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) 0.05 Am^2
(b) 1.85 x 10^-3 Nm
Explanation:
width, w = 10 cm = 0.1 m
length, l = 20 cm = 0.2 m
Current, i = 2.5 A
Magnetic field, B = 0.037 T
(A) Magnetic moment, M = i x A
Where, A be the area of loop
M = 2.5 x 0.1 x 0.2 = 0.05 Am^2
(B) Torque, τ = M x B x Sin 90
τ = 0.05 x 0.037 x 1
τ = 1.85 x 10^-3 Nm
Answer:
d). The value of y should be -32m
Vx=0.92 m/s
Explanation:
Using equation of motion uniform to y motion
So to find t that is the same time for all the motion
The value of Xf=-3.2m because the g is negative from the axis
Now in the axis 'x' to find Vx
The one that is loaded worst. The overall weight is not important; tongue weight is a matter of loading. Our 12,000 lb snow cat trailer, which has stops to position the cat properly, has under 100 lbs tongue weight. Excessive tongue weight is a Bad Thing because it reduces weight on the towing vehicle's front wheels, leading to instability.
Answer:
The magnitude of the acceleration of the car is 35.53 m/s²
Explanation:
Given;
acceleration of the truck, 
= 12.7 m/s²
mass of the truck, 
= 2490 kg
mass of the car, 
= 890 kg
let the acceleration of the car at the moment they collided = 
Apply Newton's third law of motion;
Magnitude of force exerted by the truck = Magnitude of force exerted by the car.
The force exerted by the car occurs in the opposite direction.
Therefore, the magnitude of the acceleration of the car is 35.53 m/s²
