Answer:
a) 17.20
b) 11.31
c) 14.42
d) 12.65
Explanation:
(a)
The girl is at the origin of the x,y coordinates (i.e 0,0,0 )
the position vector of the car at time 't' secs is
at t=2s, the position vector is
Therefore, the the distance between the car and the girl is
s = 17.20
(b)
The position of the car at t = 0s is
The position of the car at t = 2s is
The distance of the car traveled in the interval from t=0s to t=2 s is as follows:
(c)
The position vector of the car at time 't' secs is
The velocity of the car is
the direction of the car's velocity at t = 2s is going to be
Thus; The speed of the car is
(d) the car's acceleration is:
The magnitude of car's acceleration at t=2s is
The answer to your question is: All of the above. Hopefully this helped you!
Answer:
5
Explanation:
The sum of the digits of the number is ...
(4+1+3)+(4+6+5)+(7+8+9) = 8+15+24 = 47
The sum of those digits is 4+7=11, and those digits sum to 1+1 = 2.
That is, the value of the number mod 9 (or 3) is 2.
The ones digit is odd, so the value of the number mod 2 is 1.
This combination of modulo values tells you the mod 6 result is 5.
_____
<em>Additional comment</em>
We can look at the (mod2, mod3) values of the numbers 0 to 5:
0 ⇒ (0, 0)
1 ⇒ (1, 1)
2 ⇒ (0, 2)
3 ⇒ (1, 0)
4 ⇒ (0, 1)
5 ⇒ (1, 2) . . . . the mod {2, 3} results we have for the number of interest.
This process of adding up the digits repeatedly is referred to as "casting out 9s." The result of it is the modulo 9 value of the number (with 0 mapped to 9). Checking the mod 9 result of arithmetic operations is one quick way to spot certain kinds of errors. It can also be used as part of a divisibility test for 3 or 9.
Answer:
471 days
Explanation:
Capacity of Carvins Cove water reservoir = 3.2 billion gallons i.e. 3.2 x 10˄9 gallons
As,
1 gallon = 0.133 cubic feet (cf)
Therefore,
Capacity of Carvins Cove water reservoir in cf = 3.2 x 10˄9 x 0.133
= 4.28 x 10˄8
Applying Mass balance i.e
Accumulation = Mass In - Mass out (Eq. 01)
Here
Mass In = 0.5 cfs
Mass out = 11 cfs
Putting values in (Eq. 01)
Accumulation = 0.5 - 11
= - 10.5 cfs
Negative accumulation shows that reservoir is depleting i.e. at a rate of 10.5 cubic feet per second.
Converting depletion of reservoir in cubic feet per hour = 10.5 x 3600
= 37,800
Converting depletion of reservoir in cubic feet per day = 37, 800 x 24
= 907,200
i.e. 907,200 cubic feet volume is being depleted in days = 1 day
1 cubic feet volume is being depleted in days = 1/907,200 day
4.28 x 10˄8 cubic feet volume will deplete in days = (4.28 x 10˄8) x 1/907,200
= 471 Days.
Hence in case of continuous drought reservoir will last for 471 days before dry-up.
Answer:
departure rate k = 7.653 vehicle per minute
Explanation:
given data
parking = 6 veh/ min
arriving = 6: 00 P. M
queue reaches = 36 vehicles
total vehicle delay = 500 veh-min
solution
we know here that total vehicle in t min will be
total vehicle in t min = 6t vehicle
and
we consider here service rate is equal to departure rate
so service rate = k vehicle per min
and service is starting when reach vehicle 36
so that no of vehicle is
no of vehicle = 6t , ( 0 ≤ t ≤ 6 )
no of vehicle = 6t - k (t -6 ) for t ≥ 6
and
Queue dissipate when here 6t - k( t-6) will be zero
so
k = ........................a
we know here
total delay is equal to 500 veh min
0.5 t × 6t - 0.5 (t-6) × k × (t-6) = 500
so solve it we get
t = 27.78 min
so k will be from equation a
k =
k = 7.653
so departure rate is 7.653 veh/min