Answer:
421.83 m.
Explanation:
The following data were obtained from the question:
Height (h) = 396.9 m
Initial velocity (u) = 46.87 m/s
Horizontal distance (s) =...?
First, we shall determine the time taken for the ball to get to the ground.
This can be calculated by doing the following:
t = √(2h/g)
Acceleration due to gravity (g) = 9.8 m/s²
Height (h) = 396.9 m
Time (t) =.?
t = √(2h/g)
t = √(2 x 396.9 / 9.8)
t = √81
t = 9 secs.
Therefore, it took 9 secs fir the ball to get to the ground.
Finally, we shall determine the horizontal distance travelled by the ball as illustrated below:
Time (t) = 9 secs.
Initial velocity (u) = 46.87 m/s
Horizontal distance (s) =...?
s = ut
s = 46.87 x 9
s = 421.83 m
Therefore, the horizontal distance travelled by the ball is 421.83 m
Answer:
Final velocity v = 8.944 m/sec
Explanation:
We have given distance S = 40 meters
Time t = 10 sec
As it starts from rest so initial velocity u = 0
From second equation of motion 
Now from first equation of motion 
, here v is final velocity, u is initial velocity, a is acceleration and t is time
So 
Answer:
v = 2.928 10³ m / s
Explanation:
For this exercise we use Newton's second law where the force is the gravitational pull force
F = ma
a = F / m
Acceleration is
a = dv / dt
a = dv / dr dr / dt
a = dv / dr v
v dv = a dr
We substitute
v dv = a dr
∫ v dv = 1 / m G m M ∫ 1 / r² dr
We integrate
½ v² = G M (-1 / r)
We evaluate from the lower limit v = 0 for r = R m to the upper limit v = v for r = R + 2.73 10³, where R is the radius of Saturn's moon
v² = 2G M (- 1 / R +2.73 10³+ 1 / R)
We calculate
v² = 2 6,674 10⁻¹¹ 1.10 10²¹ (10⁻³ / 5.61 - 10⁻³ /(5.61 + 2.73))
v² = 14.6828 10⁷ (0.1783 -0.1199)
v = √8.5748 10⁶
v = 2.928 10³ m / s
Explanation:What is centripetal acceleration?
Can an object accelerate if it's moving with constant speed? Yup! Many people find this counter-intuitive at first because they forget that changes in the direction of motion of an object—even if the object is maintaining a constant speed—still count as acceleration.
Acceleration is a change in velocity, either in its magnitude—i.e., speed—or in its direction, or both. In uniform circular motion, the direction of the velocity changes constantly, so there is always an associated acceleration, even though the speed might be constant. You experience this acceleration yourself when you turn a corner in your car—if you hold the wheel steady during a turn and move at constant speed, you are in uniform circular motion. What you notice is a sideways acceleration because you and the car are changing direction. The sharper the curve and the greater your speed, the more noticeable this acceleration will become. In this section we'll examine the direction and magnitude of that acceleration.
The figure below shows an object moving in a circular path at constant speed. The direction of the instantaneous velocity is shown at two points along the path. Acceleration is in the direction of the change in velocity, which points directly toward the center of rotation—the center of the circular path. This direction is shown with the vector diagram in the figure. We call the acceleration of an object moving in uniform circular motion—resulting from a net external force—the centripetal acceleration
a
c
a
c
a, start subscript, c, end subscript; centripetal means “toward the center” or “center seeking”.
Find the amount of work that the spring does. This can be found using the equation 1/2kx^2. Then, you must set that equal to the amount of kinetic energy the car has. This is possible thanks to the work-energy theorem.
1/2kx^2 = 1/2mv^2
Solve to find velocity. Remember, the spring is displaced .15 m, not 15!
To find the acceleration, use F = ma. The force being applied to the car is kx, and you know the mass. You do the math.
For problem C I don't know, haven't done that yet in my class. Sorry!
