Answer:
The back end of the vessel will pass the pier at 4.83 m/s
Explanation:
This is purely a kinetics question (assuming we're ignoring drag and other forces) so the weight of the boat doesn't matter. We're given:
Δd = 315.5 m
vi = 2.10 m/s
a = 0.03 m/s^2
vf = ?
The kinetics equation that incorporates all these variables is:
vf^2 = vi^2 + 2aΔd
vf = √(2.1^2 + 2(0.03)(315.5))
vf = 4.83 m/s
the reign of the sun is very an important because the typically is about the size of 6,000 ceilometers
I would go with C because all the rest of the answer are sitting on something.
other helper already explains but here are two examples:
2. let N be +ve n S be -ve
N-going Zak momentum=mass*velocity=50*4=200
S-going Zak momentum=40*(-5)=-200
after collision, they get stuck together so their masses=50+40=90
by conservation of momentum:
200+(-200)=90*final vel
0=90*final vel
final vel=0m/s
3. cars have rear end collision so they were traveling in same direction be4
total momentum be4 collision=1200*20+1000*15=39000
cars continue 2 move as 1 mass after collision so final mass=1200+1000=2200
by conservation of momentum, final momentum=2200*final vel=39000
final vel=39000/2200=17.73m/s
plz post in comment if u need 4 n 5
<h3>
<u>Provided</u><u>:</u><u>-</u></h3>
- Initial velocity = 15 m/s
- Final velocity = 10 m/s
- Time taken = 2 s
<h3><u>To FinD:-</u></h3>
- Accleration of the particle....?
<h3>
<u>How</u><u> </u><u>to</u><u> </u><u>solve</u><u>?</u></h3>
We will solve the above Question by using equations of motion that are:-
- v = u + at
- s = ut + 1/2 at²
- v² = u² + 2as
Here,
- v = Final velocity
- u = Initial velocity
- a = acceleration
- t = time taken
- s = distance travelled
<h3>
<u>Work</u><u> </u><u>out</u><u>:</u></h3>
By using first equation of motion,
⇛ v = u + at
⇛ 10 = 15 + a(2)
⇛ -5 = 2a
Flipping it,
⇛ 2a = -5
⇛ a = -2.5 m/s² [ANSWER]
❍ Acclearation is negative because final velocity is less than Initial velocity.
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