Answer:
τ = 0.00203 seconds
Explanation:
The time constant τ in a R-L circuit is given by
τ = L/R
First we have to find out the equivalent resistance of the circuit.
Since there is a parallel combination of 19 Ω and 6.0 Ω resistor
Req = 19*6/19+6
Req = 4.56 Ω
Now we can find out the time constant
τ = L/R
τ = 0.0093/4.56
τ = 0.00203 seconds
Therefore, the time constant of this circuit is 0.00203 seconds.
This should include distances. Displacement is the shortest distance from point A to point B. Distance is the total length of travel. For an example I have included a simple map.
Displacement would not be an accurate representation of the driving instructions because it would not tell the total length of travel. Basically, the driver would be going through buildings if you told them displacement. Directions would tell them the roads traveled.
Answer:
a) N = 7.90 10³ N, b) N = -1.04 10⁴ N
Explanation:
a) For this exercise we can use Newton's second law
N -W = m (-a)
The relationship is centripetal, the negative sign of the acceleration is because it points towards the center of the circle
a = v² / r
we substitute
N = mg -m v² /r
N = m (g - v² /r)
let's calculate
v = 8.7 m / s
N = 1240 (9.81 - 8.7²/22)
N = 7.90 10³ N
b) v = 20 m / s
N = 1240 (9.81 - 20²/22)
N = -1.04 10⁴ N
Find the amount of work that the spring does. This can be found using the equation 1/2kx^2. Then, you must set that equal to the amount of kinetic energy the car has. This is possible thanks to the work-energy theorem.
1/2kx^2 = 1/2mv^2
Solve to find velocity. Remember, the spring is displaced .15 m, not 15!
To find the acceleration, use F = ma. The force being applied to the car is kx, and you know the mass. You do the math.
For problem C I don't know, haven't done that yet in my class. Sorry!
The acceleration of gravity in low Earth orbit is a fraction smaller than it is on the Earth's surface, like maybe 15% less.
The astronauts feel no gravity at all, because they and everything around them are constantly falling.
