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lozanna [386]
3 years ago
5

Which of the following is not an appropriate category of childrens book to include in the early childhood classroom?

Physics
2 answers:
Katarina [22]3 years ago
5 0

Answer:

A. Chapter Book is the correct answer.

Explanation:

If the desired category is children's book, it is important to take in consideration that it has to be something that attracts the child's or children's attention. A chapter book might be very long for a young person; on the other hand, a picture book, a concept book, and a counting book can help the children learn and keep their attention.

Anon25 [30]3 years ago
3 0

Answer:

Chapter books because counting, picture, and concept books all have a basic and simple theme and have more simpler words than a chapter book.

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g An electron enters a region of space containing a uniform 1.63 × 10 − 5 T magnetic field. Its speed is 121 m/s and it enters p
kolbaska11 [484]

Answer:

i. The radius 'r' of the electron's path is 4.23 × 10^{-5} m.

ii. The frequency 'f' of the motion is 455.44 KHz.

Explanation:

The radius 'r' of the electron's path is called a gyroradius. Gyroradius is the radius of the circular motion of a charged particle in the presence of a uniform magnetic field.

                 r = \frac{mv}{qB}

Where: B is the strength magnetic field, q is the charge, v is its velocity and m is the mass of the particle.

From the question, B = 1.63 × 10^{-5}T, v = 121 m/s, Θ = 90^{0} (since it enters perpendicularly to the field), q = e  = 1.6 × 10^{-19}C and m = 9.11 × 10^{-31}Kg.

Thus,

         r = \frac{mv}{qB} ÷ sinΘ

But,  sinΘ =  sin 90^{0} = 1.

So that;

          r = \frac{mv}{qB}

            = (9.11 × 10^{-31} × 121) ÷ (1.6 × 10^{-19}  × 1.63 × 10^{-5})

            = 1.10231 × 10^{-28}   ÷ 2.608 × 10^{-24}

            = 4.2266 × 10^{-5}

            = 4.23 × 10^{-5} m

The radius 'r' of the electron's path is 4.23 × 10^{-5} m.

B. The frequency 'f' of the motion is called cyclotron frequency;

           f = \frac{qB}{2\pi m}

             =  (1.6 × 10^{-19}  × 1.63 × 10^{-5}) ÷ (2 ×\frac{22}{7} × 9.11 × 10^{-31})

             =  2.608 × 10^{-24} ÷  5.7263 × 10^{-30}

             = 455442.4323

          f  = 455.44 KHz

The frequency 'f' of the motion is 455.44 KHz.

3 0
3 years ago
Read 2 more answers
Average velocity is different than average speed because calculating average velocity involves
borishaifa [10]

Answer:

average velocity include total displacement whereas average speed include total distance

6 0
3 years ago
A rock is thrown with a force of 500 N and an acceleration is 75 m/s^2. What is its mass?
artcher [175]

Answer:

We conclude that the mass of a rock with a force of 500 N and an acceleration of 75 m/s² is 6.7 kg.

Hence, option D is correct.

Explanation:

Given

  • Force F = 500 N
  • Acceleration a = 75 m/s²

To determine

Mass m = ?

Important Tip:

  • The mass of a rock can be found using the formula F = ma

Using the formula

F = ma

where

  • F is the force (N)
  • m is the mass (kg)
  • a is the acceleration (m/s²)

now substituting F = 500, and a = 75 m/s² in the formula

F = ma

500 = m(75)

switch sides

m\left(75\right)=500

Divide both sides by 75

\frac{m\cdot \:75}{75}=\frac{500}{75}

simplify

m=\frac{20}{3}

m=6.7 kg

Therefore, we conclude that the mass of a rock with a force of 500 N and an acceleration of 75 m/s² is 6.7 kg.

Hence, option D is correct.

7 0
3 years ago
What newton equal to in terms of units of mass and acceleration
Korvikt [17]
This equation is one of the most useful in classical physics. It is a concise statement of Isaac Newton's<span> Second Law of Motion, holding both the proportions and vectors of the Second Law. It translates as: The net force on an object is </span>equal<span> to the </span>mass<span>of the object multiplied by the </span>acceleration<span> of the object.</span>
4 0
3 years ago
Read 2 more answers
A redecor travelling of 94 m/s s lows at a anstant
kotykmax [81]

Answer:

638 m.

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 94 m/s

Final velocity (v) = 22 m/s

Time (t) = 11 s

Distance (s) =?

We can obtain the distance travelled by using the following formula:

s = (u + v) t /2

s = (94 + 22) × 11 /2

s = 116 × 11 /2

s = 1276 /2

s = 638 m

Thus, the distance travelled is 638 m.

3 0
3 years ago
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