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Natasha2012 [34]
3 years ago
14

Compute the kinetic energy of a proton (mass 1.67×10−27kg ) using both the nonrelativistic and relativistic expressions for spee

d of 6.00×107m/s.
Physics
1 answer:
Nimfa-mama [501]3 years ago
5 0

Answer:

KE = 5.01 \times 10^{-13}Joules

Explanation:

Kinetic energy is the energy possessed by a body by virtue of its motion. It is expressed as;

KE = \frac{1}{2}mv^2\\ where:

m is the mass of proton

v is the speed

Substitute:

KE = \frac{1}{2}(1.67\times 10 ^{-27})(6.0\times10^7)^2\\\\KE = \frac{1}{2}(1.67\times 10 ^{-27})(6.0\times10^{14})\\\\\\KE = \frac{1}{2}(1.67\times 6.0 \times 10 ^{-27}\times10^{14})\\\\KE =  \frac{1}{2}(1.67\times 6.0 \times 10 ^{-27+14})\\\\KE = 3 \times 1.67 \times 10^{-13}\\\\KE = 5.01 \times 10^{-13}Joules

Hence the kinetic energy of a proton is 5.01 \times 10^{-13}Joules

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Galina-37 [17]

Answer:

Tension= \frac{20g}{3}  (g=acceleration of gravity)

Explanation:

Given that,

A 5Kg and 10Kg are attached by a cable suspended over a pulley.

As 10Kg > 5Kg , the 10 kg mass accelerates down and the 5kg mass accelerates up, let it be a. Let the tension in the cable be T.

So, the equations of motion are

10g-T = 10a

T-5g=5a

Now adding them we get,

5g=15a

a=\frac{g}{3}

Substituting them back in the equation we get,

10g-10(\frac{g}{3} ) = T

T= \frac{20g}{3}

3 0
3 years ago
PLEASE PLEASE HELP ME!!!!!!!!!!
marysya [2.9K]
373 kelvin = 99.9 Celsius. Round makes it 100. 373 kelvin also equals 212 Fahrenheit so the correct answer is A.
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8 0
3 years ago
What is the ground-state electron configuration of a neutral atom of titanium?
Georgia [21]

Answer:

\bold {1s^22s^22p^63s^23p^64s^23d^2}

Explanation:

electronic configuration of an atom is the spatial arrangement of the electrons around the nucleus in the energy orbits

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Elevator mass 750kg tension on cable 8950n what is the net force action on the elevator
Hunter-Best [27]
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8 0
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A tensile test specimen has a gage length = 50 mm and its cross-sectional area = 100 mm2. The specimen yields at 48,000 N, and t
Shalnov [3]

Answer:

a) yield strength

   \sigma_y = \dfrac{F_y}{A} = =\dfrac{48000}{100} = 480 MPa

b) modulus of elasticity

strain calculation

\varepsilon_0=\dfrac{L-L_0}{L_0}=\dfrac{50.23-50}{50} = 0.0046

strain for offset yield point

\varepsilon_{new} = \varepsilon_0 -0.002

                              =0.0046-0.002 = 0.0026

now, modulus of elasticity

 E = \dfrac{\sigma_y}{\varepsilon_{new}}=\dfrac{480}{0.0026}

    = 184615.28 MPa = 184.615 GPa

c) tensile strength

 \sigma_u =\dfrac{F_{max}}{A}=\dfrac{87000}{100}=870MPa

d) percentage elongation

\% Elongation = \dfrac{L-L_0}{L_0}\times 100 = \dfrac{67.3-50}{50}\times 100 = 34.6\%

e) percentage of area reduction

\% Area\ reduction = \dfrac{A-A_f}{A}\times 100=\dfrac{100-53}{100}= 47 \%                            

7 0
3 years ago
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