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Natalija [7]
3 years ago
15

How did cyanobacteria affect early Earth’s biosphere? Select all that apply. Cyanobacteria allowed organisms that rely on oxygen

to evolve. Cyanobacteria stopped the formation of rust in the Earth’s crust. Cyanobacteria preceded the first photosynthetic organisms. Cyanobacteria produced excess oxygen.
Physics
1 answer:
FrozenT [24]3 years ago
7 0

Answer:

The correct statements would be

  • Cyanobacteria allowed organisms that rely on oxygen to evolve.
  • Cyanobacteria preceded the first photosynthetic organisms.
  • Cyanobacteria produced excess oxygen.

Cyanobacteria, also termed as blue-green algae are the prokaryotes which are able to perform photosynthesis.

They were the major contributors of oxygen in the atmosphere and thus helped the organisms that rely on oxygen to evolve.

By the process of endosymbiosis, they lead to the origin of plants. The chloroplasts present in green plants is considered as the cyanobacteria living in the plant cell. It helps in photosynthesis and in return plants cell provides shelter to it.

It is believed that the oxygen released from early cyanobacteria reacted with dissolved iron ions to form iron oxide.

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You need a 450 microgram sample of gold, but you only have a mass balance that measures in decigrams. Convert the amount of gold
sveta [45]

The amount of gold in decigrams if 450 micrograms is needed is 4.5 × 10-³ decigrams.

<h3>How to convert micrograms to decigrams?</h3>

According to this question, 450 micrograms of a sample of gold is needed but we only have a mass balance that measures in decigrams.

This means that we are to convert the amount of gold you need to decigrams by comparing the exponents.

The conversion factor of micrograms to decigrams is as follows:

1 micrograms = 1 × 10-⁵ decigrams

This means 450 micrograms is equivalent to 450 × 1 × 10-⁵ = 4.5 × 10-³ decigrams

Therefore, the amount of gold in decigrams if 450 micrograms is needed is 4.5 × 10-³ decigrams.

Learn more about decigrams at: brainly.com/question/6869599

#SPJ1

7 0
1 year ago
The engine of a 1520-kg automobile has a power rating of 75 kW. Determine the time required to accelerate this car from rest to
LUCKY_DIMON [66]

Answer:

t=15.68 s

Explanation:

Given that

m = 1520 kg

P =75 KW

We know that

Power  ,P = F .v

F=force

v=velocity

v= 100 km/h

v=\dfrac{1000}{3600}\times 100\ m/s

v=27.77 m/s

75 x 1000  = F x 27.77

F= \dfrac{75000}{27.77}\ N

F= 2700.75 N

F= m a

m=mass

a=acceleration

2700.75 = 1520 x a

 a=1.77 m/s²

time t given as

v= u + a t

27.77 = 0 + 1.77 x t

t=15.68 s

3 0
3 years ago
A) Determine the x and y-components of the ball's velocity at t = 0.0s, 2.0, 3.0 secs.
malfutka [58]

The kinematic relationships we can find the position, acceleration and launch angle of the body on the planet Exidor.

a) the position are

      time (s)  x (m)   y(m)

        0            0          0

        2.0         3.6        1.2

        3.0         5.4        0.9

b) The aceleration is  g = 0.6 m / s²

c) The launch angle      θ = 33.7º

given parameters

  • the initial velocity of the body vₓ = 1.8m / s and v_y = 1.2 m / s
  • the movement times t = 1.0s, 2.0s and 3.0 s

to find

    a) position

    b) acceleration

    c) launch angle

Projectile launch is an application of kinematics to the movement of the body in two dimensions where there is no acceleration on the x axis and the y axis has the planet's gravity acceleration

b) To calculate the acceleration of the plant acting on the y-axis, we use that the vertical velocity of the body at the highest point is zero.

         v_y = v_{oy} - g t

where v and v({oy}  are the velocities of the body, g the acceleration of the planet's gravity and t the time

          0 = v_{oy} - gt

           g = v_{oy} / t

from the graph we observe that the highest point occurs for t = 2.0 s

           g = 1.2 / 2.0

           g = 0.6 m / s²

 

a) The position is requested for several times

X axis

in this axis there is no acceleration so we can use the uniform motion relationships

          vₓ = x / t

          x = vₓ t

where x is the position, vx is the velocity and t is the time

we calculate for the time

t = 0.0 s

          x₀ = 0

           

t = 2.0 s

          x₂ = 1.8 2

          x₂ = 3.6 m

t = 3.0 s

          x₃ = 1.8 3

          x₃ = 5.4 m

Y axis

In this axis there is the acceleration of the planet, let us use for the position the relation

          y = v_{oy} t - ½ g t²

t = 0.0 s

          y₀ = 0

          y₀ = 0 m

t = 2.0 s

         y₂ = 1.2 2 - ½ 0.6 2²

         y₂ = 1.2 m

t = 3.0 s

        y₃ = 1.2  3 - ½  0.6  3²

        y₃ = 0.9 m

c) the launch angle use the trigonometry relation

        tan θ = \frac{v_y}{v_x}

        θ = tan⁻¹ \frac{v_y}{v_x}

        θ = tan⁻¹ \frac{1.2}{1.8}

        θ = 33.7º

measured counterclockwise from the positive side of the x-axis

With the kinematic relationships we can find the position, acceleration and launch angle of the body on the planet Exidor.

a) the position are

      time (s)  x (m)   y(m)

        0            0          0

        2.0         3.6        1.2

        3.0         5.4        0.9

b) The aceleration is  g = 0.6 m / s²

c) The launch angle      θ = 33.7ºto)

learn more about projectile launch here:

brainly.com/question/10903823

4 0
3 years ago
What voltage produces a current of 50 amps with a resistance of 20 ohms?<br>​
PIT_PIT [208]

Answer:

1000 V

Explanation:

Ohm's law:

V = IR

V = (50 A) (20 Ω)

V = 1000 V

5 0
3 years ago
A proton traveling with speed 2 × 105 m/s in the -y direction passes through a region in which there is a uniform magnetic field
stepladder [879]

Answer:

\vec{E} =  1.2\times 10^5(-i)

Explanation:

Given that

Speed ,v= 2 x 10⁵ m/s ( - y direction)

B= 0.6 T (- z direction)

The resultant force on the proton given as

\vec{F}=q.\vec{E}+ q.(\vec{v}\times \vec{B})

F= m a

For uniform motion acceleration should be zero.

F = 0

0=q.\vec{E}+ q.(\vec{v}\times \vec{B})

0=\vec{E}+ (\vec{v}\times \vec{B})

0=\vec{E}+2\times 10^5(-j) \times 0.6(-k)

\vec{E} =- 2\times 10^5(-j) \times 0.6(-k)

\vec{E} =-1.2\times 10^5(i)

\vec{E} = 1.2\times 10^5(-i)

Electric filed should be apply in the negative x direction.

5 0
3 years ago
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