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MAVERICK [17]
1 year ago
13

which planet detection method would work best for a massive planet that's close to its host star but doesn't pass between us and

the star?
Physics
1 answer:
Alona [7]1 year ago
3 0

The Doppler method is most frequently employed to find extrasolar planets, however it works best for finding extremely large planets that are near to their host star.

<h3>Which planet is the hottest?</h3>

Venus is the hottest planet within our solar system, with average global temperatures hot enough even to melt lead due to the runaway greenhouse effect caused by its thick atmosphere, which traps heat. Venus is around 700°F (390°C) hotter that it would be in the absence of the greenhouse effect.

<h3>Is it 8 or 9 planets?</h3>

Our solar system is made up of ten planetary systems, 146 moons, countless comets, meteorites, and other space objects, as well as Pluto and several dwarf planets. In the system, there is just the Sun. The eight planets are Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and finally Neptune.

To know more about planet visit:

brainly.com/question/15268075

#SPJ4

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What is the water table?
Setler [38]

The subsurface zone in which all openings of the rock are filled with water is called the zone of saturation.

8 0
4 years ago
A sprinter reaches his maximum speed in 2.6 seconds from rest with constant acceleration. He then maintains that speed and finis
Delvig [45]

Answer: maximum speed vmax = 11.42m/s

Explanation:

Given that the sprinter maintained constant acceleration during the first 2.6 seconds.

a = vmax/ta .......1

The distance covered during the acceleration period is;

da = 0.5a(ta)^2 .....2

Substituting equation 1 to 2

da = 0.5(vmax/ta)(ta)^2 = 0.5vmax(ta) .....3

The distance covered during the period of constant speed vmax is;

dv = vmax (tv) ......4

The total distance travelled is

d = da + dv = 100 (Given)

da + dv = 100 ......5

Substituting equation 3 and 4 into 5

0.5vmax(ta) + vmax(tv) = 100

vmax ( 0.5ta +tv) = 100

vmax = 100/(0.5ta + tv) ....6

But,

t = ta + tv

tv = t - ta .......7

Substituting equation 6 into equation 7

vmax = 100/(0.5ta + t - ta)

vmax = 100/(t-0.5ta)

t = 10.06 s

ta = 2.6 s

Substituting the values;

vmax = 100/(10.06 -0.5(2.6))

vmax = 11.42m/s

Note:

ta = acceleration time

tv = constant velocity vmax time

t = overall time

da , dv and d = acceleration, constant velocity and overall distance covered respectively.

8 0
4 years ago
A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 44 ft/s2. What is the dist
elena55 [62]

Answer:

Distance, d = 61.13 ft

Explanation:

It is given that,

Initial speed of the car, u = 50 mi/h = 73.34 ft/s

Finally, it stops i.e. v = 0

Deceleration of the car, a=-44\ ft/s^2

We need to find the distance covered before the car comes to a stop. Let the distance is s. It can be calculated using third law of motion as :

v^2-u^2=2as

s=\dfrac{v^2-u^2}{2a}

s=\dfrac{0-(73.34\ ft/s)^2}{2\times -44\ ft/s^2}

s = 61.13 ft

So, the distance covered by the car before it comes to rest is 61.13 ft. Hence, this is the required solution.

4 0
4 years ago
A pump is used to lift 100 KG of water from a wel 60 m deep,in 20 S If force of gravity on 1 KG is 10 N,find
Harlamova29_29 [7]

Explanation:

Given,

  • m = 100 kg
  • g = 10 N/kg¹
  • h = 60 m
  • t = 20 s

To Find:

a) Work done by the pump

b) Potential energy stored in the water

c)Power spent by the pump

d)Power rating of the pump.

Solution:

  • a) Work done by the pump

We know that,

\rm \: Work  \: done = Force * Distance  \: moved

  • f = 100 kg * 10N/kg
  • d = 60 m

\rm \: Work\; Done =(100 \: kg \times  \cfrac{10N}{kg} ) \times 60 \: m

\rm \: Work\; Done =1000 \times 60 \: joule

\boxed{\rm \: Work\; Done =60000 \: joule}

[The unit'll be joule since N×M = J]

  • b) Potential energy stored in the water

\rm \: P.E = m \cdot g \cdot  h

  • m = 100 kg
  • g = 10N/kg
  • h = 60

\rm \: P.E =100 \:kg \:  \times  \cfrac{10 \: N}{kg}  \times 60

\boxed{\rm \: P.E =60000  \: joule}

  • same condition here as well, N×M = J
  • c) Power of the Pump

\rm \: P = W/T

  • where P = Power; W = Work done & T = Time taken
  • As we got the value of work done on question (a),& ATQ time taken is 20 S.

\rm \: P =  \cfrac{60000 \: joule}{20 \: seconds}  =\boxed{\rm { 3000 \: Watts }}\: or \: \boxed{\rm 3 \: kW}

  • d) Power rating of the pump = 3 kW

Assumption: The pump is 100% efficient & works well.

6 0
2 years ago
By how much should the pressure of a litre of water be changed to compress it by 0.10% ?​
wel

Answer:

it is given that the water is to be compressed by 0.10%. Therefore, the pressure on the water should be

2.2 × 10^6 Nm ^ -2 .

8 0
3 years ago
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