Let F be the magnitude of the force. The impulse of this force while the ball is in contact with the wall is
Ft = F (0.0210 s)
and this impulse is equal to the change in the ball's momentum,
m ∆v = (1.30 kg) (6.50 m/s - (-10.5 m/s)) = (1.30 kg) (17.0 m/s)
Solve for F :
F (0.0210 s) = (1.30 kg) (17.0 m/s)
F = (1.30 kg) (17.0 m/s) / (0.0210 s)
F ≈ 1050 N
Answer: Kinetic energy is the energy of motion, observable as the movement of an object, particle, or set of particles. Any object in motion is using kinetic energy: a person walking, a thrown baseball, a crumb falling from a table, and a charged particle in an electric field are all examples of kinetic energy at work.
Explanation:
Answer:
50 degree.
Explanation:
Given that the components of vector A are given as follows: Ax = 5.6 Ay = -4.7
The angle between vector A and B in the positive direction of x-axis will be achieved by using the formula:
Tan Ø = Ay/Ax
Substitute Ay and Ax into the formula above.
Tan Ø = -4.7 / 5.6
Tan Ø = -0.839
Ø = tan^-1(-0. 839)
Ø = - 40 degree
Therefore, the angle between vector A and B positive direction of x-axis will be
90 - 40 = 50 degree.
Answer:
Explanation:
(a)
Magnitude of 
= 
= 4.47 m
Let θ be the direction of vector D
θ = 63.44°
(b)
Magnitude of 
= 
= 8.485 m
Let θ be the direction of vector D
θ = 135°
Answer: 244.05 J
Explanation:
To find speed at 30 m above the ground use equation:
V²=Vo²-2Gs
V0=31.4m/s
s=30m
G=9.81m/s²
-----------------------
V²=31.4²-2*9.81*30
V²=985.96+588.6
V²=1574.56
V=39.68m/s ---speed of arrow on 30 m obove the ground
Use equation for kinetic enrgy:
Ke=mV²/2
m=0.155kg
V=39.68m/s
-------------------------
Ke=0.155kg*(39.68m/s)²/2
Ke=0.155*1574.5/2
Ke=244.05J
