Answer:
v = 1.25 m/s
Explanation:
We have,
Distance covered by a person is 10 meters
Time taken by him to cover that distance is 8 seconds.
If we want to find the speed of a person, we must know distance covered by it and taken. In this case, we know both distance and time. His speed is given by :

So, the speed of the person is 1.25 m/s.
Answer:
281.25 J
Explanation:
We are told that the two objects with masses m and 3m.
Also that energy stored in the spring is 375 joules.
Now, initially the centre of mass of the system took place at rest, it means v1 = v and v2 = v/3
Thus, from principle of conservation of energy, we have;
½mv² + ½(3m)(v/3)² = 375J
(m + 3m/9)½v² = 375
(4/3)m × ½v² = 375
Multiply both sides by ¾ to get;
½mv² = 375 × ¾
½mv² = 281.25 J
Therefore, energy of lighter body is 281.25 J
Newton's law of universal gravitation states that every point mass in the universe attracts every other point mass with a force that is directly proportional to the product of their masses, and inversely proportional to the square of the distance between them. Newton's law of universal gravitation states that every point mass in the universe attracts every other point mass with a force that is directly proportional to the product of their masses, and inversely proportional to the square of the distance between them.
Answer:
The answer is below
Explanation:
The length of the rope is equal to the radius of the circle formed by the complete rotation of the rope. Therefore the radius = 1.50 m.
a) The distance covered by the rope when completing one rotation is the same as the perimeter of the circle. Hence:
Distance covered in one rotation = 2π * radius = 2π * 1.5 = 3π meters
The velocity of the ball = Distance / time = 3π meters / 3.4 seconds = 2.77 m/s
b) The initial velocity (u) is 0 m/s, the final velocity is 2.77 m/s during time (t) = 3.4 s. Hence acceleration (a):
v = u + at
2.77 = 3.4a
a = 0.82 m/s²
c) Force on ball = mass * acceleration = 4 * 0.82 = 3.28 N
Hi there!

Since the object is being pulled at a constant velocity, the forces must be balanced.
Since there is no movement vertically, we must take into account the horizontal forces. We can also assume a positive acceleration to be in the direction of motion.
The acceleration and force due to gravity on an incline is:
a = gsinФ
F = MgsinФ
∑F = -MgsinФ + T
Since it is getting pulled at a constant velocity, ∑F = 0. So:
0 = -MgsinФ + T
MgsinФ = T
Solve for T by plugging in values. Let g = 10 m/s²
T = (120)(10)sin(27) ≈ 545 N