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3 years ago

Explain the runaway refrigerator effect and the role it may have played in the evolution of Mars.

Physics

1 answer:

Fiesta28 [93]3 years ago

6 0

Answer:

Explained

Explanation:

The runway refrigerator effect and the role it would have played in the evolution of Mars can be summarized as follows

The weak gravity of Mars does not allow it have a gaseous atmosphere over it. The thin atmosphere would have lead to lower temperature on mars.In in turn would have lead to freezing of gases thus lowering the temperature further. The thinner atmosphere and colder temperature would lead to loss of most water in the planet.

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You glide 28 units to the +y-direction with respect to the +x-axis, then move 32 units with respect to the +x-direction. What is

NNADVOKAT [17]

This question can be solved by using Pythagora's Theorem.

The resultant magnitude of the movement is "42.5 units".

The x and y components of the movement are given. We can use Pythagora's Theorem to find the resultant of these movements. Hence, applying the Pythagora's Theorem<em>:</em>

$d = \sqrt{d_x^2+d_y^2}$

where,

d = resultant movement = ?

$d_x$ = movement in x direction = 32 units

$d_y$ = movement in y direction = 28 units

Therefore,

$d = \sqrt{(32\ units)^2+(28\ units)^2}$

<u>d = 42.5 units</u>

Learn more about Pythagora's Theorem here:

brainly.com/question/343682?referrer=searchResults

The attached picture shows Pythagora's Theorem<em>.</em>

4 0

2 years ago

Anyone know how to do this?

Gala2k [10]

Answer:

I think, (remember think) it might be 2.0 m/s

Explanation:

If it's wrong I'm truly sorry.

6 0

2 years ago

A steady current I flows through a wire of radius a. The current density in the wire varies with r as J = kr, where k is a const

grin007 [14]

Answer:

Explanation:

we can consider an element of radius r < a and thickness dr. and Area of this element is

$dA=2\pi r dr$

since current density is given

$J=kr$

then , current through this element will be,

$di_{thru}=JdA=(kr)(2\pi\,r\,dr)=2\pi\,kr^2\,dr$

integrating on both sides between the appropriate limits,

$\int_0^Idi_{thru}=\int_0^a2\pi\,kr^2\,dr
\\\\
I=\frac{2\pi\,ka^3}{3} -------------------------------(1)$

Magnetic field can be found by using Ampere's law

$\oint{\vec{B}\cdot\,d\vec{l}}=\mu_0\,i_{enc}$

for points inside the wire ( r<a)

now, consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius r.

by applying the Ampere's law, we can write

$\oint{\vec{B}_{in}\cdot\,d\vec{l}}=\mu_0\,i_{enc}
$

by symmetry $\vec{B}$ will be of uniform magnitude on this loop and it's direction will be tangential to the loop.

Hence,

$B_{in}\times2\pi\,l=\mu_0\int_0^r(kr)(2\pi\,r\,dr)=
\\\\2\pi\,B_{in} l=2\pi\mu_0k \frac{r^3}{3}
\\\\B_{in}=\frac{\mu_0kl^2}{3}
$

now using equation 1, putting the value of k,

$B_{in} = \frac{\mu_{0} l^2 }{3 } \,\,\, \frac{3I}{2 \pi a^3}
\\\\B_{in} = \frac{ \mu_{0} I l^2}{2 \pi a^3}
$

now, for points outside the wire ( r>a)

consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius l.

applying the Ampere's law

$\oint{\vec{B}_{out}\cdot\,d\vec{l}}=\mu_0\,i_{enc}
$

by symmetry $\vec{B}$ will be of uniform magnitude on this loop and it's direction will be tangential to the loop. Hence

$B_{out}\times2\pi\,r=\mu_0\int_0^a(kr)(2\pi\,r\,dr)
\\\\2\pi\,B_{out}r=2\pi\mu_0k\frac{a^3}{3}
\\\\B_{out}=\frac{\mu_0ka^3}{3r}
$

again using,equaiton 1,

$B_{out}= \mu_0 \frac{a^3}{3r} \times \frac{3 I}{2 \pi a^3}
\\\\B_{out} = \frac{ \mu_{0} I}{2 \pi r}$

8 0

3 years ago

At the top of a hill a roller coaster has gravitational potential energy due to its position. What happend to this potential ene

Anuta_ua [19.1K]

As the roller coaster speeds up on the way down the hill, the potential energy of roller coaster will be converted to kinetic energy.

<h3>What is Conservation of Energy ?</h3>

Conservation of energy state that energy is neither created nor destroy, they can only be transformed from one form to another. Energy of and object can transform from Potential energy to kinetic energy and vice versa

Given that at the top of a hill a roller coaster has gravitational potential energy due to its position. What will happen to this potential energy as the roller coaster speeds up on the way down the hill is that the potential energy to the roller coaster will start decreasing while the kinetic energy will start to increase.

The total energy of the roller coaster will be constant because of conservation of energy. As the roller coaster speeds up on the way down the hill, the potential energy will eventually reduce to zero where the total energy of the as the roller coaster will be equal to maximum kinetic energy.

Therefore, as the roller coaster speeds up on the way down the hill, the potential energy of roller coaster will be converted to kinetic energy.

Learn more about Energy here: brainly.com/question/25959744

#SPJ1

3 0

1 year ago

If only an external force can change the velocity of a body, how can the internal force of the brakes bring a car to rest? 1. It

RoseWind [281]

Answer:

4. It is the force of the road on the tires (an external force) that stops the car.

Explanation:

If there is no friction between the road and the tires, the car won't stop.

You can see this, for example, when there is ice on the road. You can still apply the brakes (internal force), but since there is no friction (external force) the car won't stop.

The force of the brakes on the wheels is not what makes the car stop, it is the friction of the road against still tires that makes it stop.

3 0

3 years ago