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DochEvi [55]
3 years ago
5

What are some of the landforms rovers have found on Mars that show it was once an active place?

Physics
2 answers:
Schach [20]3 years ago
8 0

Answer:

The Curiosity rover found that ancient Mars had the right chemistry to support living microbes. Curiosity found sulfur, nitrogen, oxygen, phosphorus and carbon-- key ingredients necessary for life--in the powder sample drilled from the "Sheepbed" mudstone in Yellowknife Bay.

Explanation:

Hope it helped

topjm [15]3 years ago
7 0

Answer:

,

Explanation:

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Answer: Many multicellular animals like fishes, mammals, birds, etc. reproduce through sexual reproduction process. Even reproduction in domestic animals also occurs through the sexual process. Domestic animals (pets or farm animals) are an essential part of our life. MARK AS BRAINLIST PLZ

Explanation:

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3 years ago
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find a magnitude of the force such that if the act at right angle there resultant is √10N but if the act of 50° the resultant is
Readme [11.4K]

Explanation:

Let magnitude of the two forces be x and y.

Resultant at right angle R1= √15N) and at

60 degrees be R2= √18N.

Now, R1 = √(x² + y²) = √15,

R2= √(x² + y² +2xycos50) = √18.

So x² + y² = 15,

and x² + y² + 1.29xy = 18,

therefore 1.29xy = 3,

y = 3/1.29x.

y = 2.33/x

Now, x2 + (2.33/x)2 = 15,

x² + 5.45/x² = 15

multiply through by x²

x⁴ + 5.45 = 15x²

x⁴ - 15x2 + 5.45 = 0

Now find the roots of the equation, and later y. The two values of x will correspond to the

magnitudes of the two vectors.

Good luck

7 0
3 years ago
Which best describes what occurs when an object takes in a wave as the wave hits it?.
statuscvo [17]

Answer:

Absorption

Explanation:

7 0
2 years ago
A projectile is launched at an angle of 36.7 degrees above the horizontal with an initial speed of 175 m/s and lands at the same
Softa [21]

Answer:

a) The maximum height reached by the projectile is 558 m.

b) The projectile was 21.3 s in the air.

Explanation:

The position and velocity of the projectile at any time "t" is given by the following vectors:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

v = (v0 · cos α, v0 · sin α + g · t)

Where:

r = position vector at time "t"

x0 = initial horizontal position

v0 = initial velocity

t = time

α = launching angle

y0 = initial vertical position

g = acceleration due to gravity (-9.80 m/s² considering the upward direction as positive).

v = velocity vector at time t

a) Notice in the figure that at maximum height the velocity vector is horizontal. That means that the y-component of the velocity (vy) at that time is 0. Using this, we can find the time at which the projectile is at maximum height:

vy = v0 · sin α + g · t

0 = 175 m/s · sin 36.7° - 9.80 m/s² · t

-  175 m/s · sin 36.7° /  - 9.80 m/s² = t

t = 10.7 s

Now, we have to find the magnitude of the y-component of the vector position at that time to obtain the maximum height (In the figure, the vector position at t = 10.7 s is r1 and its y-component is r1y).

Notice in the figure that the frame of reference is located at the launching point, so that y0 = 0.

y = y0 + v0 · t · sin α + 1/2 · g · t²

y = 175 m/s · 10.7 s · sin 36.7° - 1/2 · 9.8 m/s² · (10.7 s)²

y = 558 m

The maximum height reached by the projectile is 558 m

b) Since the motion of the projectile is parabolic and the acceleration is the same during all the trajectory, the time of flight will be twice the time it takes the projectile to reach the maximum height. Then, the time of flight of the projectile will be (2 · 10.7 s) 21.4 s. However, let´s calculate it using the equation for the position of the projectile.

We know that at final time the y-component of the vector position (r final in the figure) is 0 (because the vector is horizontal, see figure). Then:

y = y0 + v0 · t · sin α + 1/2 · g · t²

0 = 175 m/s · t · sin 36.7° - 1/2 · 9.8 m/s² · t²

0 = t (175 m/s ·  sin 36.7 - 1/2 · 9.8 m/s² · t)

0 = 175 m/s ·  sin 36.7 - 1/2 · 9.8 m/s² · t

-  175 m/s ·  sin 36.7 / -(1/2 · 9.8 m/s²) = t

t = 21.3 s

The projectile was 21.3 s in the air.

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What are two situational examples of unbalanced forces?
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An example of a balanced force is two cards leaning against each other and not falling over, or two football players blocking each other but neither overpowering the other. An example of an unbalanced force is two cards leaning on each other then falling over, or two football players blocking each other, then one tackles the other.

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3 years ago
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