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2 years ago

HELP PLEEAAAASSSEEEEEEE What is the definition of net force?

Physics

1 answer:

aleksley [76]2 years ago

8 0

Answer:

the sum of all force being applied to an object.

Explanation:

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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

2 years ago

Lightning flashes one mile (1609 m) away from you. How much time does it take the light to travel that distance?

hichkok12 [17]

Answer:

$t=5.36\times 10^{-6}\ s$

Explanation:

Given that,

Lightning flashes one mile (1609 m) away from you.

We need to find the time it take the light to travel that distance. Let the time be t. We know that,

speed = distance/time

$t=\dfrac{d}{v}\\\\t=\dfrac{1609}{3\cdot10^{8}}\\\\=5.36\times 10^{-6}\ s$

So, the required time is $5.36\times 10^{-6}\ s$

7 0

2 years ago

A beam of light strikes a mirror reflects off at the angle in which it hit the mirror. which term does the diagram below illustr

zavuch27 [327]

Answer:

B.) reflection

Explanation:

3 0

3 years ago

If all this energy is added to 50 kg of water (the amount of water in a 165-lb person) at 37 ∘C, what is the final state of the

Reika [66]

Answer:

Vapors

Explanation:

We take into account that all the energy from the lightning has been transformed into steam.

$\Delta U = Q - W\\Q = mC \Delta T\\Q = mL$

We calculate the amount of energy required by water to convert into steam.

$Q_{water} = 50 \times \times 4180 \times (100-37)\\= 1.132 \times 10^7 \ J$

$Q_{change\ to\ steam} = 50 \times 2.256 \times 106 \\= 1.128 \times 10^8 \ J$

$Q_{total} = 1.132 \times 10^7 + 1.128 \times 10^8\\= 1.126 \times 10^8 \ J$

From the lightning we received $10^{10} \ J$ of energy, out of which $1.126 \times 10^8$ has been used to convert the water into steam.

Energy left = $10^{10} - 1.126 \times 10^8 = 9.88 \times 10^9 \ J$

We use this energy to convert steam into vapors.

$Q = \Delta E$

$Q = \Delta E = mc (T_{f} - T{i})\\T_{f} = \frac {\Delta E}{mc} + T_{i}\\ \\T_{f} = 100 + \frac{9.88 \times 10^{10}}{50 \times 1970}\\T_{f} = 100 + 10^8\\T_{f} = 10 ^{ \ 8} \ {^ \circ } C$

With this temperature, we can easily interpret that the vapors will be dissociated in hydrogen and oxygen particles.

5 0

2 years ago

If this resolving power is diffraction-limited, to what effective diameter of your eye's optical system does this correspond? us

ZanzabumX [31]

"This resolving power" was obviously stated earlier, somewhere before the point where you started copying. With no resolving power specified, there's actually no question, and so no answer.

5 0

2 years ago