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solniwko [45]
3 years ago
9

The first confirmed detections of extrasolar planets occurred in ____________. The first confirmed detections of extrasolar plan

ets occurred in ____________. the mid-17th century the mid-20th century the 1990s 2009
Physics
1 answer:
nirvana33 [79]3 years ago
4 0

Answer:

1992 (Early 1990s)

Explanation:

First of all, i would like to define an extrasolar planet as a planet that orbits a start that is not our own.

The first confirmed detections of extrasolar planets occured in the early 1990s (specifically 1992, some say 1995). The name of the first extrasolar planet is widely believed to be called Dimidium or 51 Pegasi b.  

Extrasolar were searched by monitoring stars for slight dimming that might occur as unseen planets pass in front of them.

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• The average length is ________ cm. This is the mean or average. b)• Subtract the highest value from the lowest value: ________
coldgirl [10]
16, 5 , 3 = 16+5+3= 24 + 3

So at the end put 24 + 3 cm
And put 16 for the lengths
For the value 5 and for the diving thingy 3
7 0
3 years ago
The leaves of a tree lose water to the atmosphere via the process of transpiration. A particular tree loses water at the rate of
Gnoma [55]

Answer:

The speed of the sap flowing in the vessel is 1.90 mm/s

Explanation:

Given:

The rate of water loss, Q = 3 × 10 ⁻⁸ m³/s

Number of vessels contained, n = 2000

Diameter of the vessel, D = 100 Mu m

thus, the radius of the vessel, r = 50 × 10⁻⁶ m

Now, the rate of flow is given as:

Q = AV    .............(1)

where, A is the area of the cross-section

V is the velocity

Total area, A = n × (πr²)

substituting the values in the equation (1), we get

3 × 10 ⁻⁸ m³/s = [2000 × (π × (50 × 10⁻⁶)²)] × V

or

V = 1.909 × 10⁻³ m/s or 1.90 mm/s

Hence, the speed of the sap flowing in the vessel is 1.90 mm/s

7 0
3 years ago
From the top of a tall building, a gun is fired. The bullet leaves the gun at a speed of 340 m/s, parallel to the ground. As the
Ivahew [28]

Answer:

The launching point is at a distance D = 962.2m and H = 39.2m

Explanation:

It would have been easier with the drawing. This problem is a projectile launching exercise, as they give us data after the window passes and the wall collides, let's calculate with this data the speeds at the point of contact with the window.

X axis

           x = Vox t

           t = x / vox

           t = 7.1 / 340

           t = 2.09 10-2 s

In this same time the height of the window fell

           Y = Voy t - ½ g t²

Let's calculate the initial vertical speed, this speed is in the window

           Voy = (Y + ½ g t²) / t

           Voy = [0.6 + ½ 9.8 (2.09 10⁻²)²] /2.09 10⁻² = 0.579 / 0.0209

            Voy = 27.7 m / s

We already have the speed at the point of contact with the window. Now let's calculate the distance (D) and height (H) to the launch point, for this we calculate the time it takes to get from the launch point to the window; at this point the vertical speed is Vy2 = 27.7 m / s

             Vy = Voy - gt₂

             Vy = 0 -g t₂

             t₂ = Vy / g

             t₂ = 27.7 / 9.8

             t₂ = 2.83 s

This is the time it also takes to travel the horizontal and vertical distance

            X = Vox t₂

            D = 340 2.83

            D = 962.2 m

           

            Y = Voy₂– ½ g t₂²

            Y = 0 - ½ g t2

            H = Y = - ½ 9.8 2.83 2

            H = 39.2 m

The launching point is at a distance D = 962.2m and H = 39.2m

6 0
2 years ago
How does Mercury's close proximity to the sun and thin atmosphere affect its ability to maintain liquid water
alekssr [168]

Answer

A thin atmosphere does not supply much oxygen, and the heat from the sun would evaporate it, because mercury is close to the sun.

5 0
3 years ago
Two cars A and B are moving with velocities 20 m/s and 15 m/s in the direction east and west respectively. If
Sphinxa [80]

Answer:

<u>Distance</u><u> </u><u>between</u><u> </u><u>them</u><u> </u><u>is</u><u> </u><u>4</u><u>,</u><u>2</u><u>0</u><u>0</u><u> </u><u>meters</u><u>.</u>

Explanation:

Consinder car A:

{ \bf{distance =  speed \times time }}

substitute:

distance = 20 \times (2 \times 60) \\  = 2400 \: m

Consider car B:

distance = 15 \times (2 \times 60) \\  = 1800 \: m

since these cars move in opposite directions, distance between them is their summation:

distance \: between = { \sum(distance \: of \: each \: car)} \\  = 2400 + 1800 \\  = 4200 \: m

3 0
2 years ago
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