Answer:
16/22
Explanation:
you add red and blue together
Answer:
1.4 m/s
Explanation:
From the question given above, we obtained the following data:
Initial Displacement (d1) = 0.9 m
Final Displacement (d2) = 1.6 m
Initial time (t1) = 1.5 secs
Final time (t2) = 2 secs
Velocity (v) =..?
The velocity of an object can be defined as the rate of change of the displacement of the object with time. Mathematically, it can be expressed as follow:
Velocity = change of displacement /time
v = Δd / Δt
Thus, with the above formula, we can obtain the velocity of the car as follow:
Initial Displacement (d1) = 0.9 m
Final Displacement (d2) = 1.6 m
Change in displacement (Δd) = d2 – d1 = 1.6 – 0.9
= 0.7 m
Initial time (t1) = 1.5 secs
Final time (t2) = 2 secs
Change in time (Δt) = t2 – t1
= 2 – 1.5
= 0.5 s
Velocity (v) =..?
v = Δd / Δt
v = 0.7/0.5
v = 1.4 m/s
Therefore, the velocity of the car is 1.4 m/s
Answer:
Answer is 183.6 J
Explanation:
Using the Physics reference sheet the formula for Potential energy is
(mass) x (gravity) x (height)
Mass= 1.2
Gravity I used is 9.81 (use 10 to get the answer most schools use)
Height= 15.6
Answer:
The dependent variable is academic performance
The independent variable is the presence/absence of tutorial support
The control group are students who did not get the tutorial support.
The experimental group were students that got the tutorial support
Explanation:
In every experiment, there is a dependent and independent variable as well as an experimental and a control group.
The experimental group receive the treatment while the control group do not receive the treatment. The independent variable is manipulated and its impact on the dependent variable is evaluated.
The control group are students who did not receive the tutorial support while the experimental group are students that received the tutorial support.
The dependent variable in this case is academic performance. Its outcome depends on the presence or absence of tutorial support (independent variable).
Answer:
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Explanation:
Given two mass on an incline code
and
and an angle of inclination
.
. Assume that
is the weight being pulled up and
the hanging weight.
-The equations of motion from Newton's Second Law are:
where a is the acceleration.
#Substituting for
(tension) gives:
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#and solving for 
which is the system's acceleration.