3 years ago

How many words can be made with five letters

Physics

2 answers:

dangina [55]3 years ago

6 0

The last word is gear

Firdavs [7]3 years ago

4 0

Answer:

agree eager eagre ragee

Explanation:

hope this help

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If an object has a kinetic energy of 30 j and mass of 34kg how fast is the object moving ?

cricket20 [7]

Ek = (m*V^2) / 2 where m is mass and V is speed, then we can take this equation and manipulate it a little to isolate the speed.

Ek = mv^2 / 2 — multiply both sides by 2

2Ek = mv^2 — divide both sides by m

2Ek / m = V^2 — switch sides

V^2 = 2Ek / m — plug in values

V^2 = 2*30J / 34kg

V^2 = 60J/34kg

V^2 = 1.76 m/s — sqrt of both sides

V = sqrt(1.76)

V = 1.32m/s (roughly)

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3 years ago

Kenneth was preparing a meal for his family. As he was prepping the kitchen, his mother asked him, “Do you need help? I know coo

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According to e2020, the answer is reduced performance due to stereotype threat.

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3 years ago

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A power plant to represent the need for and use for energy; label it energy production.

meriva

I’m confused about this question

6 0

3 years ago

The position of a particle is given by ~r(t) = (3.0 t2 ˆi + 5.0 ˆj j 6.0 t kˆ) m

Julli [10]

Answer:

$v=(6ti+6k)\ m/s$

Explanation:

Given that,

The position of a particle is given by :

$r(t) = (3.0 t^2 i + 5.0j+ 6.0 tk) m$

Let us assume we need to find its velocity.

We know that,

$v=\dfrac{dr}{dt}\\\\=\dfrac{d}{dt}(3.0 t^2 i + 5.0j+ 6.0 tk) \\\\=(6ti+6k)\ m/s$

So, the velocity of the particle is $(6ti+6k)\ m/s$ .

5 0

3 years ago

The distance between the objective and eyepiece lenses in a microscope is 19 cm . The objective lens has a focal length of 5.5 m

kramer

Answer:

The focal length of eyepiece is 3.68 cm.

Explanation:

Given that,

Distance = 19 cm

Focal length = 5.5 mm

Magnification = 200

Object distance = -25 cm

We need to calculate the focal length

Using formula of magnification

$m=\dfrac{d}{f_{o}}+\dfrac{-25}{f_{0}f_{e}}$

Put the value into the formula

$f_{e}=\dfrac{19\times(-25)}{.55(-200-\dfrac{19}{.55})}$

$f_{e}=3.68\ cm$

Hence, The focal length of eyepiece is 3.68 cm.

3 0

4 years ago