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Natasha2012 [34]
3 years ago
14

A boy standing at one end of a floating raft that is stationary relative to the shore walks to the opposite end of the raft, awa

y from the shore. As a consequence, the raft does which of the following?
a. remains stationary
b. moves away from the shore
c. moves toward the shore.
Physics
1 answer:
lapo4ka [179]3 years ago
6 0

Answer:

The raft moves towards the shore

Explanation:

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A 0.153 kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.700 m/s. It has a head-on col
DedPeter [7]

Answer:

3.1216 m/s.

Explanation:

Given:

M1 = 0.153 kg

v1 = 0.7 m/s

M2 = 0.308 kg

v2 = -2.16 m/s

M1v1 + M2v2 = M1V1 + M2V2

0.153 × 0.7 + 0.308 × -2.16 = 0.153 × V1 + 0.308 × V2

= 0.1071 - 0.66528 = 0.153 × V1 + 0.308 × V2

0.153V1 + 0.308V2 = -0.55818. i

For the velocities,

v1 - v2 = -(V1 - V2)

0.7 - (-2.16) = -(V1 - V2)

-(V1 - V2) = 2.86

V2 - V1 = 2.86. ii

Solving equation i and ii simultaneously,

V1 = 3.1216 m/s

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3 years ago
How much elastic potential energy is stored in a bungee cord with a spring constant of 10.0 N/m when the cord is stretched 2.00
Aleonysh [2.5K]

Answer:

<em> The elastic potential energy stored in the bungee cord = 20 J</em>

Explanation:

potential energy: This is the energy possessed by a body due to its position. The S.I unit of energy is Joules. The mathematical expression for elastic potential energy is given below

E = 1/2ke²................ Equation 1

Where E = elastic potential energy of the spring, k = force constant of the spring, e = extension

<em>Given: K = 10 N/m, e = 2.00 m</em>

<em>Substituting these values into Equation 1</em>

<em>E = 1/2(10)(2)²</em>

<em>E = 5×4</em>

<em>E = 20 Joules.</em>

<em>Therefore the elastic potential energy stored in the bungee cord = 20 J</em>

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