<h3>
Answer: 130 newtons</h3>
===============================================================
Explanation:
We'll need the acceleration first.
- The initial speed (let's call that Vi) is 8.0 m/s
- The final speed (Vf) is 0 m/s since Sam comes to a complete stop at the end.
- This happens over a duration of t = 4.0 seconds
The acceleration is equal to the change in speed over change in time
a = acceleration
a = (change in speed)/(change in time)
a = (Vf - Vi)/(4 seconds)
a = (0 - 8.0)/4
a = -8/4
a = -2
The acceleration is -2 m/s^2, meaning that Sam slows down by 2 m/s every second. Negative accelerations are often associated with slowing down. The term "deceleration" can be used here.
Here's a further break down of Sam's speeds at the four points of interest
- At 0 seconds, he's going 8 m/s
- At the 1 second mark, he's slowing down to 8-2 = 6 m/s
- At the 2 second mark, he's now at 6-2 = 4 m/s
- At the 3 second mark, he's at 4-2 = 2 m/s
- Finally, at the 4 second mark, he's at 2-2 = 0 m/s
Next, we'll apply Newton's Second Law of motion
F = m*a
where,
- F = force applied
- m = mass
- a = acceleration
We just found the acceleration, and the mass is fairly easy as all we need to do is add Sam's mass with the sled's mass to get 60+5.0 = 65 kg
So the force applied must be:
F = m*a
F = 65*(-2)
F = -130 newtons
This force is negative to indicate it's pushing against the sled's momentum to slow Sam down.
The magnitude of this force is |F| = |-130| = 130 newtons
Answer:
You have the answer in your comments. I will be copying it so your question doesn't get deleted.
The answers is $0.58
$11.48
the video tape
the new shirt
The first one is C. Hope this helps!! If you need anymore help just message me and I will try to get back to you quickly and help in any way i can!
Wildlife researcher starts from a and then reaches b, he turns towards north 40 degree to move towards c.
Total displacement is ac
Total horizontal displacement = 4+4 cos40 =7.06 km
Total vertical displacement = 4 sin40 =2.57 km
Total displacement 
= 7.51 km
Answer:
When the blood and the dialysate are flowing in the same direction, as the the dialysate and the blood move away from the region of higher concentration of the urea, to a region distant from the source, the concentration of urea in the blood stream and in the dialysis reach equilibrium and diffusion across the semipermeable membrane stops within the higher filter regions such as II, III, IV or V
However, for counter current flow, as the concentration of the urea in the blood stream becomes increasingly lesser the, it encounters increasingly unadulterated dialysate coming from the dialysate source, such that diffusion takes place in all regions of the filter
Explanation:
