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: 14.7kJ, 29.4kJ, 44.1kJ
Explanation:
<em>The gravitational potential energy is the energy that a body or object possesses, due to its position in a gravitational field. </em>
<em />
In the case of the Earth, in which the gravitational field is considered constant, the value of the gravitational potential energy 
will be:
Where 
is the mass of the object, 
the acceleration due gravity and 
the height of the object.
Knowing this, let's begin with the calculaations:
For m=3kg
For m=6kg
For m=9kg
Answer:
<h3>JAWAB SECEPATNYA pliss</h3><h3 /><h3>Anda memiliki rangkaian paralel 10 volt, dengan 2 resistor di atasnya. Berapakah tegangan pada</h3><h3>resistor pertama? Di seberang kedua?</h3><h3 /><h3>(saya akan menandai tercerdas tolong bantu)</h3>
Explanation:
Hukum Ohm
= tegangan
= kuat arus
= ketahanan
Kalau kamu mau mencari tegangan listrik, kamu gunakan rumus V = I.R. Kalau ternyata kamu perlu mencari kuat arus listrik, maka gunakan rumus I = V/R. Nah, kalau yang kamu cari adalah hambatan listrik, maka gunakan rumus R = V/I.
Answer: 0°
Explanation:
Step 1: Squaring the given equation and simplifying it
Let θ be the angle between a and b.
Given: a+b=c
Squaring on both sides:
... (a+b) . (a+b) = c.c
> |a|² + |b|² + 2(a.b) = |c|²
> |a|² + |b|² + 2|a| |b| cos 0 = |c|²
a.b = |a| |b| cos 0]
We are also given;
|a+|b| = |c|
Squaring above equation
> |a|² + |b|² + 2|a| |b| = |c|²
Step 2: Comparing the equations:
Comparing eq( insert: small n)(1) and (2)
We get, cos 0 = 1
> 0 = 0°
Final answer: 0°
[Reminders: every letters in here has an arrow above on it]
