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4 years ago

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Hi my name is Lexi I'm looking for some friends on here! I I go to Baldwin arts and academics. so can you PLZ be my friend on he

re?

2 answers:

katrin [286]4 years ago

8 0

Answer:

You should use this for work related questions :/

Anit [1.1K]4 years ago

8 0

Answer:

yes sounds good friends indeed

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a car moving with 72 km/hr towards east and another car with 90km/hr towards west find relative of A w.r to B

hodyreva [135]

Answer:

please tell me what does w.r mean ? then I will try to give answer

7 0

3 years ago

At a given moment in time, instantaneous speed can be thought of as the magnitude of instantaneous

kaheart [24]

At a given moment in time, the instantaneous speed can be thought of as the magnitude of instantaneous velocity.

Instantaneous speed is the magnitude of the instantaneous velocity, the instantaneous velocity has direction but the instantaneous speed does not have any direction. Hence, the instantaneous speed has the same value as that of the magnitude of the instantaneous velocity. It doesn't have any direction.

5 0

3 years ago

Read 2 more answers

A population has a birthrate of 10/1000 , a death rate of 9/1000, an immigration rate of 3/1000, and an emigration rate of 7/100

Charra [1.4K]

Birth rate 10/1000 = +population

Death rate 9/1000 = -population

Immigration rate 3/1000 = +population

Emigration rate 7/1000 = -population

10/1000 - 9/1000 + 3/1000 - 7/1000= -3/1000

Growth rate = (-3/1000)(100%)= -0.3%

3 0

3 years ago

In a cyclotron, the orbital radius of protons with energy 300 keV is 16.0 cm . You are redesigning the cyclotron to be used inst

dem82 [27]

Answer:

$r_\alpha=16cm$

Explanation:

The radius of the circumference described by a particle in a cyclotron is given by:

$r=\frac{mv}{qB}(1)$

m is the particle's mass, v is the speed of the particle, q is the particle's charge and B is the magnitude of the magnetic field.

Kinetic energy is defined as:

$K=\frac{mv^2}{2}=\frac{m^2v^2}{2m}\\$

Solving this for mv:

$mv=\sqrt{2mK}(2)$

Replacing (2) in (1):

$r=\frac{\sqrt{2mK}}{qB}$

For protons, we have:

$r_p=\frac{\sqrt{2m_pK}}{eB}(3)$

For alpha particles, we have:

$r_\alpha=\frac{\sqrt{2m_\alpha K}}{(2e)B}(4)$

Dividing (4) in (3):

$\frac{r_\alpha}{r_p}=\frac{\frac{\sqrt{2m_\alpha K}}{(2e)B}}{\frac{\sqrt{2m_p K}}{(e)B}}\\r_\alpha=\frac{r_p}{2}\sqrt{\frac{m_\alpha}{m_p}}\\r_\alpha=\frac{16cm}{2}\sqrt{\frac{6.64*10^{-27}kg}{1.67*10^{-27}kg}}\\r_\alpha=\frac{16cm}{2}(\sqrt{3.98})\\\\r_\alpha=\frac{16cm}{2}(2)\\r_\alpha=16cm$

4 0

4 years ago

A gamma ray photon has an energy of 0.91 GeV. (1 GeV = 109 eV.) What is the wavelength of the gamma ray in fm? (1 fm = 10-15 m)?

anastassius [24]

Answer:

$\lambda=1.37 fm$

Explanation:

The Planck Eistein relation, states that the energy of a photon is proportional to its frequency:

$E=h\nu(1)$

h is the Plank constant.The frequency of a photon is defined as the speed of light over its wavelength:

$\nu=\frac{c}{\lambda}(2)$

Replacing (2) in (1):

$E=\frac{hc}{\lambda}\\\lambda=\frac{hc}{E}\\\lambda=\frac{(4.14*10^{-15}eV\cdot s)(3*10^8\frac{m}{s})}{0.91*10^{9}eV}\\\\\lambda=(1.37*10^{-15}m)*\frac{1fm}{10^{-15}m}\\\\\lambda=1.37 fm$

6 0

3 years ago