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2 years ago

Why was Mars orbiter sent to mars?

Physics

2 answers:

Dafna11 [192]2 years ago

6 0

Explanation:

to develop the technologies required for designing, planning, management and operations of an interplanetary mission. The secondary objective is to explore Mars' surface features, morphology, mineralogy and Martian atmosphere using indigenous scientific instruments

tankabanditka [31]2 years ago

6 0

Answer: Mars Climate Orbiter was lost on arrival September 23, 1999. Engineers concluded that the spacecraft entered the planet's atmosphere too low and probably burned up.

Explanation: The scientific reasons for going to Mars can be summarised by the search for life, understanding the surface and the planet's evolution, and preparing for future human exploration. Understanding whether life existed elsewhere in the Universe beyond Earth is a fundamental question of humankind.

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Determine the voltage ratings of the high-and-low voltage windings for this connection and the MVA rating of the autotransformer

SIZIF [17.4K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The High Voltage Rating for Auto - Transformer is 86kV

The Low Voltage Rating for Auto - Transformer is 78kV

The MVA rating is 268.75 $MVA$

The efficiency is 99.4%

Explanation:

From the question we are given are given that

The transformer has Mega Volt Amp rating of 25MVA

The frequency is 60-Hz

Voltage rating 8.0kV : 78kV

The short circuit test gives : 453kV,321A,77.5kW

The open circuit test gives : 8.0kV, 39.6A, 86.2kW

This can be represented on a diagram shown on the second uploaded image

From this diagram we can deduce that the The High Voltage Rating for Auto - Transformer is 86kV and the Low Voltage Rating for Auto - Transformer is 78kV

Now to obtain the current flowing through the 8kV coil in the Auto-transformer we have

$\frac{25 \ Mega \ Volt\ Ampere }{8\ Kilo Volt}$

The volt will cancel each other

$\frac{25*10^6}{8*10^3} = 3125\ A$

Now to obtain the required MVA rating we would multiply the value of Power obtained during the open circuit test by the value of the current calculated.we are making use of the power obtain during open circuit testing because the transformer at this point is not under any load.

$MVA \ rating = (86*10^3)(3125) =268.75$

We need to understand that Iron losses is due to open circuit test which has power = 86.2kW

While copper loss is due to short circuit test which has power = 77.5kW

The the current flowing through the secondary coil $I_2$ as shown in the circuit diagram can be obtained as

$I_2 = \frac{25*10^6}{78*10^3} =320.52 A \approx 321$

Now the efficiency can be obtained as thus

$\frac{(operational \ MVA )*(Power factor \pf))}{(operational\ MVA (power factor pf) + copper loss + Iron loss)}*\frac{100}{1}$

=99.941%

8 0

2 years ago

A softball has a mass of 0.18 kg. What is the gravitational force on Earth due to the ball, and what is the Earth's resulting ac

alexgriva [62]

Answer:F=1.7802

Explanation:

Since we've been given the mass to be .18kg,we are asked to find the Force of which the formulae is

F=ma where f-force,m-mass and a-acceleration due to gravity

So we can just substitute

F-?.m=.18 and a9.89

F=.18×9.89

F=1.7802N

6 0

2 years ago

An electron is isolated from an atom and exists in vacuum. A group of scientists collectively state that they can remove part of

lesya692 [45]

D) Partial charge cannot be removed, because charge is a discrete quantity that may exist only at certain values

Explanation:

The electric charge of an object is a property of the object that is related to the ability of the object to experience/exert an electric force: if the object is electrically charge, then it is attracted or repelled by other electrically charged object.

The electric charge of an object depends on the amount of charged particles it has on it. In particular, the fundamental particles that carry electric charge are:

Protons: they carry electric charge of +e
Electrons: they carry electric charge of -e

Where "e" is the fundamental charge ( $e=1.6\cdot 10^{-19}C$ ). Therefore, one proton carry a charge of +e and one electron carry a charge of -e.

An electron is a fundamental particle: this means that it cannot be divided into smaller particles. This also means that it is not possible to remove part of the charge of the electron: in fact, it is said that electric charge exists only as discrete values, being a multiple of $e$ . Therefore, the correct statement is

D) Partial charge cannot be removed, because charge is a discrete quantity that may exist only at certain values

Learn more about particles:

brainly.com/question/2757829

#LearnwithBrainly

5 0

2 years ago

A. A land speed car can decelerate at 9.8m/s. How long does it take the car to come to a complete stop from a run of 885 km/hr (

Nimfa-mama [501]

Answer:

A. 25.08 s

B. 3082.53 m

C. 3×10⁵ m/s²

Explanation:

A. Determination of the time.

This can be obtained as illustrated below:

Acceleration (a) = –9.8 m/s²

Initial velocity (u) = 245.8 m/s

Final velocity (v) = 0 m/s

Time (t) =.?

v = u + at

0 = 245.8 + (–9.8 × t)

0 = 245.8 – 9.8t

Collect like terms

0 – 245.8 = – 9.8t

– 245.8 = – 9.8t

Divide both side by –9.8

t = –245.8 / –9.8

t = 25.08 s

Therefore, it will take 25.08 s for the car to come to a complete stop.

B. Determination of the distance travelled by the car.

Acceleration (a) = –9.8 m/s²

Initial velocity (u) = 245.8 m/s

Final velocity (v) = 0 m/s

Distance (s) =?

v² = u² + 2as

0² = 245.8² + (2 × –9.8 × s)

0 = 60417.64 – 19.6s

Collect like terms

0 – 60417.64 = – 19.6s

– 60417.64 = – 19.6s

Divide both side by –19.6

s = –60417.64 / –19.6

s = 3082.53 m

Thus, the car travelled a distance of 3082.53 m before stopping completely.

C. Determination of the acceleration of the object.

Initial velocity (u) = 0 m/s

Final velocity (v) = 600 m/s

Distance (s) = 0.6 m

Acceleration (a) =?

v² = u² + 2as

600² = 0² + (2 × a × 0.6)

360000 = 0 + 1.2a

360000 = 1.2a

Divide both side by 1.2

a = 360000 / 1.2

a = 300000 = 3×10⁵ m/s²

7 0

2 years ago

According to Kepler's Third Law, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about

NARA [144]

Answer:

Orbital period, T = 1.00074 years

Explanation:

It is given that,

Orbital radius of a solar system planet, $r=4\ AU=1.496\times 10^{11}\ m$

The orbital period of the planet can be calculated using third law of Kepler's. It is as follows :

$T^2=\dfrac{4\pi^2}{GM}r^3$

M is the mass of the sun

$T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times (1.496\times 10^{11})^3$

$T^2=\sqrt{9.96\times 10^{14}}\ s$

T = 31559467.6761 s

T = 1.00074 years

So, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about 1.00074 years.

6 0

2 years ago

Why was Mars orbiter sent to mars?​

Why was Mars orbiter sent to mars?