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3 years ago

One celled organisms that reproduce by fission?

Physics

2 answers:

Varvara68 [4.7K]3 years ago

7 0

One celled organism that’s reproduced by fission is bacteria.

Answer: Bacteria

Varvara68 [4.7K]3 years ago

7 0

Answer:

Bacteria

Hope this helped! ;)

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Can somebody help please !!!!

Nataly [62]

Vector A is of magnitude 12 m and it makes an angle of 37 degree with Y axis

So here we can say that

$\frac{A_x}{A} = sin37$

$A_x = A sin37$

$A_x = 12 sin37$

$A_x = 7.22 m$

Similarly we have

$\frac{A_y}{A} = cos37$

$A_y = A cos37$

$A_y = 12 cos37$

$A_y = 9.58 m$

So here we have

$A_y = 9.58 m, A_x = 7.22 m$

option A is correct

3 0

3 years ago

how do index contours let you know if the land you are viewing is increasing or decreasing in elevation?

Aleks [24]

Index Contours are indicated by a thicker line compared to the others. Index Contours are labelled with specific elevations along it to give a better understanding of the scale of elevation.

The elevations on the Index Contour along with the legend of the map, that allows you read intermediate contour lines, gives you a clear perspective of increasing/decreasing elevation.

8 0

4 years ago

I need answers and solvings to these questions

den301095 [7]

1) The period of a simple pendulum depends on B) III. only (the length of the pendulum)

2) The angular acceleration is C) $15.7 rad/s^2$

3) The frequency of the oscillation is C) 1.6 Hz

4) The period of vibration is B) 0.6 s

5) The diameter of the nozzle is A) 5.0 mm

6) The force that must be applied is B) 266.7 N

Explanation:

The period of a simple pendulum is given by

$T=2\pi \sqrt{\frac{L}{g}}$

where

T is the period

L is the length of the pendulum

g is the acceleration of gravity

From the equation, we see that the period of the pendulum depends only on its length and on the acceleration of gravity, while there is no dependence on the mass of the pendulum or on the amplitude of oscillation. Therefore, the correct option is

B) III. only (the length of the pendulum)

The angular acceleration of the rotating disc is given by the equation

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f$ is the final angular velocity

$\omega_i$ is the initial angular velocity

t is the time elapsed

For the compact disc in this problem we have:

$\omega_i = 0$ (since it starts from rest)

$\omega_f = 300 rpm \cdot \frac{2\pi rad/rev}{60 s/min}=31.4 rad/s$ is the final angular velocity

t = 2 s

Substituting, we find

$\alpha = \frac{31.4-0}{2}=15.7 rad/s^2$

For a simple harmonic oscillator, the acceleration and the displacement of the system are related by the equation

$a=-\omega^2 x$

where

a is the acceleration

x is the displacement

$\omega$ is the angular frequency of the system

For the oscillator in this problem, we have the following relationship

$a=-100 x$

which implies that

$\omega^2 = 100$

And so

$\omega = \sqrt{100}=10 rad/s$

Also, the angular frequency is related to the frequency $f$ by

$f=\frac{\omega}{2\pi}$

Therefore, the frequency of this simple harmonic oscillator is

$f=\frac{10}{2\pi}=1.6 Hz$

When the mass is hanging on the sping, the weight of the mass is equal to the restoring force on the spring, so we can write

$mg=kx$

where

m is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

k is the spring constant

x = 8.0 cm = 0.08 m is the stretching of the spring

We can re-arrange the equation as

$\frac{k}{m}=\frac{g}{x}=\frac{9.8}{0.08}=122.5$

The angular frequency of the spring is given by

$\omega=\sqrt{\frac{k}{m}}=\sqrt{122.5}=11.1 Hz$

And therefore, its period is

$T=\frac{2\pi}{\omega}=\frac{2\pi}{11.1}=0.6 s$

According to the equation of continuity, the volume flow rate must remain constant, so we can write

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-sectional area of the hose, with $r_1 = 5 mm$ being the radius of the hose

$v_1 = 4 m/s$ is the speed of the petrol in the hose

$A_2 = \pi r_2^2$ is the cross-sectional area of the nozzle, with $r_2$ being the radius of the nozzle

$v_2 = 16 m/s$ is the speed in the nozzle

Solving for $r_2$ , we find the radius of the nozzle:

$\pi r_1^2 v_1 = \pi r_2^2 v_2\\r_2 = r_1 \sqrt{\frac{v_1}{v_2}}=(5)\sqrt{\frac{4}{16}}=2.5 mm$

So, the diameter of the nozzle will be

$d_2 = 2r_2 = 2(2.5)=5.0 mm$

According to the Pascal principle, the pressure on the two pistons is the same, so we can write

$\frac{F_1}{A_1}=\frac{F_2}{A_2}$

where

$F_1$ is the force that must be applied to the small piston

$A_1 = \pi r_1^2$ is the area of the first piston, with $r_1= 2 cm$ being its radius

$F_2 = mg = (1500 kg)(9.8 m/s^2)=14700 N$ is the force applied on the bigger piston (the weight of the car)

$A_2 = \pi r_2^2$ is the area of the bigger piston, with $r_2= 15 cm$ being its radius

Solving for $F_1$ , we find

$F_1 = \frac{F_2A_1}{A_2}=\frac{F_2 \pi r_1^2}{\pi r_2^2}=\frac{(14700)(2)^2}{(15)^2}=261 N$

So, the closest answer is B) 266.7 N.

Learn more about pressure:

brainly.com/question/4868239

brainly.com/question/2438000

#LearnwithBrainly

5 0

3 years ago

What does velocity tell about movement

Kaylis [27]

<span>Velocity tells you what speed a moving object travels at and in what direction.</span>

3 0

3 years ago

Read 2 more answers

Which equation represents the law of conservation of energy in a closed system?

Irina-Kira [14]

Answer:

KE + PE = KE + PE

Explanation:

In a closed system, the mechanical energy of the system is constant.

Mechanical energy is given by the sum of kinetic energy and potential energy; mathematically:

U = KE + PE

where

KE is the kinetic energy

PE is the potential energy

This means that if we consider two situations, one at the beginning and one at the end, the value of U will not change if the system is closed; this means that the sum KE + PE will remain the same, so we can write:

KE + PE = KE + PE

7 0

3 years ago

Read 2 more answers