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devlian [24]
3 years ago
15

A pen contains a spring with a spring constant of 257 N/m. When the tip of the pen is in its retracted position, the spring is c

ompressed 5.1 mm from its unstrained length. In order to push the tip out and lock it into its writing position, the spring must be compressed an additional 6.1 mm. How much work is done by the spring force to ready the pen for writing

Physics
1 answer:
Mandarinka [93]3 years ago
3 0

Explanation:

Below is an attachment containing the solution.

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A muon has a rest mass energy of 105.7 MeV, and it decays into an electron and a massless particle. If all the lost mass is conv
sergeinik [125]

Answer:

The electron’s velocity is 0.9999 c m/s.

Explanation:

Given that,

Rest mass energy of muon = 105.7 MeV

We know the rest mass of electron = 0.511 Mev

We need to calculate the value of γ

Using formula of energy

K_{rel}=(\gamma-1)mc^2

\dfrac{K_{rel}}{mc^2}=\gamma-1

Put the value into the formula

\gamma=\dfrac{105.7}{0.511}+1

\gamma=208

We need to calculate the electron’s velocity

Using formula of velocity

\gamma=\dfrac{1}{\sqrt{1-(\dfrac{v}{c})^2}}

\gamma^2=\dfrac{1}{1-\dfrac{v^2}{c^2}}

\gamma^2-\gamma^2\times\dfrac{v^2}{c^2}=1

v^2=\dfrac{1-\gamma^2}{-\gamma^2}\times c^2

Put the value into the formula

v^2=\dfrac{1-(208)^2}{-208^2}\times c^2

v=c\sqrt{\dfrac{1-(208)^2}{-208^2}}

v=0.9999 c\ m/s

Hence, The electron’s velocity is 0.9999 c m/s.

6 0
3 years ago
Spiders may "tune" strands of their webs to give an enhanced response at frequencies corresponding to the frequencies at which d
Bingel [31]

Answer:

0.0000167283525619 N

Explanation:

\rho = Density of silk = 1300 kg/m³

d = Diameter = 0.002 mm

r = Radius = 0.001 mm

l = Length = 16 cm

f = Frequency = 200 Hz

Mass of the string is

m=\rho V\\\Rightarrow m=\rho Al\\\Rightarrow m=1300\times \pi (0.001\times 10^{-3})^2\times 0.16\\\Rightarrow m=6.5345127195\times 10^{-10}\ kg

Frequency is given by

f=\dfrac{1}{2l}\sqrt{\dfrac{T}{m/l}}\\\Rightarrow T=4lf^2m\\\Rightarrow T=4\times 0.16\times 200^2\times 6.5345127195\times 10^{-10}\\\Rightarrow T=0.0000167283525619\ N

The tension on the string is 0.0000167283525619 N

3 0
3 years ago
The planet Krypton has a mass of 8.8 × 1023 kg and radius of 2.5 × 106 m. What is the acceleration of an object in free fall nea
djverab [1.8K]

Answer:

Acceleration, a=9.39\ m/s^2

Explanation:

Given that,

Mass of the planet Krypton, m=8.8\times 10^{23}\ kg

Radius of the planet Krypton, r=2.5\times 10^{6}\ m

Value of gravitational constant, G=6.6726\times 10^{-11}\ Nm^2/kg^2

To find,

The acceleration of an object in free fall near the surface of Krypton.

Solution,

The relation for the acceleration of the object is given by the below formula as :

a=\dfrac{Gm}{r^2}

a=\dfrac{6.6726\times 10^{-11}\times 8.8\times 10^{23}}{(2.5\times 10^{6})^2}

a=9.39\ m/s^2

So, the value of acceleration of an object in free fall near the surface of Krypton is 9.39\ m/s^2

5 0
3 years ago
Can someone help me with this question
geniusboy [140]

833.33 sec

5000m/6ms

Divide 5000 by 6 and you get your answer !!

6 0
2 years ago
Edge-nuity<br> Can anyone help?
ella [17]

Answer:

It's a site on clever I had trouble getting into it too, just login through clever.

Explanation:

5 0
3 years ago
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