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Sunitha can type 1800 words in half an hour. What is her typing speed in words per minute?

Andre45 [30]

Answer:

60words/minute

Explanation:

If Sunitha can type 1800 words in half an hour, this can be expressed as;

1800 words = 30 minutes

To get her typing speed per minute, we will use the formula

Speed = Number of words/Time used

Typing speed = 1800/30

Typing speed = 60words/minute

Hence her typing speed in words per minute is 60words/minute

6 0

2 years ago

True or False—An external force is defined as a force generated outside the system of interest that acts on an object inside the

inn [45]

Answer:

An external force is a force that acts on an object within the system from outside the system. This type of force is different than an internal force, which acts between two objects that are both within the system. The net external force combines these two definitions; it is the total combined external force

Explanation:

ig the answer is true

5 0

3 years ago

Fiber optics are an important part of our modern internet. In these fibers, two different glasses are used to confine the light

professor190 [17]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

$\theta_{max} =18.38^o$

b

New $n_{cladding} =1.491$

Explanation:

From the question we are told that

The refractive index of the core is $n_{core} = 1.497$

The refractive index of the cladding is $n_{cladding} = 1.421$

Generally according to Snell's law

$n_{core} * sin(90- \theta) = n_{cladding} * sin (90)$

Where $\theta_{max}$ is the largest angle a largest angle a ray will make with respect to the interface of the fiber and experience total internal reflection

$\theta_{max} = 90 - sin^{-1} [\frac{n_{cladding}}{n_{core}} ]$

$\theta_{max} = 90 - sin^{-1} [\frac{1.421}{1.497}} ]$

$\theta_{max} =18.38^o$

Given from the question the the largest angle is 5°

Generally the refraction index of the cladding is mathematically represented as

$n_{cladding} = n_{core} * sin (90 - 5)$

$n_{cladding} =1.491$

5 0

3 years ago

A rabbit is hopping along at an approximately constant speed of 3.9 m/s. The rabbit passes a crouched cat ready to chase the rab

Firlakuza [10]

Answer:

t = 7,8 s

Explanation:

From the instant, the rabbit passes the cat. The cat star running acceleration of 0,5 m/s² .

When the cat arrives at the speed of 3,9 m/s the cat catches the rabbit

Then for the cat arrives at 3,9 m/s nedds

v = vo + a*t vo = 0 then v = a*t

3,9 ( m/s) = 0,5 ( m/s² ) * t

t = 7,8 s

v = 3,9 m/s =

4 0

2 years ago

A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its sur

muminat

Answer:

a) E = 0

b) $E = \dfrac{k_e \cdot q}{ r^2 }$

Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) $\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}$

From which we have;

$E \cdot A = \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}} = \dfrac{0}{\varepsilon _{0}} = 0$

E = 0/A = 0

E = 0

b) $\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}$

$E \cdot A = \dfrac{+q }{\varepsilon _{0}}$

$E = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}$

By Gauss theorem, we have;

$E\oint dS = \dfrac{q}{\varepsilon _{0}}$

Therefore, we get;

$E \cdot (4 \cdot \pi \cdot r^2) = \dfrac{q}{\varepsilon _{0}}$

The electrical field outside the spherical shell

$E = \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }= \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }$

$k_e= \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }$

Therefore, we have;

$E = \dfrac{k_e \cdot q}{ r^2 }$

5 0

3 years ago