The velocity with which the jumper leaves the floor is 5.1 m/s.
<h3>
What is the initial velocity of the jumper?</h3>
The initial velocity of the jumper or the velocity with which the jumper leaves the floor is calculated by applying the principle of conservation of energy as shown below.
Kinetic energy of the jumper at the floor = Potential energy of the jumper at the maximum height
¹/₂mv² = mgh
v² = 2gh
v = √2gh
where;
- v is the initial velocity of the jumper on the floor
- h is the maximum height reached by the jumper
- g is acceleration due to gravity
v = √(2 x 9.8 x 1.3)
v = 5.1 m/s
Learn more about initial velocity here: brainly.com/question/19365526
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Answer:
The net force acting on this object is 180.89 N.
Explanation:
Given that,
Mass = 3.00 kg
Coordinate of position of 
Coordinate of position of 
Time = 2.00 s
We need to calculate the acceleration
For x coordinates
On differentiate w.r.to t
On differentiate again w.r.to t
The acceleration in x axis at 2 sec
For y coordinates
On differentiate w.r.to t
On differentiate again w.r.to t
The acceleration in y axis at 2 sec
The acceleration is
We need to calculate the net force
The magnitude of the force
Hence, The net force acting on this object is 180.89 N.
An external force that is being applied in the direction of the displacement
Answer:
Resistance of the second wire is twice the first wire.
Explanation:
Let us first see the formula of resistance;
R = pxL/A
Here L is the lenght of the wire, A the area and p is the resistivity of wire.
As we are given that the length of second wire is double than that of the first wire, hence the resistance of second wire would be double.
Since we have two loop in second case, inducing double voltage but as resistance is doubled so the current would remain same according to ohms law
I = V/R
