Which type of field is drawn with arrows pointing outward?

Calculate the Schwarzschild radius (in kilometers) for each of the following.1.) A 1 ×108MSun black hole in the center of a quas

Westkost [7]

Answer:

(I). The Schwarzschild radius is $2.94\times10^{8}\ km$

(II). The Schwarzschild radius is 17.7 km.

(III). The Schwarzschild radius is $1.1\times10^{-7}\ km$

(IV). The Schwarzschild radius is $7.4\times10^{-29}\ km$

Explanation:

Given that,

Mass of black hole $m= 1\times10^{8} M_{sun}$

(I). We need to calculate the Schwarzschild radius

Using formula of radius

$R_{g}=\dfrac{2MG}{c^2}$

Where, G = gravitational constant

M = mass

c = speed of light

Put the value into the formula

$R_{g}=\dfrac{2\times6.67\times10^{-11}\times1\times10^{8}\times1.989\times10^{30}}{(3\times10^{8})^2}$

$R_{g}=2.94\times10^{8}\ km$

(II). Mass of block hole $m= 6 M_{sun}$

We need to calculate the Schwarzschild radius

Using formula of radius

$R_{g}=\dfrac{2MG}{c^2}$

Put the value into the formula

$R_{g}=\dfrac{2\times6.67\times10^{-11}\times6\times1.989\times10^{30}}{(3\times10^{8})^2}$

$R_{g}=17.7\ km$

(III). Mass of block hole m= mass of moon

We need to calculate the Schwarzschild radius

Using formula of radius

$R_{g}=\dfrac{2MG}{c^2}$

Put the value into the formula

$R_{g}=\dfrac{2\times6.67\times10^{-11}\times7.35\times10^{22}}{(3\times10^{8})^2}$

$R_{g}=1.1\times10^{-7}\ km$

(IV). Mass = 50 kg

We need to calculate the Schwarzschild radius

Using formula of radius

$R_{g}=\dfrac{2MG}{c^2}$

Put the value into the formula

$R_{g}=\dfrac{2\times6.67\times10^{-11}\times50}{(3\times10^{8})^2}$

$R_{g}=7.4\times10^{-29}\ km$

Hence, (I). The Schwarzschild radius is $2.94\times10^{8}\ km$

(II). The Schwarzschild radius is 17.7 km.

(III). The Schwarzschild radius is $1.1\times10^{-7}\ km$

(IV). The Schwarzschild radius is $7.4\times10^{-29}\ km$

8 0

4 years ago

Name the forces acting on a plastic bucket container water held above ground level in your hand. Discuss why the forces acting o

SashulF [63]

There are two forces at play:

The gravitational force acting downward due to the mass of the bucket and the water that it contains.
The upward force that your hand exerts on the bucket.

If the magnitude of the force your hand exerts on the bucket equals the magnitude of the gravitational force, the bucket is in static equilibrium. That means the bucket is not moving and the forces acting on it balance each other out, making the net force 0.

Having 0 net force means the bucket doesn't undergo any acceleration, or change in motion.

5 0

3 years ago

Determine the image distance and image height for a 5.00 cm tall object placed 30.0 cm from a double convex lens with a focal le

vlabodo [156]

Answer:

The image distance is 30 cm

image height = - 5 cm

Explanation:

The formula for calculating the image distance is expressed as

1/f = 1/u + 1/v

where

f is the focal length

u is the object distance

v is the image distance

From the information given,

u = 30

f = 15

By substituting these values into the formula,

1/15 = 1/30 + 1/v

1/v = 1/15 - 1/30 = (2 - 1)/30 = 1/30

Taking the reciprocal of both sides,

v = 30

The image distance is 30 cm

magnification = image height/object height = - v/u

Given that object height = 5 cm, then

image height/5 = - 30/30 = - 1

image height = - 5 * 1

image height = - 5 cm

8 0

1 year ago

What will happen to the Sun after it loses so much mass that its gravity

Natali5045456 [20]

Answer:

The answer is C

Explanation:

7 0

3 years ago

Read 2 more answers

I will mark you as brainliest if you answer correctly

aleksklad [387]

(A) We can solve the problem by using Ohm's law, which states:
$V=IR$
where
V is the potential difference across the electrical device
I is the current through the device
R is its resistance
For the heater coil in the problem, we know $V=220 V$ and $R=220 \Omega$ , therefore we can rearrange Ohm's law to find the current through the device:
$I= \frac{V}{R}= \frac{220 V}{220 \Omega}=1 A$

(B) The resistance of a conductive wire depends on three factors. In fact, it is given by:
$R= \rho \frac{L}{A}$
where
$\rho$ is the resistivity of the material of the wire
L is the length of the wire
A is the cross-sectional area of the wire
Basically, we see that the longer the wire, the larger its resistance; and the larger the section of the wire, the smaller its resistance.

6 0

3 years ago