Momentum (p) = mass × velocity
so, 480×40 = 19,200 kg km/hr
so the answer is C !!
Answer:
d = 84 m
Explanation:
As we know that when an object moves with uniform acceleration or deceleration then we can use equation of kinematics to find the distance moved by the object
here we know that
initial speed 
final speed 
time taken by the car to stop

now the distance moved by the car before it stop is given as

now we have


Answer:
circuit sketched in first attached image.
Second attached image is for calculating the equivalent output resistance
Explanation:
For calculating the output voltage with regarding the first image.

![Vout = 5 \frac{2000}{5000}[/[tex][tex]Vout = 5 \frac{2000}{5000}\\Vout = 5 \frac{2}{5} = 2 V](https://tex.z-dn.net/?f=Vout%20%3D%205%20%5Cfrac%7B2000%7D%7B5000%7D%5B%2F%5Btex%5D%3C%2Fp%3E%3Cp%3E%5Btex%5DVout%20%3D%205%20%5Cfrac%7B2000%7D%7B5000%7D%5C%5CVout%20%3D%205%20%5Cfrac%7B2%7D%7B5%7D%20%3D%202%20V)
For the calculus of the equivalent output resistance we apply thevenin, the voltage source is short and current sources are open circuit, resulting in the second image.
so.

Taking into account the %5 tolerance, with the minimal bound for Voltage and resistance.
if the -5% is applied to both resistors the Voltage is still 5V because the quotient has 5% / 5% so it cancels. to be more logic it applies the 5% just to one resistor, the resistor in this case we choose 2k but the essential is to show that the resistors usually don't have the same value. applying to the 2k resistor we have:




so.

Answer:
The frequency is 
Explanation:
From the question we are told that
The time taken for it to decay to half its original size is
Let the voltage of the capacitor when it is fully charged be
Then the voltage of the capacitor at time t is said to be 
Now this voltage can be mathematical represented as

Where RC is the time constant
substituting values





Generally the cross-over frequency for a low pass filter is mathematically represented as

substituting values


Kepler's third law states that, for a planet orbiting around the Sun, the ratio between the cube of the radius of the orbit and the square of the orbital period is a constant:

(1)
where
r is the radius of the orbit
T is the period
G is the gravitational constant
M is the mass of the Sun
Let's convert the radius of the orbit (the distance between the Sun and Neptune) from AU to meters. We know that 1 AU corresponds to 150 million km, so

so the radius of the orbit is

And if we re-arrange the equation (1), we can find the orbital period of Neptune:

We can convert this value into years, to have a more meaningful number. To do that we must divide by 60 (number of seconds in 1 minute) by 60 (number of minutes in 1 hour) by 24 (number of hours in 1 day) by 365 (number of days in 1 year), and we get