The speed of both cars is the same ... 80 km per hour.
But their velocities are different, because DIRECTION is part of velocity, and their directions are different.
Answer:
.05
Explanation:
wavelength/frequency = wave speed
45 = 2.5/x
x = 0.05
Answer:
Right shoe
Explanation:
Let the mass and velocity of incoming puck be m and v respectively.
Momentum of the colliding puck will be mv
In case of first case , the momentum of puck becomes zero so change in momentum after collision with left shoe
= mv - 0 = mv
If time duration of collision be t
rate of change of momentum
= mv / t
This is the force exerted by puck on the left shoe .
Now let us consider collision with right shoe
momentum after collision with right shoe
- mv
change in momentum
= mv - ( - mv ) = 2mv
If time duration of collision be t
rate of change of momentum
= 2mv / t
This is the force exerted by puck on the right shoe .
Since the force on the right shoe is more , this shoe will have greater speed
after collision.
Answer: a) 0.13*τ ; b) 2.08*τ
Explanation: In order to explain the discharg of a capacitor through a resistor, we have to consider the following:
Q(t)=Qo* exp(-t/τ) for a lose of 1/8-th of its charge
in this case, Q(t)=7/8*Qo=7/8*exp(-t/τ)
ln(7/8)*τ=-t
then, t= -ln(7/8)*τ =0.13
For a lose of 7/8 th of its charge , we have
Q(t)=1/7*Qo*exp(-t/τ)
t=-ln(1/8)*τ=2.08
Answer:
V(t) = (q0/C) * e^(−t/RC
)
Explanation:
If there were a battery in the circuit with EMF E , the equation for V(t) would be V(t)=E−(RC)(dV(t)/dt) . This differential equation is no longer homogeneous in V(t) (homogeneous means that if you multiply any solution by a constant it is still a solution). However, it can be solved simply by the substitution Vb(t)=V(t)−E . The effect of this substitution is to eliminate the E term and yield an equation for Vb(t) that is identical to the equation you solved for V(t) . If a battery is added, the initial condition is usually that the capacitor has zero charge at time t=0 . The solution under these conditions will look like V(t)=E(1−e−t/(RC)) . This solution implies that the voltage across the capacitor is zero at time t=0 (since the capacitor was uncharged then) and rises asymptotically to E (with the result that current essentially stops flowing through the circuit).