Work = Force x Distance
Assuming that this work is being done parallel to the displacement that is, but under that assumption:
W = (50)(10)
W = 500 J
Momentum, p = m.v
m of the girl = 60.0 kg
m of the boat = 180 kg
v of the girl = 4.0 m/s
A) Momentum of the girl as she is diving:
p = m.v = 60.0 kg * 4.0 m/s = 24.0 N/s
B) momentum of the raft = - momentum of the girl = -24.0 N/s
C) speed of the raft
p = m.v ; v = p/m = 24.0N/s / 180 kg = -0.13 m/s [i.e. in the opposite direction of the girl's velocity]
Answer:
the magnitude of the average contact force exerted on the leg is 3466.98 N
Explanation:
Given the data in the question;
Initial velocity of hand v₀ = 5.25 m/s
final velocity of hand v = 0 m/s
time interval t = 2.65 ms = 0.00265 s
mass of hand m = 1.75 kg
We calculate force on the hand F
using equation for impulse in momentum
F
× t = m( v - v₀ )
we substitute
F
× 0.00265 = 1.75( 0 - 5.25 )
F
× 0.00265 = 1.75( - 5.25 )
F
× 0.00265 = -9.1875
F
= -9.1875 / 0.00265
F
= -3466.98 N
Next we determine force on the leg F
Using Newton's third law of motion
for every action, there is an equal opposite reaction;
so, F
= - F
we substitute
F
= - ( -3466.98 N )
F
= 3466.98 N
Therefore, the magnitude of the average contact force exerted on the leg is 3466.98 N
Answer:
82 degrees
Explanation:
consider your staying point to be the center of a circle. this center has the coordinates (0, 0).
the radius of the circle is the distance you walked East (14 miles).
I assume your teacher means as "angle of displacement" the angle between the East-West line going through your starting point and the direct line from your starting point to your current position.
then the 100 miles North is tan(displacement angle)×14.
as it is the same, if you first went North and then East, or the other way around. you end up at the same point, with the same coordinates.
so, again.
100 = 14×tan(angle)
tan(angle) = 100/14 = 50/7 = 7.142857...
the displacement angle is then 82 degrees.