Answer:
PAPER CLIPS ON NOSE OF A PAPER AIRPLANE
Purpose: To determine if the number of paperclips on the nose of a paper airplane affects the velocity and speed, measured in meters per seconds.
Make a Hypothesis Based on the Learning Thus Far: If the number of paperclips on the nose of a paper airplane increases, then the speed will _increase______ (increase, decrease, stay the same) in a __linear_______ (linear, exponential, logarithmic) mathematical relationship, and the velocity will (increase, decrease, stay the same) in a __exponential____ (linear, exponential, logarithmic) mathematical relationship. (Fill in the appropriate words for your hypothesis.)
Pictures: Insert at least 3 pictures of yourself conducting the experiment into this lab report. At least 2 pictures must show your face as you conduct the investigation. You may need to ask someone to help take these photos.
Explanation:
Recall that average velocity is equal to change in position over a given time interval,
so that the <em>x</em>-component of 
is
and its <em>y</em>-component is
Solve for 
and 
, which are the <em>x</em>- and <em>y</em>-components of the copter's position vector after <em>t</em> = 1.60 s.
Note that I'm reading the given details as
so if any of these are incorrect, you should make the appropriate adjustments to the work above.
Answer: 2.94×10^8 J
Explanation:
Using the relation
T^2 = (4π^2/GMe) r^3
Where v= velocity
r = radius
T = period
Me = mass of earth= 6×10^24
G = gravitational constant= 6.67×10^-11
4π^2/GMe = 4π^2 / [(6.67x10^-11 x6.0x10^24)]
= 0.9865 x 10^-13
Therefore,
T^2 = (0.9865 × 10^-13) × r^3
r^3 = 1/(0.9865 × 10^-13) ×T^2
r^3 = (1.014 x 10^13) × T^2
To find r1 and r2
T1 = 120min = 120*60 = 7200s
T2 = 180min = 180*60= 10800s
Therefore,
r1 = [(1.014 x 10^13)7200^2]^(1/3) = 8.07 x 10^6 m
r2 = [(1.014 x 10^13)10800^2]^(1/3) = 10.57 x 10^6 m
Required Mechanical energy
= - GMem/2 [1/r2 - 1/r1]
= (6.67 x 10^-11 x 6.0 x 10^24 * 50)/2 * [(1/8.07 × 10^-6 )- (1/10.57 × 10^-6)]
= (2001 x 10^7)/2 * (0.1239 - 0.0945)
= (1000.5 × 10^7) × 0.0294
= 29.4147 × 10^7 J
= 2.94 x 10^8 J.
Answer:
The first part of the question is asking about BUOYANT FORCE or UPTHRUST.
Upthrust =TRUE WEIGHT-APPARENT WEIGHT
TRUE WEIGHT=mg
TRUE weight=50kg×10m/s²
=500N
upthrust=500N-380N
FB=120N
volume of the rock=mass/density.
since the granite is completely submerged, the volume of the displaced liquid will be equal to the volume of the body.
upthrust=Vdg
120N=V×1000kg/m³×10m/s²
120N=V×10000kg/m²s²
120/10000=V
v=0.012m³
please mark brainliest, hope it helped
