The short answer is that the displacement is equal tothe area under the curve in the velocity-time graph. The region under the curve in the first 4.0 s is a triangle with height 10.0 m/s and length 4.0 s, so its area - and hence the displacement - is
1/2 • (10.0 m/s) • (4.0 s) = 20.00 m
Another way to derive this: since velocity is linear over the first 4.0 s, that means acceleration is constant. Recall that average velocity is defined as
<em>v</em> (ave) = ∆<em>x</em> / ∆<em>t</em>
and under constant acceleration,
<em>v</em> (ave) = (<em>v</em> (final) + <em>v</em> (initial)) / 2
According to the plot, with ∆<em>t</em> = 4.0 s, we have <em>v</em> (initial) = 0 and <em>v</em> (final) = 10.0 m/s, so
∆<em>x</em> / (4.0 s) = (10.0 m/s) / 2
∆<em>x</em> = ((4.0 s) • (10.0 m/s)) / 2
∆<em>x</em> = 20.00 m
Answer:
u = 10.63 m/s
h = 1.10 m
Explanation:
For Take-off speed ..
by using the standard range equation we have

R = 9.1 m
θ = 26º,
Initial velocity = u
solving for u



u = 10.63 m/s
for Max height
using the standard h(max) equation ..



h = 1.10 m
Answer:

Explanation:
First of all, we need to calculate the total energy supplied to the calorimeter.
We know that:
V = 3.6 V is the voltage applied
I = 2.6 A is the current
So, the power delivered is

Then, this power is delivered for a time of
t = 350 s
Therefore, the energy supplied is

Finally, the change in temperature of an object is related to the energy supplied by

where in this problem:
E = 3276 J is the energy supplied
C is the heat capacity of the object
is the change in temperature
Solving for C, we find:
