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madreJ [45]
3 years ago
9

An athlete stretches a spring an extra 40.0 cm beyond its initial length. How much energy has he transferred to the spring, if t

he spring constant is 52.9 N/cm?
a) 4230 kJ
b) 4230 J
c) 423 kJ
d) 423 J
Physics
1 answer:
notsponge [240]3 years ago
4 0

Answer:

<h2>b) 4230 J </h2>

Explanation:

Step one:

given data

extension= 40cm

Spring constant K= 52.9N/cm

Step two:

Required

the Kinetic Energy KE

the expression to find the kinetic energy is

KE= 1/2ke^2

substituting our data we have

KE= 1/2*52.9*40^2

KE=0.5*52.9*1600

KE= 42320Joules

<u>The answer is b) 4230 J </u>

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A wave pulse travels down a slinky. The mass of the slinky is m = 0.87 kg and is initially stretched to a length L = 6.8 m. The
Ber [7]

Answer:

1. v=14.2259\ m.s^{-1}

2. F_T=25.8924\ N

3. \lambda=29.6373\ m

Explanation:

Given:

  • mass of slinky, m=0.87\ kg
  • length of slinky, L=6.8\ m
  • amplitude of wave pulse, A=0.23\ m
  • time taken by the wave pulse to travel down the length, t=0.478\ s
  • frequency of wave pulse, f=0.48\ Hz=0.48\ s^{-1}

1.

\rm Speed\ of\ wave\ pulse=Length\ of\ slinky\div time\ taken\ by\ the\ wave\ to\ travel

v=\frac{6.8}{0.478}

v=14.2259\ m.s^{-1}

2.

<em>Now, we find the linear mass density of the slinky.</em>

\mu=\frac{m}{L}

\mu=\frac{0.87}{6.8}\ kg.m^{-1}

We have the relation involving the tension force as:

v=\sqrt{\frac{F_T}{\mu} }

14.2259=\sqrt{\frac{F_T}{\frac{0.87}{6.8}} }

202.3774=F_T\times \frac{6.8}{0.87}

F_T=25.8924\ N

3.

We have the relation for wavelength as:

\lambda=\frac{v}{f}

\lambda=\frac{14.2259}{0.48}

\lambda=29.6373\ m

8 0
3 years ago
A meter stick is suspended vertically at a pivot point 22 cm from the top end. It is rotated on the pivot until it is horizontal
suter [353]

Answer:

5.82812 rad/s

Explanation:

L = Length of meter stick = 1 m = 100 cm

m_c = The center of mass of the stick = \frac{L}{2}-0.22=0.5-0.22=0.28\ m

\omega = Angular velocity

Moment of inertia of the system is given by

I=I_c+mr^2\\\Rightarrow I=\frac{mL^2}{12}+mr^2\\\Rightarrow I=\frac{m1^2}{12}+m0.28^2\\\Rightarrow I=m(\frac{1}{12}+0.0784)

As the energy in the system is conserved

mgh=I\frac{\omega^2}{2}\\\Rightarrow mgh=m(\frac{1}{12}+0.0784)\frac{\omega^2}{2}\\\Rightarrow gh=(\frac{1}{12}+0.0784)\frac{\omega^2}{2}\\\Rightarrow \omega=\sqrt{\frac{2gh}{\frac{1}{12}+0.0784}}\\\Rightarrow \omega=\sqrt{\frac{2\times 9.81\times 0.28}{\frac{1}{12}+0.0784}}\\\Rightarrow \omega=5.82812\ rad/s

The maximum angular velocity is 5.82812 rad/s

4 0
3 years ago
Where were scientific advancements being made during the dark ages
Leto [7]

Answer:

The period saw major technological advances, including the adoption of gunpowder, the invention of vertical windmills, spectacles, mechanical clocks, and greatly improved water mills, building techniques (Gothic architecture, medieval castles), and agriculture in general (three-field crop rotation).

Explanation:

Hope this helps :)

7 0
3 years ago
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Solve: A car travels 2 km North , 10 km East, then 3 km West. pythagorean theorem
Evgen [1.6K]

________b_____ 7 km east

|

| 2km north.

|a

|

°

pythagorean theorem : ✓a² + b² = c²

c² = a² + b² = 4 + 49 = 53

c = ✓53 km

displacement = c = ✓53 km

distance = 10 + 3 + 2 = 15 km

5 0
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A crew on a spacecraft watches a movie that is two hours long. The spacecraft is moving at high speed through space. An Earth-ba
laiz [17]

Answer:

The duration of the movie is longer than 2 hrs.

Explanation:

Given:

The duration of the movie observed by the crew on the spacecraft is 2 hrs.

According to time-dilation formula:

t = \dfrac{t_{0}}{\sqrt{1 - \dfrac{v^{2}}{c^{2}}}}

Here, t is the required time, t_{0} is the original time, v is the velocity of the spacecraft and c is the velocity of light.

Since v, so t_{0} > t.

So the time required will be large.

4 0
3 years ago
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