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Helen [10]
3 years ago
5

What would happen if mass is continually added to a 1.4 solar mass neutron star?

Physics
1 answer:
andrey2020 [161]3 years ago
8 0
Eventually wouldn't it collapse in on itself and create black hole.
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An ion with charge of Q = +3.2 x 10-19 C is in a region where a uniform electric field of magnitude E = 5.0 X 105 V/m is perpend
san4es73 [151]

Answer:

option C

Explanation:

given,

Q = +3.2 x 10⁻¹⁹ C

E = 5.0 X 10⁵ V/m

B = 0.80 T

ion's acceleration is zero  

when acceleration is zero the magnitude of both the forces becomes equal.

q E = q V B

v = \dfrac{E}{B}

v= \dfrac{5 \times 10^5}{0.80}

v = 6.25 × 10⁵ m/s ≈ 6.3 × 10⁵ m/s

hence, the correct answer is option C

8 0
2 years ago
Green sea turtles eat sea grass that grows on the sea floor. Seagrass, a foundation species, provides an important breeding grou
d1i1m1o1n [39]
The answer is D. <span> There would be a decrease in the population of marine organisms. </span>
3 0
2 years ago
Read 2 more answers
What is the unit of frequency
Phantasy [73]

Answer:

hertz (Hz)

The number of periods or cycles per second is called frequency. The SI unit for frequency is the hertz (Hz).

6 0
2 years ago
Nitroglycerin flows through a pipe of diameter 3.0 cm at 2.0 m/s. If the diameter narrows to 0.5 cm, what will the velocity be?
Korolek [52]

Answer:

72 m/s

Explanation:

D1 = 3 cm, v1 = 2 m/s

D2 = 0.5 cm,

Let the velocity at narrow end be v2.

By use of equation of continuity

A1 v1 = A2 v2

3.14 × 3 × 3 × 2 = 3.14 × 0.5 ×0.5 × v2

v2 = 72 m/s

8 0
3 years ago
A 460 W heating unit is designed to operate with an applied potential difference of 120 V (a) By what percentage will its heat o
dybincka [34]

Answer:

(a) = -0.16%

(b) = smaller

Explanation:

given

power = 460 W

potential difference = 120 V

(a) what percentage will   its heat output drop if the applied potential difference drops to 110 V ?

we know p = \frac{v^2}{R} .....................(i)

we need to find change in power

\Delta P = \frac{\Delta (V^2)}{R}  

\Delta P = \frac{2 V \Delta V}{R}..............(ii)

from equations we get

\frac{\Delta P}{P} =  \frac{2 \Delta V}{V}

\frac{\Delta P}{P} = 2 \frac{110 -120}{120}

\frac{\Delta P}{P} =  -2(\frac{10}{120})

\frac{\Delta P}{P} = - 0.16 %

(b)

if we increase temperature resistance will increase and decrease with decrease in temperature and we know power is inversely proportional to resistance so if potential decrease and it would cause drop in power

and due to this increment of heating power resistance will decrease so actual drop in the power would  be smaller

7 0
3 years ago
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