A team exerts a force of 10 N towards the south in a tug of war. The other, opposite team, exerts a force of 17 N towards the no
1 answer:
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Answer:
Explanation:
The capacitance of a parallel plate capacitor is given by
where A is the area of plates and d is the separation between the plates.
Now the area is halved and the separation is tripled.
The new capacitance is
C' = C / 6
Initial potential energy is given by
here the charge is constant
The new energy is given by
U' = 6 U
The energy becomes six times the initial value.
Answer:
The initial velocity should be greater than 23.9 m/s. This velocity is underestimated because it does not take into account the resistance of the air.
Explanation:
Hi there!
The final position vector of the car (denoted as "r" in the figure) is
r = ( 120, -100) m
(placing the origin of the frame of reference at the launching point).
The position vector at a time "t" is calculated by the following equation:
r = (x0 + v0 · t · cos θ, y0 + v0 · t · sin θ + 1/2 · g · t²)
Where:
x0 = initial vertical position.
v0 = initial velocity.
t = time.
θ = launching angle.
y0 = initial vertical position.
g = acceleration due to gravity.
At final time:
120 m = x0 + v0 · t · cos θ
-100 m = y0 + v0 · t · sin θ + 1/2 · g · t²
Since the origin is placed at the launching point, x0 and y0 = 0, then:
120 m = v0 · t · cos θ
-100 m = v0 · t · sin θ + 1/2 · g · t²
We have a system with two equations and two unknowns, so, we can solve it.
Solving the first equation for "v0":
120 m = v0 · t · cos θ
v0 = 120 m / (cos θ · t)
Replacing v0 in the second equation:
-100 m = v0 · t · sin θ + 1/2 · g · t²
-100 m = (120 m/cos 15° · t) · t · sin 15° - 1/2 · (9.8 m/s²) · t²
-100 m = 120 m · tan 15° - 4.9 m/s² · t²
(-100 m - 120 m · tan 15°) / (-4.9 m/s²) = t²
t = 5.2 s
Now we can calculate the initial velocity:
v0 = 120 m / cos 15° · 5.2 s
v0 = 23.9 m/s
The initial velocity should be greater than 23.9 m/s
This velocity is underestimated because it does not take into account the resistance of the air.
