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3 years ago

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nata0808 [166]3 years ago

5 0

Answer:

Its sure 2nd gare beaciuse

Explanation:

of tobi was obito was tobi his som came and start doing enginner but he faild so i gave him answer he repiled oh baby baby oh

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A proton moves along the x-axis in the laboratory with velocity uy = 0.6c. An observer moves with a velocity of v=0.8c along the

Dima020 [189]

Explanation:

Given that,

Velocity of the proton in lab frame $u_{x} = 0.6c$

Velocity of the observer v= 0.8c

We need to calculate the velocity of the proton with respect to the observer

Using formula of velocity

$u'=\dfrac{v-u_{x}}{1-\dfrac{u_{x}v}{c^2}}$

$u'=\dfrac{0.8c-0.6c}{1-\dfrac{0.6c\times0.8c}{c^2}}$

$u'=c(\dfrac{0.8-0.6}{1-(0.8\times0.6)})$

$u'=0.385c$

(a). We need to calculate the total energy of the proton in the lab frame

Using formula of kinetic energy

$K.E=m_{0}c^2(\dfrac{1}{\sqrt{1-\dfrac{u_{x}^2}{c^2}}}-1)$

Where, Proton mass energy = m₀c²

Put the value into the formula

$K.E=938.28\times(\dfrac{1}{\sqrt{1-\dfrac{(0.6c)^2}{c^2}}}-1)$

$K.E=938.28\times(\dfrac{1}{\sqrt{1-(0.6)^2}}-1)$

$K.E=234.57\ MeV$

(b). We need to calculate the kinetic energy of the proton in the observer

Using formula of kinetic energy

$K.E=m_{0}c^2(\dfrac{1}{\sqrt{1-\dfrac{u'^2}{c^2}}}-1)$

$K.E=938.28\times(\dfrac{1}{\sqrt{1-\dfrac{(0.385)^2}{c^2}}}-1)$

$K.E=938.28\times(\dfrac{1}{\sqrt{1-(0.385)^2}}-1)$

$K.E=78.366\ MeV$

(c). We need to calculate the momentum of the proton with respect to observer

Using formula of momentum

$P_{obs}=\dfrac{m_{0}u'^2}{\sqrt{1-\dfrac{u'^2}{c^2}}}$

We know that,

Proton mass energy = m₀c²

$m_{0}=\dfrac{938.28}{c^2}$

$P_{obs}=\dfrac{\dfrac{938.28}{c^2}\times(0.385c)^2}{\sqrt{1-\dfrac{(0.385c)^2}{c^2}}}$

$P_{obs}=\dfrac{938.28\times(0.385)^2}{\sqrt{1-0.385^2}}$

$P_{obs}=150.69\ MeV/c$

Hence, This is required solution.

3 0

3 years ago

A bicycle of mass m requires 50 J of work to move from rest to a final speed v. If the same amount of work is performed during t

insens350 [35]

Answer:

The final speed of the second bicycle is (v·√2)/2

Explanation:

The mass of the given bicycle = m

The amount of work required to move the bicycle from rest to speed v = 50 J

The final speed of the first bicycle = v

The mass of the second bicycle = 2m

Therefore, from conservation of energy, we have;

Work required by the first bicycle = Kinetic energy gained by the bicycle

The kinetic energy = 1/2·m·v²

∴ Energy required by the first bicycle = 50 J = 1/2·m·v²

Given that the same amount of work is performed on the second bicycle, we have;

Work performed on the second bicycle = 50 J = kinetic energy of second bicycle = 1/2·(2·m)·v₂²

Also, given that 50 J = 1/2·m·v², we have;

Work performed on the second bicycle = 50 J = 1/2·m·v²= 1/2·(2·m)·v₂²

1/2·m·v²= 1/2·(2·m)·v₂²

m·v² = 2·m·v₂²

v² = 2·v₂²

v₂ = √(v²/2) = v/√2 = (v·√2)/2

v₂ = (v·√2)/2

The final speed of the second bicycle = v₂ = (v·√2)/2.

3 0

3 years ago

a roller coaster has a mass of 275kg. it sits at the top of a hill with a height 85m if it drops from this hill, how fast is it

prohojiy [21]

Hope this helps you.

3 0

4 years ago

Marisa does 3.2J of work to lower the window shade in

Pani-rosa [81]

Work is considered as the Force performed on a body to move it a certain distance, that is

W = Fd

Here

W = Work

F = Force

d = Distance

In this case we have the values of work and distance, therefore clearing for the Force we would have to

$F = \frac{W}{d}$

Replacing,

$F = \frac{3.2J}{8m}$

$F = 0.4N$

Therefore the force that Marissa must exert on the windows shade is 0.4N

5 0

3 years ago

Some hydrogen gas is enclosed within a chamber being held at 200^\ { C} with a volume of 0.025 \rm m^3. The chamber is fitted wi

vlada-n [284]

Answer:

The final volume is $0.039 m^3$

Explanation:

<u>Data:</u>

Initial temperature: $T1=200C$

Final temperature: $T2=200C$

Initial pressure: $P1=1.50 \times10^6 Pa$

Final pressure: $P2=0.950 \times10^6 Pa$

Initial volume: $V1=0.025m^{3}$

Final volume: $V2=?$

Assuming hydrogen gas as a perfect gas it satisfies the perfect gas equation:

$\frac{PV}{T}=nR$ (1)

With P the pressure, V the volume, T the temperature, R the perfect gas constant and n the number of moles. If no gas escapes the number of moles of the gas remain constant so the right side of equation (1) is a constant, that allows to equate:

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$

Subscript 2 referring to final state and 1 to initial state.

solving for V2:

$V_{2}=\frac{P_{1}V_{1}T_{2}}{T_{1}P_{2}}=\frac{(1.50 \times10^6)(0.025)(200)}{(200)(0.950 \times10^6)}$

$V_{2}=0.039 m^3$

3 0

4 years ago