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notka56 [123]
3 years ago
6

A bungee cord with a spring constant of 800 StartFraction N over m EndFraction stretches 6 meters at its greatest displacement.

How much elastic potential energy does the bungee cord have? The bungee cord has J of elastic potential energy.
Physics
1 answer:
zalisa [80]3 years ago
8 0

Explanation:

EE = ½ kx²

EE = ½ (800 N/m) (6 m)²

EE = 14,400 J

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Which color of light is diffracted at a greater angle from a diffraction grating, red or yellow light?
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3 years ago
Which is an si metric unit of measurement that is used to record the heat transfer of a solution in a classroom investigation?
kumpel [21]
The SI unit for heat energy is joule
4 0
3 years ago
ben walks 2 m from his desk to the teachers desk. From the teachers desk he then walks 3 m in the same direction to the classroo
vaieri [72.5K]
Distance is the total length covered = 2m + 3m = 5m

Displacement is his distance from original position.

Displacement =  2m + (-3)m.               Representing the 3m walked back as -3.

Displacement = 2m - 3m = -1m.

So his displacement  is 1m behind his original starting point.
4 0
3 years ago
A person of mass 70 kg stands at the center of a rotating merry-go-round platform of radius 2.9 m and moment of inertia 900 kg⋅m
Cloud [144]

Explanation:

It is given that,

Mass of person, m = 70 kg

Radius of merry go round, r = 2.9 m

The moment of inertia, I_1=900\ kg.m^2

Initial angular velocity of the platform, \omega=0.95\ rad/s

Part A,

Let \omega_2 is the angular velocity when the person reaches the edge. We need to find it. It can be calculated using the conservation of angular momentum as :

I_1\omega_1=I_2\omega_2

Here, I_2=I_1+mr^2

I_1\omega_1=(I_1+mr^2)\omega_2

900\times 0.95=(900+70\times (2.9)^2)\omega_2

Solving the above equation, we get the value as :

\omega_2=0.574\ rad/s

Part B,

The initial rotational kinetic energy is given by :

k_i=\dfrac{1}{2}I_1\omega_1^2

k_i=\dfrac{1}{2}\times 900\times (0.95)^2

k_i=406.12\ rad/s

The final rotational kinetic energy is given by :

k_f=\dfrac{1}{2}(I_1+mr^2)\omega_1^2

k_f=\dfrac{1}{2}\times (900+70\times (2.9)^2)(0.574)^2

k_f=245.24\ rad/s

Hence, this is the required solution.

5 0
3 years ago
The graph above shows the position x as a function of time for the center of mass of a system of particles of total mass 6. 0 kg
kumpel [21]

The resulting change in momentum of the system will be +18.6 Ns. The momentum is conserved.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

The given data in the problem is;

m is the mass =6.0 kg

t is the time interval=2 second

From Newton's second law;

\rm \triangle P =m \triangle V \\\\ \triangle P= m(\frac{\triangle x}{\triangle t} )\\\\

From the graph;

\rm \triangle t = 2sec\\\\ \triangle x = (12-8) m

The change in the momentum is;

\rm \triangle P = m\tr(\frac{v-u}{t}) \\\\ \triangle P =9.3 \times \frac{12-8}{2} \\\\ \triangle P= +18.6 \  N.s

Hence, the resulting change in momentum of the system will be +18.6 Ns.

To learn more about the law of conservation of momentum, refer;

brainly.com/question/1113396

#SPJ1

6 0
2 years ago
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