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1 year ago

The tallest volcano on Mars

Physics

2 answers:

Nonamiya [84]1 year ago

8 0

Answer: Is three times the height of Mt. Everest

Explanation:

Using the process of elimination we can rule out “D” Because its a volcano, C because it says we have visited mars 44 times not the volcano, B because in the passage is literally says “3 times the size of Mt. Everest”.

pychu [463]1 year ago

8 0

Answer A is correct

Mars features the tallest volcano in the solar system, which is three times the
height of Mt. Everest and approximately the size of New Mexico. And It's a dry, rocky, cold planet situated
fourth from the Sun. It's called the red planet because of its red hue. Mars has two moons
called Phobos and Deimos.
The tallest volcano on Mars

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Two parallel conducting plates are separated by 10.0 cm, and one of them is taken to be at zero volts. (a) What is the magnitude

aivan3 [116]

Answer:

$E=6.8Kv/m$

Explanation:

From the question we are told that

Distance b/w plate $d=10cm=>0.1m$

P_1 Potential at 7.35 $V=533v$

Generally the equation for electric field at a distance is mathematically given as

$E=\frac{v}{d}$

$E=\frac{533}{7.85*10^-^2}$

$E=6789.808917$

$E=6.8*10^3$

$E=6.8Kv/m$

7 0

2 years ago

A very powerful vacuum cleaner which has a hose of circular cross section can lift a brick of mass 12 kg when the hose is placed

valkas [14]

Answer:

A). 1.9 cm

Explanation:

m = Mass of brick = 12 kg

g = Acceleration due to gravity = 9.81 m/s²

r = Radius of hose

A = Area = $\pi r^2$

F = Force = $mg$

Let us assume that the pressure required to lift the brick would be atmospheric pressure

$P=\dfrac{F}{A}\\\Rightarrow P=\dfrac{F}{\pi r^2}\\\Rightarrow r=\sqrt{\dfrac{F}{\pi P}}\\\Rightarrow r=\sqrt{\dfrac{12\times 9.81}{\pi\times 101325}}\\\Rightarrow r=0.01923\ m=1.9\ cm$

The radius of the hose should be 1.9 cm

6 0

2 years ago

A baseball player leads off the game and hits a long home run. The ball leaves the bat at an angle of 70.0 from the horizontal w

professor190 [17]

Answer: 211.059 m

Explanation:

We have the following data:

$\theta=70\°$ The angle at which the ball leaves the bat

$V_{o}=55 m/s$ The initial velocity of the ball

$g=-9.8 m/s^{2}$ The acceleration due gravity

We need to find how far (horizontally) the ball travels in the air: $x$

Firstly we need to know this velocity has two components:

Horizontally:

$V_{ox}=V_{o}cos \theta$ (1)

$V_{ox}=55 m/s cos(70\°)=18.811 m/s$ (2)

Vertically:

$V_{oy}=V_{o}sin \theta$ (3)

$V_{oy}=55 m/s sin(70\°)=51.683 m/s$ (4)

On the other hand, when we talk about parabolic movement (as in this situation) the ball reaches its maximum height just in the middle of this parabola, when $V=0$ and the time $t$ is half the time it takes the complete parabolic path.

So, if we use the following equation, we will find $t$ :

$V=V_{o}+gt=0$ (5)

Isolating $t$ :

$t=\frac{-V_{o}}{g}$ (6)

$t=\frac{-55 m/s}{-9.8 m/s^{2}}$ (7)

$t=5.61 s$ (8)

Now that we have the time it takes to the ball to travel half of is path, we can find the total time $T$ it takes the complete parabolic path, which is twice $t$ :