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3 years ago

The sun will probably end up as a white dwarf. How will it be different then from what it is today

Physics

1 answer:

just olya [345]3 years ago

6 0

Answer:

The differences that will be observed are;

1) The Sun will become faint and will no longer be yellow but rather appear white and will no longer be visible (become invisible) by unassisted vision as we can see the Sun today

2) The size of the Sun will shrink to a size comparable to the size of the Earth

3) The Sun will cool down and will no longer radiate as much heat

4) The nuclear reactions that generate energy on the Sun's will seize and the and the heat from the Sun will be from residual thermal energy

5) The core, which is the hottest part of the Sun will no longer be hydrogen but carbon and oxygen

Explanation:

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A speedboat is moving at a constant speed, and the force propelling it forward is balanced by the force of the water pulling it

Kipish [7]

The correct answer is C: the speedboat's direction of motion will change. This is In fact, the new force of 500 N from the side of the boat is not balanced by any forces from the other side, and therefore this force will cause an acceleration in the direction of the force itself, which is not the same direction of motion. Therefore, the speedboat will acquire a movement on the lateral direction as well.

6 0

3 years ago

Read 2 more answers

CHEGG 42 mT magnetic field points due west. If a proton of kinetic energy 9 x 10-12 J enters this field in an upward direction,

alexdok [17]

Answer:

The magnitude of the Force is $F = 697 *10^{-15}N$ and the direction is South

Explanation:

From the question we are told that

The magnetic field point due west and since East point toward the positive x -axis $(i)$ then this magnetic field would be mathematically represented as

$\= B = 42(-i)mT = 42*10^{-3} (-i) T$

Now from the question we are told that the kinetic energy is

$KE = 9*10^{-12}J$

Now this kinetic energy can be mathematically represented as

$KE = \frac{1}{2}mv^2$

Where m is the mass of proton which has a general value of

$m = 1.67*10^{-27}kg$

Now making the subject of the formula

$v = \sqrt{\frac{KE}{0.5 * m} }$

Substituting values we have

$v = \sqrt{\frac{9*10^{-12}}{0.5 * 1.67*10^{-27}} }$

$= 10.37*10^7m/s$

Now from the question we are told that proton is moving upward which is in the positive z direction so the velocity of the proton would be in the positive

So the velocity would be

$\= v = 10.37*10^{7} \r k \ m/s$

Now the magnetic Force can be mathematically represented as

$\= F = q \= v * \= B$

Where q is the charge on the proton which has a general value of $q =1.6*10^{-19}C$

Now substituting the value

$\= F = 1.6*10^{-19 } * (10.37 *10^7) \r k * (42 *10^{-3})(-i)$

$= 697*10^{-15} J$

Now according to Fleming's left hand rule the direction of the magnetic force is south toward the negative Y - direction $(-j)$

So the force can be denoted as

$\= F = 697*10^{-15}(-j) N$

6 0

3 years ago

|-7|<|7| is this true?

Temka [501]

It is FALSE.

The absolute value of both numbers is 7, therefore they are equal!

Hope this helps!

3 0

3 years ago

What do counties get their power from

SOVA2 [1]

Money, or dictatiorship, trading, something that involes with money

8 0

4 years ago

An electron accelerated from rest through a voltage of 780 v enters a region of constant magnetic field. part a part complete if

maxonik [38]

The electron is accelerated through a potential difference of $\Delta V=780 V$ , so the kinetic energy gained by the electron is equal to its variation of electrical potential energy:
$\frac{1}{2}mv^2 = e \Delta V$
where
m is the electron mass
v is the final speed of the electron
e is the electron charge
$\Delta V$ is the potential difference

Re-arranging this equation, we can find the speed of the electron before entering the magnetic field:
$v= \sqrt{ \frac{2 e \Delta V}{m} } = \sqrt{ \frac{2(1.6 \cdot 10^{-19}C)(780 V)}{9.1 \cdot 10^{-31} kg} }=1.66 \cdot 10^7 m/s$

Now the electron enters the magnetic field. The Lorentz force provides the centripetal force that keeps the electron in circular orbit:
$evB=m \frac{v^2}{r}$
where B is the intensity of the magnetic field and r is the orbital radius. Since the radius is r=25 cm=0.25 m, we can re-arrange this equation to find B:
$B= \frac{mv}{er}= \frac{(9.1 \cdot 10^{-31}kg)(1.66 \cdot 10^7 m/s)}{(1.6 \cdot 10^{-19}C)(0.25 m)} =3.8 \cdot 10^{-4} T$

3 0

4 years ago