C.) Pedestrians yielding to cross traffic.
The distance traveled by the particle at the given time interval is 0.28 m.
<h3>
Position of the particle at time, t = 0</h3>
The position of the particle at the given time is calculated as follows;
x = 2 sin2(t)
y = 2 cos2(t)
x(0) = 2 sin2(0) = 0
y(0) = 2 cos2(0) = 2(1) = 2
<h3>
Position of the particle at time, t = 4</h3>
x = 2 sin2(t)
y = 2 cos2(t)
x(4) = 2 sin2(4) = 0.28
y(4) = 2 cos2(4) = 2(1) = 1.98
<h3>Distance traveled by the particle at the given time interval</h3>
d = √[(x₄ - x₀)² + (y₄ - y₀)²]
d = √[(0.28 - 0)² + (1.98 - 2)²]
d = 0.28 m
Thus, the distance traveled by the particle at the given time interval is 0.28 m.
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Answer:
Vi = 32 [m/s]
Explanation:
In order to solve this problem we must use the following the two following kinematics equations.

The negative sign of the second term of the equation means that the velocity decreases, as indicated in the problem.
where:
Vf = final velocity = 8[m/s]
Vi = initial velocity [m/s]
a = acceleration = [m/s^2]
t = time = 5 [s]
Now replacing:
8 = Vi - 5*a
Vi = (8 + 5*a)
As we can see we have two unknowns the initial velocity and the acceleration, so we must use a second kinematics equation.

where:
d = distance = 100[m]
(8^2) = (8 + 5*a)^2 - (2*a*100)
64 = (64 + 80*a + 25*a^2) - 200*a
0 = 80*a - 200*a + 25*a^2
0 = - 120*a + 25*a^2
0 = 25*a(a - 4.8)
therefore:
a = 0 or a = 4.8 [m/s^2]
We choose the value of 4.8 as the acceleration value, since the zero value would not apply.
Returning to the first equation:
8 = Vi - (4.8*5)
Vi = 32 [m/s]
Answer:
below
Explanation: When a liquid changes into a gas vaporization has occurred. The process can either occur due to boiling or evaporation. Boiling occurs when the vapor pressure of the liquid is raised (by heating) to the point where it is equal to the atmospheric pressure.
The tape is charged by friction and her hand has an induced opposite charge. They therefore attract.