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3 years ago

What is a star? What is our biggest and closest star? (Btw i wasn't sure if this was Physics.)

Physics

2 answers:

sweet-ann [11.9K]3 years ago

8 0

Answer:

A star is an astronomical object consisting of a luminous spheroid of plasma held together by its own gravity. The nearest star to Earth is the Sun. Many other stars are visible to the naked eye at night, but due to their immense distance from Earth they appear as fixed points of light in the sky

Explanation:

Georgia [21]3 years ago

3 0

Answer: Proxima Centauri, the closest star to our own, is still 40,208,000,000,000 km away. (Or about 268,770 AU.)

You might be interested in

Atoms are the smallest portions of an element that retain the original characteristics of the element. True or false

kozerog [31]

Answer:

True

Explanation:

Going even smaller than atoms would get you to subatomic particles such as quarks. From there, it is impossible to distinguish elements. So, yes, atoms are the smallest portions of an element that retains the original characteristic of the element.

4 0

3 years ago

Two charges are placed on the x axis. One of the charges (q1 = +7.7 µC) is at x1 = +3.1 cm and the other (q2 = -19 µC) is at x2

Alinara [238K]

Answer:

a) $E=50.53\times 10^{6}\ N/C$

The direction will be negative direction.

b) $E=268.22\times 10^{6}\ N/C$

The direction will be positive direction.

Explanation:

Given that

q1 = +7.7 µC is at x1 = +3.1 cm

q2 = -19 µC is at x2 = +8.9 cm

We know that electric filed due to a charge given as

$E=K\dfrac{q}{r^2}$

$E_1=K\dfrac{q_1}{r_1^2}$

$E_2=K\dfrac{q_2}{r_2^2}$

Now by putting the va;ues

$E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.031^2}\ N/C$

$E_1=72.11\times 10^{6}\ N/C$

$E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.089^2}\ N/C$

$E_2=21.58\times 10^{6}\ N/C$

The net electric field

$E=E_1-E_2$

$E=50.53\times 10^{6}\ N/C$

The direction will be negative direction.

As we know that electric filed line emerge from positive charge and concentrated at negative charge.

Now

distance for charge 1 will become =5.5 - 3.1 = 2.4 cm

distance for charge 2 will become =8.9-5.5 = 3.4 cm

$E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.024^2}\ N/C$

$E_1=120.3\times 10^{6}\ N/C$

$E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.034^2}\ N/C$

$E_2=147.92\times 10^{6}\ N/C$

The net electric field

$E=E_1+E_2$

$E=268.22\times 10^{6}\ N/C$

The direction will be positive direction.

7 0

4 years ago

Two masses, each weighing 1.0 × 103 kilograms and moving with the same speed of 12.5 meters/second, are approaching each other.

juin [17]

A perfectly elastic<span> collision is defined as one in which there is no loss of </span>kinetic energy<span> in the collision. Therefore, we just add the kinetic energies of each system. We calculate as follows:

KE = 0.5(</span>1.0 × 10^3)(12.5 )^2 + 0.5(1.0 × 10^3)(12.5 )^2
KE = 156250 J = 1.6 x 10^5 J -------> OPTION A

5 0

3 years ago

A system absorbs 194 kj of heat and the surroundings do 120 kj of work on the system. internal eneergy change

notsponge [240]

We can solve the problem by using the first law of thermodynamics, which states that:
$\Delta U = Q-W$
where
$\Delta U$ is the change in internal energy of the system
Q is the heat absorbed by the system
W is the work done by the system

In our problem, the heat absorbed by the system is Q=+194 kJ, while the work done is W=-120 kJ, where the negative sign means the work is done by the surroundings on the system. Therefore, the variation of internal energy is
$\Delta U= Q-W=+194 kJ - (-120 kJ)=+314 kJ$

6 0

3 years ago

Electric charge is distributed over the disk x2 + y2 ≤ 4 so that the charge density at (x, y) is rho(x, y) = 4x + 4y + 4x2 + 4y2

maw [93]

Answer:

Q=185.84C

Explanation:

We have to take into account the integral

$Q=\int \rho dV$

In this case we have a superficial density in coordinate system.

Hence, we have for R: x2 + y2 ≤ 4

$Q=\int_{-2}^2\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\rho dydx$

but, for symmetry:

$Q=4\int_0^2\int_0^{\sqrt{4-x^2}}\rho dydx\\\\Q=4\int_0^2\int_0^{\sqrt{4-x^2}}(4x+4y+4x^2+4y^2) dydx\\\\Q=4\int_0^{2}[4x\sqrt{4-x^2}+2(4-x^2)+4x^2\sqrt{4-x^2}+\frac{4}{3}(4-x^2)^{3/2}]dx\\\\Q=4[46.46]=185.84C$

HOPE THIS HELPS!!

8 0

3 years ago