A student performs an experiment that involves the motion of a pendulum. The student attaches one end of a string to an object o
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Answer:
The rate of change of the distance between the helicopter and yourself (in ft/s) after 5 s is ft/ sec
Explanation:
Given:
h(t) = 25 ft/sec
x(t) = 10 ft/ sec
h(5) = 25 ft/sec . 5 = 125 ft
x(5) = 10 ft/sec . 5 = 50 ft
Now we can calculate the distance between the person and the helicopter by using the Pythagorean theorem
Lets find the derivative of distance with respect to time
Substituting the values of h(t) and x(t) and simplifying we get,
=
=
ft / sec
Answer:
q₁ = + 1.25 nC
Explanation:
Theory of electrical forces
Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.
Known data
q₃=5 nC
q₂=- 3 nC
d₁₃= 2 cm
d₂₃ = 4 cm
Graphic attached
The directions of the individual forces exerted by q1 and q₂ on q₃ are shown in the attached figure.
For the net force on q3 to be zero F₁₃ and F₂₃ must have the same magnitude and opposite direction, So, the charge q₁ must be positive(q₁+).
The force (F₁₃) of q₁ on q₃ is repulsive because the charges have equal signs ,then. F₁₃ is directed to the left (-x).
The force (F₂₃) of q₂ on q₃ is attractive because the charges have opposite signs. F₂₃ is directed to the right (+x)
Calculation of q1
F₁₃ = F₂₃
We divide by (k * q3) on both sides of the equation
q₁ = + 1.25 nC
Answer:
75 m
Explanation:
The horizontal motion of the projectile is a uniform motion with constant speed, since there are no forces acting along the horizontal direction (if we neglect air resistance), so the horizontal acceleration is zero.
The horizontal component of the velocity of the projectile is
and it is constant during the motion;
the total time of flight is
t = 5 s
Therefore, we can apply the formula of the uniform motion to find the horizontal displacement of the projectile: