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2 years ago

Tell something about the series of miraculous tales of ladybug and catnoir

2 answers:

8 0

Answer:Coccinellidae is a widespread family of small beetles ranging in size from 0.8 to 18 mm. The family is commonly known as ladybugs in North America and ladybirds in Great Britain and other parts of the English-speaking world.

Explanation: Uh yea.

MrRissso [65]2 years ago

7 0

Answer:

powers of magical creatures Kwami, they transform into the superheroes Ladybug and Cat Noir. With a mission to protect the city of Paris from the evil Hawk Moth, the two of them must learn to cope with their new responsibilities, as well as ordinary troubles involving love, school, family, and friends.

And Yes I like Miraculous :D

The last question I search it up it says:

On 7 September 2019, it was confirmed by Jeremy on his Insta that seasons 4 and 5 are on the way and air date for season 4 was late 2020, but it was pushed to 2021, due to the effects of the corana . On 13 October 2019, Thomas announced that scripts for season 4 have been done.

The sixth season of Miraculous: Tales of Ladybug & Cat Noir was confirmed to be planned. It will consist of 26 episodes and is set to premiere in 2023.

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1. A Faraday ice pail similar to the one you will use in the experiment is used in this question. The principle of its operation

algol [13]

Answer:

your mommy

Explanation:

YOUR MAMMA

I LOVE MY DADDY

HE DOES ME REAL HARD

8 0

3 years ago

Please I need 14, 15, and 16

Finger [1]

15:) using more force in your muscle will increase the force used to bounce the basketball

16:) the pulling of gravity livitation does not allow the ball to go back up with the hieght it was dropped from on the scientifical drop point

14:) <span>a weight hung from a fixed point so that it can swing freely backward and forward, especially a rod with a weight at the end that regulates the mechanism of a clock that is the deffinition of to which of the word pendulum read it do not plagarize and i hope ii helped and have a great day bye.)::</span>

8 0

3 years ago

If a 100-N net force acts on a 50-kg car, what will the acceleration of the car be?

Tema [17]

Newton's 2nd law of motion:

   Force = (mass) x (acceleration)

Divide each side by (mass):

   Acceleration = (force) / (mass)

   = (100 N) / (50 kg)

   = 2 m/s²

5 0

3 years ago

The curve below shows the percentage of population of aquatic species that die in response to doses of pollutant A:

Elis [28]

Based on the trend produced by the dose - response graph, it would be best to evacuate the residents in other to prevent the increasing percentage of deaths due to the rising level of pollutant A.

The curve shows that the pollutant level in mg/kg of pollutant A is still increasing, hence, groundwater monitoring alone won't be the best decision to make.

Since the pollutant level is still increasing, then the spill level still need effective monitoring.

Evacuation of residents seems to be the best decision that should be taken based on the information interpreted on the graph.

Therefore, Evacuating residents to prevent rising death percentage is required as the pollutant level is yet to subside.

Learn more :brainly.com/question/24844489

8 0

2 years ago

A 4.9-MeV (kinetic energy) proton enters a 0.28-T field, in a plane perpendicular to the field. Part APart complete What is the

BartSMP [9]

Answer:

$r=1.14m$

Explanation:

$\theta$ is the angle between the velocity and the magnetic field. So, the magnetic force on the proton is:

$F_m=qvBsen\theta\\F_m=qvBsen(90^\circ)\\F_m=qvB$

A charged particle describes a semicircle in a uniform magnetic field. Therefore, applying Newton's second law to uniform circular motion:

$F_m=F_c\\qvB=F_c(1)$

$F_c$ is the centripetal force and is defined as:

$F_c=m\frac{v^2}{r}$

Here $v$ is the proton's speed and $r$ is the radius of the circular motion. Replacing this in (1) and solving for r:

$qvB=\frac{mv^2}{r}\\r=\frac{mv^2}{qvB}\\r=\frac{mv}{qB}$

Recall that 1 J is equal to $6.242*10^{12}MeV$ , so:

$4.9MeV*\frac{1J}{6.242*10^{12}MeV}=7.85*10^{-13}J$

We can calculate $v$ from the kinetic energy of the proton:

$K=\frac{mv^2}{2}\\\\v=\sqrt{\frac{2K}{m}}\\v=\sqrt{\frac{2(7.85*10^{-13}J)}{1.67*10^{-27}kg}}\\v=3.06*10^{7}\frac{m}{s}$

Finally, we calculate the radius of the proton path:

$r=\frac{mv}{qB}\\r=\frac{1.67*10^{-27}kg(3.06*10^{7}\frac{m}{s})}{1.6*10^{-19}C(0.28T)}\\r=1.14m$

8 0

3 years ago