Answer:
F = -4567.40 N
Explanation:
Given that,
The power developed by the engine, P = 196 hp
1 hp = 746 W
196 hp = 146157 W
Speed of the car, v = 32 m/s
Let F is the total friction force acting on the car. The product of force and velocity is called the power developed by the engine. It is given by :



F = -4567.40 N
So, the total frictional force acting on the car is 4567.40 N. Hence, this is the required solution.
Answer:
A
Explanation:
because u are subtracting if this is from flvs that is what i did and it was right
Answer:
The minimum value of width for first minima is λ
The minimum value of width for 50 minima is 50λ
The minimum value of width for 1000 minima is 1000λ
Explanation:
Given that,
Wavelength = λ
For D to be small,
We need to calculate the minimum width
Using formula of minimum width


Where, D = width of slit
= wavelength
Put the value into the formula

Here,
should be maximum.
So. maximum value of
is 1
Put the value into the formula


(b). If the minimum number is 50
Then, the width is


(c). If the minimum number is 1000
Then, the width is


Hence, The minimum value of width for first minima is λ
The minimum value of width for 50 minima is 50λ
The minimum value of width for 1000 minima is 1000λ
Answer
B. F=ma
Explanation
The Newton's laws of motion tries to explain the how bodies behave and the energy changes when theys are in motion. For the 3 of them to hold, the bodies must be moving in a straight line and with constant velocity.
The second one states that, "the change of momentum of a moving body is directly proportional to the force producing it and it takes place to the direction of force."
From the choices given, the appropriate answer is B. F=ma
Answer:
The pickup truck and hatchback will meet again at 440.896 m
Explanation:
Let us assume that both vehicles are at origin at the start means initial position is zero i.e.
= 0. Both the vehicles will cross each other at same time so we will make equations for both and will solve for time.
Truck:
= 33.2 m/s, a = 0 (since the velocity is constant),
= 0
Using 
s = 33.2t .......... eq (1)
Hatchback:
,
= 0 m/s (since initial velocity is zero),
= 0
Using 
putting in the data we will get

now putting 's' value from eq (1)

which will give,
t = 13.28 s
so both vehicles will meet up gain after 13.28 sec.
putting t = 13.28 in eq (1) will give
s = 440.896 m
So, both vehicles will meet up again at 440.896 m.