All categories

2 years ago

Facts about conolizing into mars?

Physics

1 answer:

Leto [7]2 years ago

7 0

Answer:

xxxx

Explanation:

Mars also has a thin atmosphere. Because of this it has potential to host humans and other organic life. That makes Mars the best choice for thriving colony off the earth. The moon has also been proposed as the first location for human colonization but it not known to have air or water.

Hope this helps!

Have an amazing day/night

You might be interested in

Scientific knowledge

emmasim [6.3K]

Answer:

Explanation:

As we progress and learn more, scientists make new discoveries that can contradict earlier discoveries, and since are technology is better today we are able to discover a lot more to add to or change our previous theories.

6 0

3 years ago

Read 2 more answers

Which layer of the earths atmosphere contains the ozone layer?

Burka [1]

The answer to that will be the Troposphere.

8 0

3 years ago

Read 2 more answers

A pizza delivery driver must make three stops on her route. She will first leave the restaurant and travel 4 km due north to the

topjm [15]

Explanation:

This question involves continuous displacement in various directions. When it becomes difficult to imagine, vector analysis becomes handy.

Let us denote each of the individual displacements by a vector. Consider the unit vectors $\vec{i}\textrm{ and }\vec{j}$ as the unit vectors in the direction of East and North respectively.

By simple calculations, we can derive the unit vectors $\vec{j},\frac{-\vec{i}-\vec{j}}{2}\textrm{ and }\frac{-\frac{1}{2}\vec{i}+\frac{\sqrt{3}}{2}\vec{j}}{2}$ in the directions North, $45^{o}$ South of West and $60^{o}$ North of West respectively.

So Total displacement vector = Sum of individual displacement vectors.

Displacement vector = $4(\vec{j})+6(\frac{-\vec{i}-\vec{j}}{2})+5(\frac{-\frac{1}{2}\vec{i}+\frac{\sqrt{3}}{2}\vec{j}}{2})=-4.25\vec{i}+3.165\vec{j}$

Magnitude of Displacement = $|-4.25\vec{i}+3.165\vec{j}|=5.3km$

∴ Total displacement = $5.3km$

4 0

3 years ago

A uniform beam with mass M and length L is attached to the wall by a hinge, and supported by a cable. A mass of value 3M is susp

Jobisdone [24]

Answer:

The tension is $T= \frac{11}{2\sqrt{3} } Mg$

The horizontal force provided by hinge $Fx= \frac{11}{4\sqrt{3} } Mg$

Explanation:

From the question we are told that

The mass of the beam is $m_b =M$

The length of the beam is $l = L$

The hanging mass is $m_h = 3M$

The length of the hannging mass is $l_h = \frac{3}{4} l$

The angle the cable makes with the wall is $\theta = 60^o$

The free body diagram of this setup is shown on the first uploaded image

The force $F_x \ \ and \ \ F_y$ are the forces experienced by the beam due to the hinges

Looking at the diagram we ca see that the moment of the force about the fixed end of the beam along both the x-axis and the y- axis is zero

$\sum F =0$

Now about the x-axis the moment is

$F_x -T cos \theta = 0$

=> $F_x = Tcos \theta$

Substituting values

$F_x =T cos (60)$

$F_x= \frac{T}{2} ---(1)$

Now about the y-axis the moment is

$F_y + Tsin \theta = M *g + 3M *g ----(2)$

Now the torque on the system is zero because their is no rotation

So the torque above point 0 is

$M* g * \frac{L}{2} + 3M * g \frac{3L}{2} - T sin(60) * L = 0$

$\frac{Mg}{2} + \frac{9 Mg}{4} - T * \frac{\sqrt{3} }{2} = 0$

$\frac{2Mg + 9Mg}{4} = T * \frac{\sqrt{3} }{2}$

$T = \frac{11Mg}{4} * \frac{2}{\sqrt{3} }$

$T= \frac{11}{2\sqrt{3} } Mg$

The horizontal force provided by the hinge is

$F_x= \frac{T}{2} ---(1)$

Now substituting for T

$F_{x} = \frac{11}{2\sqrt{3} } * \frac{1}{2}$

$Fx= \frac{11}{4\sqrt{3} } Mg$

4 0

3 years ago

The SAME amount of current I passes through three different resistors. R2 has twice the cross-sectional area and the same length

Snezhnost [94]

Answer:

resistor R₂ has the lowest current density

Explanation:

The current density is

j = I / A

now let's analyze each case

a) R₂ has an area 2A₀ and a length L₀ that R₁

b) R₃ has an area Ao and a length 3L₀ what R₁

we can see that all the area is given in relation to the resistance R₁

the current density in R₁ is

j₁ = I / A₀

the current density in R₂

j₂ = I / 2A₀

j₂ 2 = ½ I/A₀

the current density in R₃

j₃ = I / A₀

j₂ < j₁ = j₃

therefore resistor R₂ has the lowest current density

5 0

3 years ago

Facts about conolizing into mars?​

Facts about conolizing into mars?