I think its true i dont kno for sure
Let's look at Newton's second law
Force is directly proportional towards mass
If mass is more force will be more.
Between baseball and bowling ball Bowling ball has higher mass
So it would expert most force
Option D
Answer:
Velocity.
Explanation:
Projectile motion is characterized as the motion that an object undergoes when it is thrown into the air and it is only exposed to acceleration due to gravity.
As per the question, 'any change in the initial velocity of the projectile(object having gravity as the only force) would lead to a change in the range as well as the maximum height of the projectile.' To illustrate numerically:
Horizontal range: As per expression:
R= (
*sin2θ)/g
the range depending on the square of the initial velocity.
Maximum height: As per expression:
H= (
*
θ
)/2g
the maximum distance also depends upon square of the initial velocity.
Answer:
V = 2.87 m/s
Explanation:
The minimum speed required would be that at which the acceleration due to gravity is negated by the centrifugal force on the water.
Thus, we simply need to set the centripetal acceleration equal to gravity and solve for the speed V using the following equation:
Centripetal acceleration = V^2 / r
where r is the distance of water from the pivot or shoulder.
For our case, r will be 0.65 + 0.19 = 0.84 m
and solving the above equation we get:
9.81 = V^2 / 0.84
V^2 = 8.2404
V = 2.87 m/s
Answer:
V = 10.88 m/s
Explanation:
V_i =initial velocity = 0m/s
a= acceleration= gsinθ-
cosθ
putting values we get
a= 9.8sin25-0.2cos25= 2.4 m/s^2
v_f= final velocity and d= displacement along the inclined plane = 10.4 m
using the equation


v_f= 7.04 m/s
let the speed just before she lands be "V"
using conservation of energy
KE + PE at the edge of cliff = KE at bottom of cliff
(0.5) m V_f^2 + mgh = (0.5) m V^2
V^2 = V_f^2 + 2gh
V^2 = 7.04^2 + 2 x 9.8 x 3.5
V = 10.88 m/s