The work done occurs only in the direction the block was moved - horizontally. Work is given by:
W = F(h) * d
Where F(h) is the force applied in that direction (horizontal) and d is the distance in that direction. In this case, F(h) is the horizontal component of the applied force, F(app). However, the question doesn't give us F(app), so we need to find it some other way.
Since the block is moving at a constant speed, we know the horizontal forces must be balanced so that the net force is 0. This means that F(h) must be exactly balanced by the friction force, f. We can express F(h) as a function of F(app):
F(h) = F(app)cos(23)
Friction is a little trickier - since the block is being PUSHED into the ground a bit by the vertical component of the applied force, F(v), the normal force, N, is actually a bit more than mg:
N = mg + F(v) = mg + F(app)sin(23)
Now we can get down to business and solve for F(app) - as mentioned above:
F(h) = f
F(h) = uN
F(h) = u * (mg + F(v))
F(app)cos(23) = 0.20 * (33 * 9.8 + F(app)sin(23))
F(app) = 76.8
Now that we have F(app), we can find the exact value of F(h):
F(h) = F(app)cos(23)
F(h) = 76.8cos(23)
F(h) = 70.7
And now that we have F(h), we can find W:
W = F(h) * d
W = 70.7 * 6.1
W = 431.3
Therefore, the work done by the worker's force is 431.3 J. This also represents the increase in thermal energy of the block-floor system.
85)
The bone structure from outside would be periosteum, cortical, cancellous then medulla. Periosteum is the area outside the bone that will supply nutrition into the outer part of the bone. Cortical is the layer where bone mineral deposition is intense. In this part, the bone is compact and hard. This is the part of the bone that has great strength.
Cancellous is part of the bone where it is not too hard but not too soft. Mineralization is not so dense like cortical layer. That makes this part looks spongy.
In the medulla, most part is made from connective tissue and blood vessels. This part is responsible for the bone vascularization, which means the supply of mineral to the outer part is coming from the medulla. Mineralization is not much in the medulla, makes it not strong. Medulla or marrow also makes blood cells.
86)
The region of the spine would be cervical(neck), thoracal(chest), lumbar(back), sacrum then coccyx. The vertebrae in the neck are smaller since it did not need much strength but need more mobility. Lower part on the thoracal and lumbar is bigger and have a bigger process that will further stabilize the vertebrae. Thoracal vertebrae have a part on their side where the ribs attached. Sacrum shape was a kinda weird because it needs to be able to connect with the pelvis to make buttock. Coccyx the tailbone look like just a small remnant and doesn't seem to have a function in human.
87)
Axis is the name of cervical 2nd vertebral which was located below the atlas, 1st cervical vertebrae. Both of them doesn't have spinal processes that were stabilizing the side of vertebrae Axis also has an odontoid peg which will make the joint with atlas more flexible.
This effect makes the neck can turn to side easily. It also makes neck more mobile vertically, make a nodding movement is possible.
88)
Thre kind of joints would be: Synarthrosis, amphiarthrosis, and diarthrosis.
In synarthrosis, the joint mostly made by fibrous so that it can move. The example of this joint would be suture in the skull. In children, the skull joint is not closed to permit the head to grow but in the adult, it is closed.
Amphiarthrosis joint permits a small movement. This included the intervertebral disc.
Diarthrosis permit a wide degree of movement. This joint is more complex because it has the synovial membrane. The complex structure makes the joints more durable to shock. This was vital because organ with this join used frequently. The example of this joints would be femur and pelvic(hip) joints. Joints in arm and leg mostly diarthrosis joint.
Answer:
The correct answer is t = 0.92s
Explanation:
Initial velocity v0 = 3.0 m/s
Displacement Δy = ?
Acceleration a = -9.8m/s2
Final velocity v = -6.0m/s
Time t=? Target unknown
We can use the kinematic formula missing Δy to solve for the target unknown t:
V=v0+at
We can rearrange the equation to solve to t:
V-v0=at
t= v-v0/a
Substituting the known value into the kinematic formula gives:
t= (-6.0m/s)-(3.0m/s)
————————————
-9.8m/s2
= -9m/s
—————-
-9.8m/s2
=0.92s
