0 and y0=0, hence:
x = v0x t
y = - 1/2 g t^2
Given value are: x=1.50m and y=-1.15m (y is
negative since mug is going down)
<span>1.50m = v0x t
----> v0x= 1.50/t</span>
<span>-1.15m = -(1/2) (9.81) t^2 -----> t =0.4842 s</span>
Calculating for v0x:
v0x = 3.10 m/s
<span>B. </span>v0x
is constant since there are no other horizontal forces so, v0x=vx=3.10m/s
vy can be calculated from the formula:
<span>vy = v0y + at where a=-g
(negative since going down)</span>
vy = -gt = -9.81 (0.4842)
vy = -4.75 m/s
Now to get the angle below the horizontal, tan(90-Ø) = -vx/vy
tan(90-Ø )= 3.1/4.75
Ø =
56.87˚<span> below the horizontal</span>
Answer:
(a) The horizontal ground reaction force 
(b) The vertical ground reaction force 
(c) The resultant ground reaction force 
Explanation:
Given
John mass , m = 65 kg
Horizontal acceleration , 
Vertical acceleration , 
(a) Using Newton's 2nd law in horizontal direction
=>
Thus the horizontal ground reaction force
(b) Using Newton's 2nd law in vertical direction
=>
=>
Thus the vertical ground reaction force
(c) Resultant ground reaction force is
=>
=>
Thus the resultant ground reaction force
Acceleration a=3m/s^2
time t= 4.1seconds
Final velocity V= 55km/h
initial velocity U= ?
First convert V to m/s
36km/h=10m/s
55km/h= 55*10/36=15.28m/s
Using the formula V= U+at
U= V-at
U= 15.28-3*4.1=15.28-12.3=2.98m/s
Initial velocity U= 2.98m/s or 10.73km/h (Using the conversion rate 36km/h=10m/s)
Answer:
Explanation:
The acceleration of gravity is 9.8m/s^2.
So to calculate the time it will take to make the ball stop(which btw means the ball now reach its greatest height), use the formula V1=V0+at. V1 is the final velocity(which is 0), V0 is the starting velocity(which is 30m/s), and the a(cceleration) is 9.8m/s^2.
(You can ignore the fact "at" is -30 instead 30, it's because the directions two velocity travel are opposite. )
We can now know the time it takes to make the ball stop just by the gravitational force is about 3 sec.
Use another formula S=1/2at^2, to find out the S(height) is 1/2*9.8*3^2=44.1, which is approximately D.45m .
The banking angle of the curved part of the speedway is determined as 32⁰.
<h3>
Banking angle of the curved road</h3>
The banking angle of the curved part of the speedway is calculated as follows;
V(max) = √(rg tanθ)
where;
- r is radius of the path
- g is acceleration due to gravity
V² = rg tanθ
tanθ = V²/rg
tanθ = (34²)/(190 x 9.8)
tanθ = 0.62
θ = arc tan(0.62)
θ = 31.8
θ ≈ 32⁰
Learn more about banking angle here: brainly.com/question/8169892
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