Answer:
Action and reaction are the same
Explanation:
Let us carefully analyze the situation in the context of Newton's third law, when the car hits the mosquito exerts a force that acts on the mosquito, ACTION and the mosquito responds to the force that is being applied with a force of equal magnitude and direction or opposite applied to the car (Reaction).
Consequently, the correct answer is: The force of the car on the mosquito is of the same magnitude as that of the mosquito on the car
Mendeleev produced the first orderly arrangement of known elements.
Mendeleev used patterns to predict undiscovered elements.
Answer:
T = 480.2N
Explanation:
In order to find the required force, you take into account that the sum of forces must be equal to zero if the object has a constant speed.
The forces on the boxes are:
(1)
T: tension of the rope
M: mass of the boxes 0= 49kg
g: gravitational acceleration = 9.8m/s^2
The pulley is frictionless, then, you can assume that the tension of the rope T, is equal to the force that the woman makes.
By using the equation (1) you obtain:
The woman needs to pull the rope at 480.2N
The optimal angle of 45° for maximum horizontal range is only valid when initial height is the same as final height.
<span>In that particular situation, you can prove it like this: </span>
<span>initial velocity is Vo </span>
<span>launch angle is α </span>
<span>initial vertical velocity is </span>
<span>Vv = Vo×sin(α) </span>
<span>horizontal velocity is </span>
<span>Vh = Vo×cos(α) </span>
<span>total time in the air is the the time it needs to fall back to a height of 0 m, so </span>
<span>d = v×t + a×t²/2 </span>
<span>where </span>
<span>d = distance = 0 m </span>
<span>v = initial vertical velocity = Vv = Vo×sin(α) </span>
<span>t = time = ? </span>
<span>a = acceleration by gravity = g (= -9.8 m/s²) </span>
<span>so </span>
<span>0 = Vo×sin(α)×t + g×t²/2 </span>
<span>0 = (Vo×sin(α) + g×t/2)×t </span>
<span>t = 0 (obviously, the projectile is at height 0 m at time = 0s) </span>
<span>or </span>
<span>Vo×sin(α) + g×t/2 = 0 </span>
<span>t = -2×Vo×sin(α)/g </span>
<span>Now look at the horizontal range. </span>
<span>r = v × t </span>
<span>where </span>
<span>r = horizontal range = ? </span>
<span>v = horizontal velocity = Vh = Vo×cos(α) </span>
<span>t = time = -2×Vo×sin(α)/g </span>
<span>so </span>
<span>r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) </span>
<span>r = -(Vo)²×sin(2α)/g </span>
<span>To find the extreme values of r (minimum or maximum) with variable α, you must find the first derivative of r with respect to α, and set it equal to 0. </span>
<span>dr/dα = d[-(Vo)²×sin(2α)/g] / dα </span>
<span>dr/dα = -(Vo)²/g × d[sin(2α)] / dα </span>
<span>dr/dα = -(Vo)²/g × cos(2α) × d(2α) / dα </span>
<span>dr/dα = -2 × (Vo)² × cos(2α) / g </span>
<span>Vo and g are constants ≠ 0, so the only way for dr/dα to become 0 is when </span>
<span>cos(2α) = 0 </span>
<span>2α = 90° </span>
<span>α = 45° </span>
Answer:
0.647 nC
Explanation:
The force experienced by a charge due to the presence of an electric field is given by
where
q is the charge
E is the magnitude of the electric field
In this problem, each antenna is modelled as it was a single point charge, experiencing a force of
Therefore, if the electric field magnitude is
Then the charge on each antenna would be
