Force = mass x acceleration
force = 2500kg x (20m/s / 10m/s)
force = 2500kg x 2m/s^2
force = 5000kg m/s^2 = 5kN
i hope this is right (^^)
Answer:
A fundamental theory that provides a description of the physical properties of nature at the scale of atoms and subatomic particles.
Explanation:
Answer:
V' = 0.84 m/s
Explanation:
given,
Linear speed of the ball, v = 2.85 m/s
rise of the ball, h = 0.53 m
Linear speed of the ball, v' = ?
rotation kinetic energy of the ball
I of the moment of inertia of the sphere
v = R ω
using conservation of energy
Applying conservation of energy
Initial Linear KE + Initial roational KE = Final Linear KE + Final roational KE + Potential energy
V'² = 0.7025
V' = 0.84 m/s
the linear speed of the ball at the top of ramp is equal to 0.84 m/s
Answer:
570 N
Explanation:
Draw a free body diagram on the rider. There are three forces: tension force 15° below the horizontal, drag force 30° above the horizontal, and weight downwards.
The rider is moving at constant speed, so acceleration is 0.
Sum of the forces in the x direction:
∑F = ma
F cos 30° - T cos 15° = 0
F = T cos 15° / cos 30°
Sum of the forces in the y direction:
∑F = ma
F sin 30° - W - T sin 15° = 0
W = F sin 30° - T sin 15°
Substituting:
W = (T cos 15° / cos 30°) sin 30° - T sin 15°
W = T cos 15° tan 30° - T sin 15°
W = T (cos 15° tan 30° - sin 15°)
Given T = 1900 N:
W = 1900 (cos 15° tan 30° - sin 15°)
W = 570 N
The rider weighs 570 N (which is about the same as 130 lb).
Answer:
128.9 N
Explanation:
The force exerted on the golf ball is equal to the rate of change of momentum of the ball, so we can write:
where
F is the force
is the change in momentum
is the time interval
The change in momentum can be written as
where
m = 0.04593 kg is the mass of the ball
u = 0 is the initial velocity of the ball
is the final velocity of the ball
Substituting into the original equation, we find the force exerted on the golf ball: