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3 years ago

How big is the sun ☺️☺️

Physics

2 answers:

Over [174]3 years ago

6 0

Radius of a <span>695,800 km </span>and diameter of a 1,391,684 km

Klio2033 [76]3 years ago

4 0

The sun is 1.989E30 kg. <em>It's radius is 695, 800 Km.</em>

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What is the newton unit of measure?

ikadub [295]

"Newton" is a unit of force in the SI/metric system.

One newton is the force that accelerates 1 kilogram of mass at the rate of 1 m/s² .

One pound of force is about 4.448 newtons of force. (rounded)
One ounce of force is about 0.278 newton of force. (rounded)

3 0

3 years ago

A solid conducting sphere of radius 2.00 cm has a charge of 6.88 μC. A conducting spherical shell of inner radius 4.00 cm and ou

zepelin [54]

Explanation:

Given that,

Radius R= 2.00

Charge = 6.88 μC

Inner radius = 4.00 cm

Outer radius = 5.00 cm

Charge = -2.96 μC

We need to calculate the electric field

Using formula of electric field

$E=\dfrac{kq}{r^2}$

(a). For, r = 1.00 cm

Here, r<R

So, E = 0

The electric field does not exist inside the sphere.

(b). For, r = 3.00 cm

Here, r >R

The electric field is

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times6.88\times10^{-6}}{(3.00\times10^{-2})^2}$

$E=6.88\times10^{7}\ N/C$

The electric field outside the solid conducting sphere and the direction is towards sphere.

(c). For, r = 4.50 cm

Here, r lies between R₁ and R₂.

So, E = 0

The electric field does not exist inside the conducting material

(d). For, r = 7.00 cm

The electric field is

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times(-2.96\times10^{-6})}{(7.00\times10^{-2})^2}$

$E=5.43\times10^{6}\ N/C$

The electric field outside the solid conducting sphere and direction is away of solid sphere.

Hence, This is the required solution.

6 0

3 years ago

The current shape of the solar system is *

vodomira [7]

B: A spherical, with the objects in the solar system orbiting the suns in all directions

3 0

3 years ago

Approximately ____ days have passed when 20% of Cobalt (Co) remains?

brilliants [131]

The half-life of Cobalt 60 is 5.27 years. This means that after this time, only 50 % of the initial amount of Cobalt is left. The relationship between the number of nuclei left at time t and the time, for a radioactive decay, is given by
$N(t) = N_0 ( \frac{1}{2})^{ \frac{t}{t_{1/2} }$
where $N(t)$ is the number of nuclei left at time t, $N_0$ is the number of nuclei at t=0, and $t_{1/2}$ is the half-life. For cobalt, $t_{1/2}=5.27~y$ .

We can re-arrange the formula as
$\frac{N(t)}{N_0} = ( \frac{1}{2} )^{ \frac{t}{t_{1/2}} }$
and then
$t=t_{1/2} log_{1/2}( \frac{N(t)}{N_0} )$
the problem asks for the time t at which only 20% of cobalt is left, that means the time t at which
$\frac{N(t)}{N_0} = 0.20$
therefore, using this into the previous equation, we get
$t=5.27~y \cdot log_{1/2} (0.20) = 12.24~y$
so, after a time of 12.24 years, only 20% of cobalt is left.

4 0

3 years ago

What us the difference in the ways objects move at a speed of a car and an object mkvinf close to the speed of light?

Charra [1.4K]

Answer:

The difference is in who or what is observing the speed.

Explanation:

Giving that speed is relative between the objects and the reference point from which it is being observed.

It is concluded that speed alone has no direct effect on a moving object, hence it is just a determining unit for the difference in distance between two objects.

Therefore, in this case, the difference is in who or what is observing the speed.

5 0

3 years ago