Answer:
34.3 m/s
Explanation:
Newton's Second Law states that the resultant of the forces acting on the car is equal to the product between the mass of the car, m, and the centripetal acceleration 
(because the car is moving of circular motion). So at the top of the hill the equation of the forces is:
where
(mg) is the weight of the car (downward), with m being the car's mass and g=9.8 m/s^2 is the acceleration due to gravity
R is the normal reaction exerted by the road on the car (upward, so with negative sign)
v is the speed of the car
r = 0.120 km = 120 m is the radius of the curve
The problem is asking for the speed that the car would have when it tires just barely lose contact with the road: this means requiring that the normal reaction is zero, R=0. Substituting into the equation and solving for v, we find:
That's electric "current", usually described in 'Amperes'.
1 Ampere = 1 Coulomb per second ... "the rate at which
charge passes a point in the circuit".
Answer:
Explanation:
350 N force stretches the spring by 30 cm
spring constant K = 350 / 0.30 = (350 / 0.3) N / m
To calculate work done by a spring force we proceed as follows
spring force when the spring is stretched by x = Kx
This force is variable so work done by it can be calculated by integration
Work done by it in stretching from x₁ to x₂
W = ∫ F dx
= ∫ Kx dx with limit from x₁ to x ₂
= 1/2 K ( x₂² - x₁² )
Putting the given values of x₁ = 0.50 m , x₂ = 0.8 m
Work done
= 1/2 x (350 / 0.3)x ( 0.80² - 0.50² )
= 227.50 J
I think it's B ,go to physicsclassroom.com Newton's second law
Answer:
0.25 m
Explanation:
We can solve the problem by using the lens equation:
where
f is the focal length
p is the distance of the object from the lens
q is the distance of the image from the lens
In this problem, we have
f = +20 cm=+0.20 m (the focal length is positive for a converging lens)
q = +1.0 m (the image distance is positive for a real image)
Solving the equation for p, we find
