Answer:
(N-1) × (L/2R) = (N-1)/2
Explanation:
let L is length of packet
R is rate
N is number of packets
then
first packet arrived with 0 delay
Second packet arrived at = L/R
Third packet arrived at = 2L/R
Nth packet arrived at = (n-1)L/R
Total queuing delay = L/R + 2L/R + ... + (n - 1)L/R = L(n - 1)/2R
Now
L / R = (1000) / (10^6 ) s = 1 ms
L/2R = 0.5 ms
average queuing delay for N packets = (N-1) * (L/2R) = (N-1)/2
the average queuing delay of a packet = 0 ( put N=1)
Answer:
t = 30.1 sec
Explanation:
If the ant is moving at a constant speed, the velocity vector will have the same magnitude at any point, and can be decomposed in two vectors, along directions perpendicular each other.
If we choose these directions coincident with the long edge of the paper, and the other perpendicular to it, the components of the velocity vector, along these axes, can be calculated as the projections of this vector along these axes.
We are only interested in the component of the velocity across the paper, that can be calculated as follows:
vₓ = v* sin θ, where v is the magnitude of the velocity, and θ the angle that forms v with the long edge.
We know that v= 1.3 cm/s, and θ = 61º, so we can find vₓ as follows:
vₓ = 1.3 cm/s * sin 61º = 1.3 cm/s * 0.875 = 1.14 cm/s
Applying the definition of average velocity, we can solve for t:
t =
= 
⇒ t = 30.1 sec
Answer:

Explanation:
<u>Projectile Motion</u>
In projectile motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration (assuming no friction), and the acceleration in the vertical direction is always the acceleration of gravity. The basic formulas are shown below:

Where
is the angle of launch respect to the positive horizontal direction and Vo is the initial speed.

The horizontal and vertical distances are, respectively:


The total flight time can be found as that when y = 0, i.e. when the object comes back to ground (or launch) level. From the above equation we find

Using this time in the horizontal distance, we find the Range or maximum horizontal distance:

Let's solve for 

This is the general expression to determine the angles at which the projectile can be launched to hit the target. Recall the angle can have to values for fixed positive values of its sine:


Or equivalently:

Given Vo=37 m/s and R=70 m


And

Answer:
Rate of heat transfer is 0.56592 kg/hour
Explanation:
Q = kA(T2 - T1)/t
Q is rate of heat transfer in Watts or Joules per second
k is thermal conductivity of the styrofoam = 0.035 W/(mK)
A is area of the cubical picnic chest = 6L^2 = 6(0.5)^2 = 6×0.25 = 1.5 m^2
T1 is initial temperature of ice = 0 °C = 0+273 = 273 K
T2 is temperature of the styrofoam = 25 °C = 25+273 = 298 K
t is thickness of styrofoam = 0.025 m
Q = 0.035×1.5(298-273)/0.025 = 1.3125/0.025 = 52.5 W = 52.5 J/s
Mass flow rate = rate of heat transfer ÷ latent heat of melting of ice = 52.5 J/s ÷ 3.34×10^ 5 J/kg = 1.572×10^-4 kg/s = 1.572×10^-4 kg/s × 3600 s/1 hr = 0.56592 kg/hr