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TEA [102]
3 years ago
6

An electron with a speed of 0.965 c is emitted by a supernova, where c is the speed of light. What is the magnitude of the momen

tum of this electron?
Physics
2 answers:
guajiro [1.7K]3 years ago
4 0

Answer:

6.91 x 10^-23 kg m/s

Explanation:

velocity, v = 0.965 c

mass of electron, m = 9.1 x 10^-31 kg

The formula for the momentum is given by

p=\frac{mv}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

p=\frac{9.1\times 10^{-31}\times 0.965\times 3\times 10^{8}}{\sqrt{1-0.965^{2}}}

p = 6.91 x 10^-23 kg m/s

Thus, the momentum of electron is 6.91 x 10^-23 kg m/s.

Jlenok [28]3 years ago
3 0

Answer:

1.0047\times 10^{-21}kg.m/s

Explanation:

Relativistic momentum is given by:

p=\gamma m u

where, \gamma=\frac{1}{\sqrt{1-\frac{u^2}{c^2}}}

Speed of the electron, u = 0.965 c (given)

Mass of the electron, m = 9.1 ×10⁻³¹ kg

\gamma=\frac{1}{\sqrt{1-\frac{(0.965c)^2}{c^2}}}=\frac{1}{\sqrt{0.2622}}=3.81

p=3.81\times 9.1\times 10^{-31}\times 0.965\times 3\times 10^8=1.0047\times 10^{-21}kg.m/s

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medium

Explanation:

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6 0
3 years ago
A block-and-tackle pulley hoist is suspended in a warehouse by ropes of lengths 2 m and 3 m. the hoist weighs 430 n. the ropes,
ICE Princess25 [194]
Refer to the diagram shown below.

For horizontal equilibrium,
T₃ cos38 = T₂ cos 50
0.788 T₃ = 0.6428 T₂
T₃ = 0.8157 T₂                (1)

For vertical equilibrium,
T₂ sin 50 + T₃ sin 38 = 430
0.766 T₂ + 0.6157 T₃ = 430
1.2441 T₂ + T₃ = 698.392        (2)

Substitute (1) into (2).
(1.2441 + 0.8157) T₂ = 698.392
T₂ = 339.058 N
T₃ = 0.8157(399.058) = 276.571 N

Answer:
T₂ = 339.06 N
T₃ = 276.57 N

7 0
3 years ago
Which statement below best reflects the energy of the rock shown in the diagram?
Natali [406]

Answer:

32

Explanation:

32

5 0
2 years ago
Jupiter's satellite Europa orbits Jupiter with a period of 3.55 d at an orbital radius of 6.71 108 m (assume the orbit to be cir
yawa3891 [41]

Answer:

(a)

M = 1.898 x 10^27 kg

(b)

v = 13.74 km/s

(c) E = 0.28 N/kg

Explanation:

Time period, T = 3.55 days = 3.55 x 24 x 3600 second = 306720 s

Radius, r = 6.71 x 10^8 m

G = 6.67 x 10^-11 Nm^2/kg^2

(a) T=2\pi \sqrt{\frac{r^{3}}{GM}}

M=\frac{4\pi ^{2}r^{3}}{GT^{2}}

M=\frac{4\times3.14^{2}\times 6.71^{3}\times 10^{24}}{6.67\times 10^{-11}\times 306720^{2}}

M = 1.898 x 10^27 kg

(b) Let v be the orbital velocity

v=\frac{2\pi r}{T}

v=\frac{2\times 3.14\times 6.71\times 10^{8}}{306720}

v = 13739.5 m/s

v = 13.74 km/s

(b) The gravitational field E is given by

E = \frac{GM}{r^{2}}

E = \frac{6.67\times10^{-11}\times 1.898\times 10^{27}}{6.71^{2}\times 10^{16}}

E = 0.28 N/kg

6 0
2 years ago
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