Answer:
4750000 J
Explanation:
Kinetic Energy= 1/2* mass* velocity²
1/2*950*100²=4750000
Momentum = 0.5 * 4 = 2
to conclude the man’s velocity after he throws the piece of equipment, divide
this number by the man’s mass.
v = 2/90
This is about 0.0222 m/s. To know if he can move 6 meters at velocity in
4minutes, use the following equation.
d = v * t, t = 4 * 60 = 240 s
d = 2/90 * 240 = 5⅓ meters.
This is ⅔ of a meter from the spaceship. To know the velocity that he must have
to move 6 meter, use the same equation.
6 = v * 240
v = 6/240
This is about 0.00416 m/s.
His final momentum = 90 * 6/240 = 2.25
To know the velocity of the package, divide this number by the mass of the
package.
v = 2.25/0.5 = 4.5 m/s
Answer:
4334.4 J
Explanation:
Work done equals to kinetic energy change
KE=½mv²
Change in KE is given by
∆KE=½m(v²-u²)
Where m is mass of water-skier, KE is kinetic energy, ∆KE is the change in kinetic energy, v is final velocity and u is initial velocity.
Substituting 72 kg for m, 12.1 m/s for v and 5.10 m/s for u then
∆KE=½*72(12.1²-5.10²)=4334.4J
Therefore, the work done by the net external force acting on the skier is equal to 4334.4 J
Complete Question
The complete question(reference (chegg)) is shown on the first uploaded image
Answer:
The magnitude of the resultant force is 
The direction of the resultant force is
from the horizontal plane
Explanation:
Generally when resolving force, if the force (F )is moving toward the angle then the resolve force will be
while if the force is moving away from the angle then the resolved force is 
Now from the diagram let resolve the forces to their horizontal component
So


Now resolving these force into their vertical component can be mathematically evaluated as


Now the resultant force is mathematically evaluated as

substituting values


The direction of the resultant force is evaluated as
![\theta = tan^{-1}[\frac{F_y}{F_x} ]](https://tex.z-dn.net/?f=%5Ctheta%20%20%3D%20%20tan%5E%7B-1%7D%5B%5Cfrac%7BF_y%7D%7BF_x%7D%20%5D)
substituting values
![\theta = tan^{-1}[\frac{ 14.3}{199.128} ]](https://tex.z-dn.net/?f=%5Ctheta%20%20%3D%20%20tan%5E%7B-1%7D%5B%5Cfrac%7B%2014.3%7D%7B199.128%7D%20%5D)
from the horizontal plane