An example of a hypothesis for an experiment might be: “A basketball will bounce higher if there is more air it”
Step one would be to make an observation... “hey, my b-ball doesn’t have much air in it, and it isn’t bouncing ver high”
Step two is to form your hypothesis: “A basketball will bounce higher if there is more air it”
Step three is to test your hypothesis: maybe you want to drop the ball from a certain height, deflate it by some amount and then drop it from that same height again, and record how high the ball bounced each time.
Here the independent variable is how much air is in the basketball (what you want to change) and the dependent variable is how high the b-ball will bounce (what will change as a result of the independent variable)
Step four is to record all of your results and step five is to analyze that data. Does your data support your hypothesis? Why or why not?
You should only test one variable at a time because it is easier to tell why the results are how they are; you only have one cause.
Hope this helps!
Beginning when the bottom of the object first touches the water,
and as it descends and more and more of it goes under, the
buoyant force on it increases during that time.
As soon as the object is completely underwater, it doesn't matter
how deep under it is, the buoyant force on it remains the same.
Correct temperature is 80°F
Answer:
T_f = 38.83°F
Explanation:
We are given;
Volume; V = 8 ft³
Initial Pressure; P_i = 100 lbf/in² = 100 × 12² lbf/ft²
Initial temperature; T_i = 80°F = 539.67 °R
Time for outlet flow; t_o = 90 s
Mass flow rate at outlet; m'_o = 0.03 lb/s
Final pressure; P_f = 30 lbf/in² = 30 × 12² lbf/ft²
Now, from ideal gas equation,
Pv = RT
Where v is initial specific volume
R is ideal gas constant = 53.33 ft.lbf/°R
Thus;
v = RT/P
v_i = 53.33 × 539.67/(100 × 12²)
v_i = 2 ft³/lb
Formula for initial mass is;
m_i = V/v_i
m_i = 8/2
m_i = 4 lb
Now change in mass is given as;
Δm = m'_o × t_o
Δm = 0.03 × 90
Δm = 2.7 lb
Now,
m_f = m_i - Δm
Thus; m_f = 4 - 2.7
m_f = 1.3 lb
Similarly in above;
v_f = V/m_f
v_f = 8/1.3
v_f = 6.154 ft³/lb
Again;
Pv = RT
Thus;
T_f = P_f•v_f/R
T_f = (30 × 12² × 6.154)/53.33
T_f = 498.5°R
Converting to °F gives;
T_f = 38.83°F
I believe the answer is 153.8 m.
<span>5.7 km/h north and 5.8 km/h west are instantaneous velocities, while 8.1 km/h is the average velocity.
This is because each value has a magnitude and direction so it is a velocity. Moreover, the 8.1 km/h is the resultant of the two velocities so it is the average while the other two are instantaneous.</span>
