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Answer:
The algebraic equation is:
Explanation:
Given information:
mb = book's mass
vb = tangential speed
R = radius of the path
Question: Derive an algebraic equation for the vertical force, Fv = ?
To derive the equation, we need to draw a force diagram for this case, please, see the attached diagram. As you can see, there are three types of forces acting on the system. Two up and one of the weight acting down. Therefore, the algebraic equation is as follows:
The variables were defined above and g is the gravity.
Answer:
a)Initial speed of the projectile = 196.2 m/s
b)Maximum altitude = 490.5 m
c) Range of projectile = 3398.28 m
d) Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m
Explanation:
Time of flight of a projectile is given by the expression,
Here θ = 30° and t = 20 s
a)
Initial speed of the projectile = 196.2 m/s
b) Maximum altitude is given by
Maximum altitude = 490.5 m
c) Range of projectile is given by
Range of projectile = 3398.28 m
d) Horizontal velocity = ucosθ = 196.2 x cos 30 = 169.91 m/s
Vertical velocity = usinθ = 196.2 x sin 30 = 98.1 m/s
We have equation of motion s = ut + 0.5 at²
Horizontal motion
u = 169.91 m/s
a = 0 m/s²
t = 15 s
Substituting
s = 169.91 x 15 + 0.5 x 0 x 15² = 2548.71 m
Vertical motion
u = 98.1 m/s
a = -9.81 m/s²
t = 15 s
Substituting
s = 98.1 x 15 + 0.5 x -9.81 x 15² = 367.88 m
Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m
Answer:
Explanation:
Let m is the mass of both cars. The first car is moving with speed v and the other car is moving with speed 2v. The only force acting on both cars is the centripetal force.
For faster car on the road,
v = 2v
..........(1)
For the slower car on the road,
............(2)
Equation (1) becomes,
So, the frictional force required to keep the slower car on the road without skidding is one fourth of the faster car.