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Scrat [10]
3 years ago
9

at maximum extension a bungee cord stores 2.0 X 10^6 J of energy. A 10 kg mass extends the bungee cord 1.3m. what is the maximum

extension of the bungee cord.​
Physics
1 answer:
Gnoma [55]3 years ago
4 0

Explanation:

A k E increases While P E Decreases

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a 1.5 kg ball is thrown vertically upward with an initial speed of 15 m/s. if the initial potential energy is taken as zero, fin
trapecia [35]

Answer:

a) E_{p} = 0

E_{k} = 168.7 J

E_{m} = 168.7 J

b) E_{p} = 73.6 J

E_{k} = 95.8 J

E_{m} = 169.4 J

c) E_{p} = 169.2 J

E_{k} = 0

E_{m} = 169.2 J

Explanation:

We have:

m: is the ball's mass = 1.5 kg

v₀: is the initial speed = 15 m/s

g: is the gravity acceleration = 9.81 m/s²

a) In the initial position we have:

h: is the height = 0

The potential energy is given by:

E_{p} = mgh = 0

The kinetic energy is:

E_{k} = \frac{1}{2}mv^{2} = \frac{1}{2}*1.5*(15)^{2} = 168.7 J

And the mechanical energies:

E_{m} = E_{p} + E_{k} = 0 + 168.7 J = 168.7 J

b) At 5 m above the initial position we have:

h = 5 m

The potential energy is:

E_{p} = mgh = 1.5*9.81*5 = 73.6 J

Now, to find the kinetic energy we need to calculate the speed at 5 m:

v_{f}^{2} = v_{0}^{2} - 2gh = (15)^{2} - 2*9.81*5 = 126.9

v_{f} = \sqrt{126.9} = 11.3 m/s

E_{k} = \frac{1}{2}mv^{2} = \frac{1}{2}*1.5*(11.3)^{2} = 95.8 J

And the mechanical energies:

E_{m} = E_{p} + E_{k} = 73.6 + 95.8 J = 169.4 J

c) At its maximum height:

v_{f}: is the final speed = 0

h = \frac{v_{0}^{2}}{2g} = \frac{(15)^{2}}{2*9.81} = 11.5 m

Now, the potential, kinetic and mechanical energies are:

E_{p} = mgh = 1.5*9.81*11.5 = 169.2 J

E_{k} = \frac{1}{2}mv^{2} = 0

E_{m} = 169.2 J + 0 = 169.2 J

I hope it helps you!    

7 0
2 years ago
What is robot??????????
S_A_V [24]

Answer:

a machine capable of carrying out a complex series of actions automatically, especially one programmable by a computer.

6 0
3 years ago
A 3.9 kg block is pushed along a horizontal floor by a force of magnitude 30 N at a downward angle θ = 40°. The coefficient of k
Luba_88 [7]

Answer:

The frictional force  F_{fri} = 6.446 N

The acceleration of the block a = 6.04 \frac{m}{s^{2} }

Explanation:

Mass of the block = 3.9 kg

\theta = 40°

\mu = 0.22

(a). The frictional force is given by

F_{fri} = \mu R_{N}

R_{N} = mg \cos \theta

R_{N} = 3.9 × 9.81 × \cos 40

R_{N} = 29.3 N

Therefore the frictional force

F_{fri} = 0.22 × 29.3

F_{fri} = 6.446 N

(b). Block acceleration is given by

F_{net} = F - F_{fri}

F = 30 N

F_{fri} = 6.446 N

F_{net} = 30 - 6.446

F_{net} = 23.554 N

The net force acting on the block is given by

F_{net}  = ma

23.554 = 3.9 × a

a = 6.04 \frac{m}{s^{2} }

This is the acceleration of the block.

8 0
2 years ago
A uniform disk with a 25 cm radius swings without friction about a nail through the rim. If it is released from rest from a posi
ValentinkaMS [17]

Answer:

Explanation:

During the swing , the center of mass will go down due to which disc will lose potential energy which will be converted into rotational kinetic energy

mgh = 1/2 I ω² where m is mass of the disc , h is height by which c.m goes down which will be equal to radius of disc , I is moment of inertia of disc about the nail at rim , ω is angular velocity .

mgr  = 1/2 x ( 1/2 m r²+ mr²) x ω²

gr  = 1/2 x 1/2  r² x ω² + 1/2r² x ω²

g = 1 / 4 x ω² r + 1 / 2 x ω² r

g = 3  x ω² r/ 4

ω² = 4g /3 r

= 4 x 9.8 /  3 x  .25

= 52.26

ω = 7.23  rad / s .

6 0
3 years ago
A 6.99-g bullet is moving horizontally with a velocity of +341 m/s, where the sign + indicates that it is moving to the right (s
Ratling [72]

Answer:

a). 1.218 m/s

b). R=2.8^{-3}

Explanation:

m_{bullet}=6.99g*\frac{1kg}{1000g}=6.99x10^{-3}kg

v_{bullet}=341\frac{m}{s}

Momentum of the motion the first part of the motion have a momentum that is:

P_{1}=m_{bullet}*v_{bullet}

P_{1}=6.99x10^{-3}kg*341\frac{m}{s} \\P_{1}=2.3529

The final momentum is the motion before the action so:

a).

P_{2}=m_{b1}*v_{fbullet}+(m_{b2}+m_{bullet})*v_{f}}

P_{2}=1.202 kg*0.554\frac{m}{s}+(1.523kg+6.99x10^{-3}kg)*v_{f}

P_{1}=P_{2}

2.529=0.665+(1.5299)*v_{f}\\v_{f}=\frac{1.864}{1.5299}\\v_{f}=1.218 \frac{m}{s}

b).

kinetic energy

K=\frac{1}{2}*m*(v)^{2}

Kinetic energy after

Ka=\frac{1}{2}*1.202*(0.554)^{2}+\frac{1}{2}*1.523*(1.218)^{2}\\Ka=1.142 J

Kinetic energy before

Kb=\frac{1}{2}*mb*(vf)^{2}\\Kb=\frac{1}{2}*6.99x10^{-3}kg*(341)^{2}\\Kb=406.4J

Ratio =\frac{Ka}{Kb}

R=\frac{1.14}{406.4}\\R=2.8x10^{-3}

3 0
3 years ago
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