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2 years ago

The Gaia hypothesis is an example of _____

1 answer:

4 0

A complex entity involving the Earth's biosphere, atmosphere, oceans, and soil; the totality constituting a feedback or cybernetic system which seeks an optimal physical and chemical environment for life on this planet

You might be interested in

Why are we unable to work long without food

nataly862011 [7]

Answer:

Without food your not able to produce energy thats why you cannot work so long

7 0

1 year ago

A ball is thrown with a velocity of 35 meters per second at an angle of 30° above the horizontal. which quantity has a magnitude

enot [183]

The quantity that has a magnitude of zero when the ball is at the highest point in its trajectory is the vertical velocity.

In fact, the motion of the ball consists of two separate motions:
- the horizontal motion, on the x-axis, which is a uniform motion with constant velocity $v_x=v_0 cos 30^{\circ}$ , where $v_0=35 m/s$
- the vertical motion, on the y-axis, which is a uniformly accelerated motion with constant acceleration $g=9.81 m/s^2$ directed downwards, and with initial velocity $v_y=v_= sin 30^{\circ}$ . Due to the presence of the acceleration g on the vertical direction (pointing in the opposite direction of the initial vertical velocity), the vertical velocity of the ball decreases as it goes higher, up to a point where it becomes zero and it reverses its direction: when the vertical velocity becomes zero, the ball has reached its maximum height.

5 0

1 year ago

for a given initial projectile speed, you observe that the projectile has a certain range R at a launch angle of a = 30. For wha

VLD [36.1K]

Answer:

The other angle is 30 degrees.

Explanation:

The range of projectile is given by :

$R=\dfrac{u^2\ \sin2\theta}{g}$

Here,

u is the speed of launch of projectile

Here, $\theta=30^{\circ}$

We need to find the other launch angle when the projectile have the same range, such that,

$\dfrac{u^2\ \sin(60)}{g}=\dfrac{u^2\ \sin2\alpha}{g}$

$\sin(60)=\sin2\alpha$

$\dfrac{\sqrt3}{2}=\sin2\alpha$

$\alpha =30^{\circ}$

So, the other angle is 30 degrees. Hence, this is the required solution.

3 0

2 years ago

Official (Closed) - Non Sensitive

Pavlova-9 [17]

Answer:

The minimum running time is 319.47 s.

Explanation:

First we find the distance covered and time taken by the train to reach its maximum speed:

We have:

Initial Speed = Vi = 0 m/s (Since, train is initially at rest)

Final Speed = Vf = 29.17 m/s

Acceleration = a = 0.25 m/s²

Distance Covered to reach maximum speed = s₁

Time taken to reach maximum speed = t₁

Using 1st equation of motion:

Vf = Vi + at₁

t₁ = (Vf - Vi)/a

t₁ = (29.17 m/s - 0 m/s)/(0.25 m/s²)

t₁ = 116.68 s

Using 2nd equation of motion:

s₁ = (Vi)(t₁) + (0.5)(a)(t₁)²

s₁ = (0 m/s)(116.68 s) + (0.5)(0.25 m/s²)(116.68 s)²

s₁ = 1701.78 m = 1.7 km

Now, we shall calculate the end time and distance covered by train, when it comes to rest on next station.

We have:

Final Speed = Vf = 0 m/s (Since, train is finally stops)

Initial Speed = Vi = 29.17 m/s (The train must maintain max. speed for min time)

Deceleration = a = - 0.7 m/s²

Distance Covered to stop = s₂

Time taken to stop = t₂

Using 1st equation of motion:

Vf = Vi + at₂

t₂ = (Vf - Vi)/a

t₂ = (0 m/s - 29.17 m/s)/(- 0.7 m/s²)

t₂ = 41.67 s

Using 2nd equation of motion:

s₂ = (Vi)(t₂) + (0.5)(a)(t₂)²

s₂ = (29.17 m/s)(41.67 s) + (0.5)(- 0.7 m/s²)(41.67 s)²

s₂ = 607.78 m = 0.6 km

Since, we know that the rest of 7 km, the train must maintain the maximum speed to get to the next station in minimum time.

The remaining distance is:

s₃ = 7 km - s₂ - s₁

s₃ = 7 km - 0.6 km - 1.7 km

s₃ = 4.7 km

Now, for uniform speed we use the relation:

s₃ = vt₃

t₃ = s₃/v

t₃ = (4700 m)/(29.17 m/s)

t₃ = 161.12 s

So, the minimum running time will be:

t = t₁ + t₂ + t₃

t = 116.68 s + 41.67 s + 161.12 s

5 0

2 years ago

When numbers are very small or very large, it is convenient to either express the value in scientific notation and/or by using a

Oxana [17]

Answer:

5 mg, $5\cdot 10^{-3}g$

Explanation:

First of all, let's rewrite the mass in grams using scientific notation.

we have:

m = 0.005 g

To rewrite it in scientific notation, we must count by how many digits we have to move the dot on the right - in this case three. So in scientific notation is

$m=5\cdot 10^{-3}g$

If we want to convert into milligrams, we must remind that

1 g = 1000 mg

So we can use the proportion

$1 g : 1000 mg = 0.005 g : x$

and we find

$x=\frac{(1000 mg)(0.005 g)}{1 g}=5 mg$

4 0

2 years ago