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coldgirl [10]
3 years ago
15

What is the magnitude of the relativistic momentum of a proton with a relativistic total energy of 2.7 3 10210 j?

Physics
1 answer:
Hoochie [10]3 years ago
8 0

The equation that is applicable to this problem is expressed as <span>

p = γmu 

where p is the relativistic momentum, m is equal to the proton mass and u is the speed of proton. γ is equal to 1/√(1 - u²/c²) where c is the speed of light. 

<span>Another equation to determine p fully is to use this relativistic equation:
E = γmc² 

where E is the total kinetic energy including the gamma energy.</span></span>

<span>We are given with E = 2.7 × 10^(-10) J and from a proton’s data, mc^2 = 938.27201 MeV. Using the conversion from MeV to J which is 1 eV = 1.60218 * 10^(-19) J, we compute for γ</span>

γ=E/mc^2 =2.7 × 10^(-10) J / 938.27201x10^6 eV *(1.60218 * 10^(-19) J/eV) = 1.80

we use this number to determine u from the second equation γ = 1/√(1 - u²/c²)

we square both sides and cross multiply, resulting to <span>
γ² (1 - u²/c²) = 1 

u²/c² = (γ² -1)/γ² </span>

u = c/y sq rt of (γ² -1)

u = 0.83<span>

Substituting eventually to the first equation, we get

p = γmu 

p = 1.80 * 1.67262 * 10^(-27) kg * 0.83 * 2.99792 * 10^8 m/s 

<span>p = 7.481 * 10^(-19) kg.m/s</span></span>

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A 400-n block is dragged along a horizontal surface by an applied force as shown. the coefficient of kinetic friction is uk = 0.
gulaghasi [49]
The block moves with constant velocity: for Newton's second law, this means that the resultant of the forces acting on the block is zero, because the acceleration is zero.

We are only concerned about the horizontal direction, and there are only two forces acting along this direction: the force F pushing the block and the frictional force F_f acting against the motion. Since their resultant must be zero, we have:
F-F_f = 0
The frictional force is
F_f = \mu mg
where
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mg=400 N is the weight of the block. 

Substituting these values, we find the magnitude of the force F:
F=F_f = \mu mg=(0.4 )(400 N)=160 N
4 0
3 years ago
The magnification of a microscope is increased when_________.
azamat

Answer:

Option B

Explanation:

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M = M_o \times M_e  

Mo= Magnification of objective lens and  

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The correct answer is Option B

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8 0
3 years ago
Questions E6 a&amp; b and E7 a&amp;b?
erica [24]

Explanation:

6a) Work = force × distance

W = Fd

W = (60 N) (10 m)

W = 600 J

6b) Change in energy = work

ΔKE = 600 J

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W = ΔKE

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W = -125 J

8 0
3 years ago
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