Answer:
while using brainly app. there is an option of uploading image of diagram, graph or plot. plz use that feature to upload image of the standing wave so that your question can be properly answeres
Answer:
In the beginning, I was not familiar to assess assessments of the other students. Ifelt a little bit weird that is it possible to check assignments while having an instructor.I was also a bit frustrated, to be honest, that why do we have to assess thoseassessments. It was kind of extra burden for me. But after few weeks assessingmore assignments, my feeling had changed because I was learning lots of thingsthat were changing my perspectives. I was gaining extra knowledge from my peersin the form of assessments. Yes, I am comfortable with assessing assessments,because I got to learn many vocabularies and making structures of the sentencecorrectly by improving grammatically as I am not a native English speaker. Thus, inthis way, I was learning something new in each and every assessment.
Answer:
1.2 x 10¹¹ kgm²/s
Explanation:
m = mass of the airplane = 39043.01
r = altitude of the airplane = 9.2 km = 9.2 x 1000 m = 9200 m
v = speed of airplane = 335 m/s
L = Angular momentum of airplane
Angular momentum of airplane is given as
L = m v r
Inserting the values
L = (39043.01 ) (335) (9200)
L = (39043.01 ) (3082000)
L = 1.2 x 10¹¹ kgm²/s
Answer:
Mass is 152.95743
weight in ibs 3304.6914215069355 ibs
EXPLANATION:
1500 N = 152.95743 KG
152.95743 KG × 9.8 = 1498.982814
1 KG is 2.2 IBS
so
14o8.982814 is 3304.6914215069355
Answer:
The speed of the clay before the impact was 106.35 m/s.
Explanation:
the only force doing work on the system is the frictional force, f, the work done by f is given by:
Wf = ΔK = Kf - Ki
The clay and the block will come to rest after sliding Δx = 7.50 m, if their intial speed is v and the combined mass is m and μ is the coefficient of friction and g is gravity,then:
f×Δx = Ki
m×g×Δx×μ = 1/2×m×v^2
v^2 = 2×g×Δx×μ
= 2×(9.8)×(7.50)×(0.650)
= 95.55
v = 9.78 m/s
This is the veloty of clay and block after the clay hit the block.
if the velocity the clay and block attains after the impact is v and the initial speed of the clay is v1 and the mass is m and the speed of the block initially is V = 0 m/s and the mass is M, then according to the conservation of linear momentum:
m×v1 +M×V = v(m + M)
m×v1 = v(m + M)
v1 = v(m + M)/m
v1 = (9.78)(8.3×10^-3 + 82×10^-3)/(8.3×10^-3)
v1 = 106.35 m/s
Therefore, the speed of the clay before the impact was 106.35 m/s.
