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3 years ago

9

While looking at bacteria under a microscope you decide to take a break when u return the amount of bacteria has doubled in size

your break probably only took

1 answer:

VikaD [51]3 years ago

4 0

Your break probably only took 20-30 minutes. Bacteria can double every 20-30 minutes. Hope this helped!
~ Sincerely,
Cutenerd1234
XOXO

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If the Mass Number is the sum of the protons and neutrons in an atom, what is<br> the mass number of

gogolik [260]

Answer:

It's 31

Explanation:

add 15 and 16

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3 years ago

A 600kg car is moving at 5 m/s to the right and elastically collided with a stationary 900 kg car. What is the velocity of the 9

11111nata11111 [884]

Answer:

$\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}$

Explanation:

It says “Momentum before the collision is equal to momentum after the collision.” Elastic Collision formula is applied to calculate the mass or velocity of the elastic bodies.

$m_{1} v_{1}=m_{2} v_{2}$

$\mathrm{m}_{1} \text { and } \mathrm{m}_{2} \text { are masses of the object }$

$\mathrm{v}_{1} \text { velocity before the collision }$

$\mathrm{v}_{2} \text { velocity after the collision }$

$\mathrm{m}_{1}=600 \mathrm{kg}$

$\mathrm{m}_{2}=900 \mathrm{kg}$

$\text { Velocity before the collision } v_{1}=5 \mathrm{m} / \mathrm{s}$

$600 \times 5=900 \times v_{2}$

$3000=900 \times v_{2}$

$\mathrm{v}_{2}=\frac{3000}{900}$

$\mathrm{v}_{2}=3.3 \mathrm{m} / \mathrm{s}$

$\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}$

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4 years ago

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the s

yawa3891 [41]

Hello!

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the spring ?

Data:

$E_{pe}\:(elastic\:potential\:energy) = 5184\:J$

$K\:(constant) = 16200\:N/m$

$x\:(displacement) =\:?$

For a spring (or an elastic), the elastic potential energy is calculated by the following expression:

$E_{pe} = \dfrac{k*x^2}{2}$

Where k represents the elastic constant of the spring (or elastic) and x the deformation or displacement suffered by the spring.

Solving:

$E_{pe} = \dfrac{k*x^2}{2}$

$5184 = \dfrac{16200*x^2}{2}$

$5184*2 = 16200*x^2$

$10368 = 16200\:x^2$

$16200\:x^2 = 10368$

$x^{2} = \dfrac{10368}{16200}$

$x^{2} = 0.64$

$x = \sqrt{0.64}$

$\boxed{\boxed{x = 0.8\:m}}\end{array}}\qquad\checkmark$

Answer:

The displacement of the spring = 0.8 m

_______________________________

I Hope this helps, greetings ... Dexteright02! =)

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4 years ago

Dr. Matthews has submitted a proposal to the institutional review board (IRB) of a university. At this university, she intends t

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3 years ago

Joe drove at the speed of 45 miles per hour for a certain distance. He then drove at the speed of 55 miles per hour for the same

Snowcat [4.5K]

Answer:

$v_{avg} = 49.5 mph$

Explanation:

Let the distance moved by Joe is "d"

so the time taken by him to drove it by speed 45 mph is given as

$t_1 = \frac{d}{v_1}$

$t_1 = \frac{d}{45}$

now the same distance is traveled by him with speed 55 mph

so the time taken by him

$t_2 = \frac{d}{55}$

so total time taken by him for complete distance 2d

$t = t_1 + t_2$

$t = \frac{d}{45} + \frac{d}{55}$

$t = 0.0404 d$

now the average speed is given as

$v_{avg} = \frac{2d}{t}$

$v_{avg} = \frac{2d}{0.0404d}$

$v_{avg} = 49.5 mph$

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3 years ago

Read 2 more answers