Answer:
the ball travelled approximately 60 m towards north before stopping
Explanation:
Given the data in the question;
First course : 
= 0.75 m/s², 
= 20 m, 
= 10 m/s
now, form the third equation of motion;
v² = u² + 2as
we substitute
² = (10)² + (2 × 0.75 × 20)
² = 100 + 30
² = 130
= √130
= 11.4 m/s
for the Second Course:
= 11.4 m/s, 
= -1.15 m/s², 
= 0
Also, form the third equation of motion;
v² = u² + 2as
we substitute
0² = (11.4)² + (2 × (-1.15) × 
)
0 = 129.96 - 2.3
2.3 = 129.96
= 129.96 / 2.3
= 56.5 m
so;
|d| = √( ² + ² )
we substitute
|d| = √( (20)² + (56.5)² )
|d| = √( 400 + 3192.25 )
|d| = √( 3592.25 )
|d| = 59.9 m ≈ 60 m
Therefore, the ball travelled approximately 60 m towards north before stopping
a) we can answer the first part of this by recognizing the player rises 0.76m, reaches the apex of motion, and then falls back to the ground we can ask how
long it takes to fall 0.13 m from rest: dist = 1/2 gt^2 or t=sqrt[2d/g] t=0.175
s this is the time to fall from the top; it would take the same time to travel
upward the final 0.13 m, so the total time spent in the upper 0.15 m is 2x0.175
= 0.35s
b) there are a couple of ways of finding thetime it takes to travel the bottom 0.13m first way: we can use d=1/2gt^2 twice
to solve this problem the time it takes to fall the final 0.13 m is: time it
takes to fall 0.76 m - time it takes to fall 0.63 m t = sqrt[2d/g] = 0.399 s to
fall 0.76 m, and this equation yields it takes 0.359 s to fall 0.63 m, so it
takes 0.04 s to fall the final 0.13 m. The total time spent in the lower 0.13 m
is then twice this, or 0.08s
Answer:
The speed of the car, v = 19.997 m/s
Explanation:
Given,
The centripetal acceleration of the car, a = 13.33 m/s²
The radius of the curve, r = 30 m
The centripetal force acting on the car is given by the formula
F = mv²/r
Where v²/r is the acceleration component of the force
a = v²/r
Substituting the values in the above equation
13.33 = v²/30
v² = 13.33 x 30
v² = 399.9
v = 19.997 m/s
Hence, the speed of the car, v = 19.997 m/s
(5 mi/hr) x (1hr/60min) x (10min) = 5 x 10 / 60 = <em>5/6 mile</em>
(5/6 mile) x (1,760 yd/mile) = <em>1,466 and 2/3 yards</em>
Answer: 1.d) The acceleration of an object is always less than the acceleration due to gravity, g (9.81m/s^-2)
2.a)acceleration decreases
Explanation:
Newton's second law:
Newton's second law states that the acceleration of an object is defined by two variables which is the total force acting on the object and the mass of that object. The acceleration is directly proportional to the net force that is applied on an object and inversely proportional to the mass of that object.
When the force applied on an object is increased so does the acceleration of an object however if the mass increase the acceleration decreases.
This can be felt when you look at the truck which usually carry heavy loads they seem to drive slow due to the load hence their acceleration is decreased by the mass that these truck carry .
