The person should start to slow down but if close enough or in the intersection go threw. Otherwise come to a complete stop until the light turns green again
Answer: 6.47m/s
Explanation:
The tangential speed can be defined in terms of linear speed. The linear speed is the distance traveled with respect to time taken. The tangential speed is basically, the linear speed across a circular path.
The time taken for 1 revolution is, 1/3.33 = 0.30s
velocity of the wheel = d/t
Since d is not given, we find d by using formula for the circumference of a circle. 2πr. Thus, V = 2πr/t
V = 2π * 0.309 / 0.3
V = 1.94/0.3
V = 6.47m/s
The tangential speed of the tack is 6.47m/s
Answer:
Maximum height reached by the ball is 32 meters.
Explanation:
It is given that,
If a baseball is project upwards from the ground level with an initial velocity of 32 feet per second, then it's height is a function of time. The equation is given as :
...........(1)
t is the time taken
s is the height attained as a function of time.
Maximum height achieved can be calculated as :


-16 t + 32 = 0
t = 2 seconds
Put the value of t in equation (1) as :

s = 32 meters
So, the maximum height reached by the ball is 32 meters. Hence, this is the required solution.
Answer:
From the previous explanation Student No. 1 has the correct explanation
Explanation:
When the fluorescent lamp emits a light it has the shape of its emission spectrum, this light collides with the atoms of Nitrogen and excites it, so these wavelengths disappear, lacking in the spectrum seen by the observed, for which we would see an absorption spectrum
The nitrogen that was exited after a short time is given away in its emission lines, in general there are many lines, so the excitation energy is divided between the different emission lines, which must be weak
From the previous explanation Student No. 1 has the correct explanation
Answer:
q₃=5.3nC
Explanation:
First, we have to calculate the force exerted by the charges q₁ and q₂. To do this, we use the Coulomb's Law:

Since we know the net force, we can use this to calculate q₃. As q₁ is at the right side of q₃ and q₁ and q₃ have opposite signs, the force F₁₃ points to the right. In a similar way, as q₂ is at the left side of q₃, and q₂ and q₃ have equal signs, the force F₂₃ points to the right. That means that the resultant net force is the sum of these two forces:

In words, the value of q₃ must be 5.3nC.