Force applied on the car due to engine is given as
towards right
Also there is a force on the car towards left due to air drag
towards left
now the net force on the car will be given as

now we can say that since the two forces are here opposite in direction so here the vector sum of two forces will be the algebraic difference of the two forces.
So we can say



So here net force on the car will be 150 N towards right and hence it will accelerate due to same force.
Answer:
As we are converting 220V AC into a 5V DC, first we need a step-down transformer to reduce such high voltage. Here we have used 9-0-9 1A step-down transformer, which convert 220V AC to 9V AC. In transformer there are primary and secondary coils which step up or step down the voltage according to the no of turn in the coils.
Selection of proper transformer is very important. Current rating depends upon the Current requirement of Load circuit (circuit which will use the generate DC). The voltage rating should be more than the required voltage. Means if we need 5V DC, transformer should at least have a rating of 7V, because voltage regulator IC 7805 at least need 2V more i.e. 7V to provide a 5V voltage.
Answer:
<h2>C. maintenance </h2>
Explanation:
I personally believe one key disadvantages is the cost of maintaining the equipment unlike the gym where you have to subscribe for the month or the year and forget about anything, owning the gym equipment comes with the extra cost and responsibilities of maintenance to the owners.
Answer: a) vcar= 7 m/s ; b) a train= 0.65 m/s^2
Explanation: By using the kinematic equation for the car and the train we can determine the above values of the car velocity and the acceletarion of the train, respectively.
We have for the car
distance = v car* t, considering the length of train (81.1 m) travel by the car during the first 11.6 s
the v car = distance/time= 81.1 m/11.6s= 7 m/s
In order to calculate the acceleration we have to use the kinematic equation for the train from the rest
distance train = (a* t^2)/2
distance train : distance travel by the car at constant speed
so distance train= (vcar*36.35)m=421 m
the a traiin= (2* 421 m)/(36s)^2=0.65 m/s^2
The velocity when function p(t)=11 is 8 .
According to the question
The position of a car at time t represented by function :
Now,
When function p(t) = 11 , t will be
11 = t²+2t-4
0 = t² + 2t - 15
or
t² +2t-15 = 0
t² +(5-3)t-15 = 0
t² +5t-3t-15 = 0
t(t+5)-3(t+5) = 0
(t-3)(t+5) = 0
t = 3 , -5
as t cannot be -ve as given ( t≥0)
so,
t = 3
Now,
the velocity when p(t)=11
As we know velocity =
therefore to get the value of velocity from function p(t)
we have to differentiate the function with respect to time
v(t) = 2t + 2
where v(t) = velocity at that time
as t = 3 for p(t)=11
so ,
v(t) = 2t + 2
v(t) = 2*3 + 2
v(t) = 8
Hence, the velocity when function p(t)=11 is 8 .
To know more about function here:
brainly.com/question/12431044
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