How much tension must a rope withstand if it is used to accelerate a 960 kg car horizontally along a frictionless track at 1.20
2 answers:
Answer:
<em>The tension of the rope is 1152 N</em>
Explanation:
<u>Force</u>
Newton's second law states that the acceleration of an object is directly proportional to the net force and inversely related to its mass.
The law can be written as:
F = m.a
The rope must apply a force F to accelerate the car of m=960 Kg at . Given the track is frictionless, all the force applied produces acceleration, thus:
F = 1152 N
The tension of the rope is 1152 N
You might be interested in
Answer:
D. the masses of the objects and the distance between them
Explanation:
Gravitation is a force, a force doesn't care about the shape or density of objects, only about their masses... and distances.
And you can get it using the following equation:
Where :
G is the universal gravitational constant : G = 6.6726 x 10-11N-m2/kg2
m represent the mass of each of the two objects
d is the distance between the centers of the objects.
<h2>5.3 km</h2>
Explanation:
This question involves continuous displacement in various directions. When it becomes difficult to imagine, vector analysis becomes handy.
Let us denote each of the individual displacements by a vector. Consider the unit vectors as the unit vectors in the direction of East and North respectively.
By simple calculations, we can derive the unit vectors in the directions North,
South of West and
North of West respectively.
So Total displacement vector = Sum of individual displacement vectors.
Displacement vector =
Magnitude of Displacement =
∴ Total displacement =