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3 years ago

A student needs to use the metric system while measuring the length of a box. Which units should he use?

2 answers:

8 0

It depends on the size of the box.

-- If it's one of the boxes printed on a sheet of graph paper,
then he should report the measurement in millimeters.

-- If it's the box that his new smartphone arrived in, then
he should probably use centimeters.

-- If it's the box that the new tractor was shipped to his
grandfather's farm in, then meters is the best unit.

Elden [556K]3 years ago

6 0

Answer:

centimeters

Explanation:

I took the test :)

hope this helps! have a good day! :))

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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

One of the most important properties of materials in many applications is strength. Two of the qualitative measures of the stren

katrin2010 [14]

To solve this exercise it is necessary to take into account the concepts related to Tensile Strength and Shear Strenght.

In Materials Mechanics, generally the bodies under certain loads are subject to both Tensile and shear strenghts.

By definition we know that the tensile strength is defined as

$\sigma = \frac{F}{A}$

Where,

$\sigma =$ Tensile strength

F = Tensile Force

A = Cross-sectional Area

In the other hand we have that the shear strength is defined as

$\sigma_y = \frac{F_y}{A}$

where,

$\sigma_y =$ Shear strength

$F_y =$ Shear Force

$A_0 =$ Parallel Area

PART A) Replacing with our values in the equation of tensile strenght, then

$311*10^6 = \frac{F}{(15*10^{-6})(30*10^{-2})}$

Resolving for F,

$F= 1399.5N$

PART B) We need here to apply the shear strength equation, then

$\sigma_y = \frac{F_y}{A}$

$210*10^6 = \frac{F_y}{15*10^{-6}30*10^{-2}}$

$F_y = 945N$

In such a way that the material is more resistant to tensile strength than shear force.

6 0

4 years ago

In Figure 21-4 B, does the magnetic field -due to the current carrying wire- curve in a clockwise or counter-clockwise direction

weeeeeb [17]

The magnetic field between the poles of the horseshoe magnet in fig 21.4 B curves in the <u>counter clockwise direction.</u>

The magnetic field lines due to a straight current carrying conductor is given by the right hand thumb rule. This rule is stated as follows: Imagine holding the current carrying conductor in your right hand with the thumb pointing in the direction of current, then the direction in which your fingers curve around the wire gives the direction of the magnetic field lines. The direction of the magnetic field lines at a point gives the direction of magnetic field at that point.

On applying this rule, it can be seen that the magnetic field due to the conductor shown in Fig 21.4 B curves in the counter clockwise direction.

8 0

3 years ago

Read 2 more answers

What is the focal length of 2.30 D reading glasses found on the rack in a drugstore?

irakobra [83]

Answer:

Focal length, f = 0.43 meters

Explanation:

It is given that,

Power of the reading glasses, P = 2.3 D

We need to calculate the focal length of the reading glasses. The relationship between the power and the focal length inverse i.e.