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Makovka662 [10]
3 years ago
8

You compress a spring by a distance of 0.2 m. The spring has a spring constant of 37 N/m. When you release the spring, it snaps

back. What is the kinetic energy of the spring as it reaches its natural length?
Physics
2 answers:
Serggg [28]3 years ago
7 0
At that point it is no longer trying to uncompress nor is it trying to stretch.  This is the same thing as a pendulum at the bottom of its swing, no longer falling but not yet rising against gravity.  Thus the kinetic energy there is the same as the potential energy when it is compressed.  The energy of compression is
\frac{1}{2}k x^{2}
This gives E=0.5(37)(0.2)²=0.74J
This is the same as the kinetic energy when it is at natural length
Angelina_Jolie [31]3 years ago
7 0

Answer:

0.74 J

Explanation:

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8 0
2 years ago
75kg man climbs a mountain 1000m high in 3hrs and uses 4100 joulse/min. (a) calculate the power consumption in watt, (b) what is
mr Goodwill [35]

Answer:

Explanation:

a) Power consumption is 4100 J/min / 60 s/min = 68.3 W(atts)

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3 0
3 years ago
A nonconducting sphere of diameter 10.0 cm carries charge distributed uniformly inside with charge density of +5.50 µC/m3 . A pr
VLD [36.1K]

Answer:

t = 2.58*10^-6 s

Explanation:

For a nonconducting sphere you have that the value of the electric field, depends of the region:

rR:\\\\E=k\frac{Q}{r^2}

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

R: radius of the sphere = 10.0/2 = 5.0cm=0.005m

In this case you can assume that the proton is in the region for r > R. Furthermore you use the secon Newton law in order to find the acceleration of the proton produced by the force:

F=m_pa\\\\qE=m_pa\\\\k\frac{qQ}{r^2}=m_pa\\\\a=k\frac{qQ}{m_pr^2}

Due to the proton is just outside the surface you can use r=R and calculate the acceleration. Also, you take into account the charge density of the sphere in order to compute the total charge:

Q=\rho V=(5.5*10^{-6}C/m^3)(\frac{4}{3}\pi(0.05m)^3)=2.87*10^{-9}C\\\\a=(8.98*10^9Nm^2/C^2)\frac{(1.6*10^{-19}C)(2.87*10^{-9}C)}{(1.67*10^{-27}kg)(0.05m)^2}=9.87*10^{11}\frac{m}{s^2}

with this values of a you can use the following formula:

a=\frac{v-v_o}{t}\\\\t=\frac{v-v_o}{a}=\frac{2550*10^3m/s-0m/s}{9.87*10^{11}m/s^2}=2.58*10^{-6}s

hence, the time that the proton takes to reach a speed of 2550km is 2.58*10^-6 s

3 0
2 years ago
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zaharov [31]

Answer:

c. is more than that of the fluid.

Explanation:

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heat= m s \Delta T&#10;

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s= specific heat

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let s1= specific heat of solid and s2= specific heat of liquid

then

Heat lost by solid= 20(s_1)(70-30)=800s_1&#10;

Heat gained by fluid=100(s_2)(30-20)=1000s_2&#10;

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1000 S_2=800 S_1

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so the specific heat of solid is more than that of the fluid.

8 0
3 years ago
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