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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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I have a combination of myopia and presbyopia—overall, the power of my visual system is too large, but I also have a very limite

e-lub [12.9K]

Answer:

The range of powers is $- 5 \ D \le P \le - 2.667\ D$

Explanation:

From the question we are told that

The far point of the left eye is $n_f = 20 cm$

The near point of the left eye is $n = 15cm$

The near point with the glasses on is $n_g =25 \ cm$

From these parameter we can see that with the glass on that for near point the

Object distance would be $u = -25 \ cm$

Image distance would be $v = -15 \ cm$

To obtain the focal length we would apply the lens formula which is mathematically represented as

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

substituting values

$\frac{1}{f} = \frac{1}{-15} - \frac{1}{-25}$

$f = - \frac{75}{2} cm$

converting to meters

$f = - \frac{75}{2} * \frac{1}{100}$

$f = - \frac{75}{200} \ m$

Generally the power of the lens is mathematically represented as

$P = \frac{1}{f}$

Substituting values

$P = - \frac{200}{75} m$

$P = - 2.667 \ D$

From these parameter we can see that with the glass on that for far point the

Object distance would be $u_f = - \infty \ cm$

Image distance would be $v_f = -20 \ cm$

To obtain the focal length of the lens we would apply the lens formula which is mathematically represented as

$\frac{1}{f_f} = \frac{1}{v_f} - \frac{1}{u_f}$

substituting values

$\frac{1}{f} = \frac{1}{-20} - \frac{1}{- \infty}$

$\frac{1}{f} = \frac{1}{-20} - 0$

$f_f = \frac{20}{1} \ cm$

converting to meters

$f_f = - \frac{20}{1} * \frac{1}{100}$

Generally the power of the lens is mathematically represented as

$P = \frac{1}{f_f}$

Substituting values

$P = - \frac{100}{20} m$

$P = - 5 \ D$

This implies that the range of powers of the lens in his glass is

$- 5 \ D \le P \le - 2.667\ D$

3 0

3 years ago

A stretched string is 2.11 m long and has a mass of 19.5 g. When the string is oscillated at 440 Hz, which is the frequency of t

Katarina [22]

Answer:

efsfefsfsdf

Explanation:

efsfesef

7 0

3 years ago

A ball is thrown straight up with enough speed so that it is in the air for several seconds. Assume the positive direction is up

Andrew [12]

Answer:

a) v= 0 m/s b) v= 6.86 m/s

Explanation:

a) When the ball reaches to its highest point, under the influence of gravity, before starting to fall down, it momentarily comes to an stop (this is needed prior to change direction in any movement), so, applying the definition of acceleration, and replacing the acceleration a by g, we have:

vf = v₀ - g*t (1)

The minus sign means that the acceleration due to gravity is always downward, so if we assume that the positive direction is upwards it must be negative.

At the highest point, vf= 0.

b) Prior to solve this point, we need to know which is the time when the ball reaches to its highest point.

As we know vf=0, we can solve (1) for t, as follows:

th = v₀/g

Now, for a time that is 0.7 s before this time, applying the acceleration definition and solving for v again, we have:

v = v₀ -(g *(th-0.7 s)), but th= v₀/g, so we get:

v= v₀ -g((v₀/g)-0.7 s) = v₀ - v₀ + g*0.7 s

⇒ v=g*0.7 s = 9.8 m/s²*0.7 s

⇒ v = 6.86 m/s

6 0

3 years ago

In a historical movie, two knights on horseback start from rest 86 m apart and ride directly toward each other to do battle. Sir

Harlamova29_29 [7]

Answer:

Relative to Sir George's starting point, the knights collide at a distance of 38.43 m from Sir George's starting point.

Explanation:

Let the distance covered by Sir George be $S_{1}$

and the distance covered by Sir Alfred be $S_{2}$

Since the knights collide, hence they must have traveled for the same amount of time just before collision

From one of the equations of motion for linear motion

$S = ut + \frac{1}{2}at^{2}$

Where $S$ is the distance traveled

$u$ is the initial velocity

$a$ is the acceleration

and $t$ is the time

For Sir George,

$S = S_{1}$

$u$ = 0 m/s (Since they start from rest)

$a =$ 0.21 m/s²

Hence,

$S = ut + \frac{1}{2}at^{2}$ becomes

$S_{1} = (0)t + \frac{1}{2}(0.21)t^{2}\\S_{1} = 0.105 t^{2}\\$

$t^{2} = \frac{S_{1}}{0.105}$

Now, for Sir Alfred

$S = S_{2}$

$u$ = 0 m/s (Since they start from rest)

$a =$ 0.26 m/s²

Hence,

$S = ut + \frac{1}{2}at^{2}$ becomes

$S_{2} = (0)t + \frac{1}{2}(0.26)t^{2}\\S_{2} = 0.13 t^{2}\\$

$t^{2} = \frac{S_{2}}{0.13}$

Since, they traveled for the same time, $t$ just before collision, we can write

$\frac{S_{1}}{0.105}= \frac{S_{2}}{0.13}$

Since, the two nights are 86 m apart, that is, the sum of the distances covered by the knights just before collision is 86 m. Then we can write that

$S_{1} + S_{2} = 86$ m

Then, $S_{2} = 86 - S_{1}$

Then,

$\frac{S_{1}}{0.105}= \frac{S_{2}}{0.13}$ becomes

$\frac{S_{1}}{0.105}= \frac{86 -S_{1}}{0.13}$

$0.13{S_{1}}= 0.105({86 -S_{1}})\\0.13{S_{1}}= 9.03 - 0.105S_{1}}\\0.13{S_{1}} + 0.105S_{1}}= 9.03 \\0.235{S_{1}} = 9.03\\{S_{1}} =\frac{9.03}{0.235}$