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4 years ago

13

Derek reads a rate of 25 words per min. How long would it take him to read a 225 work article if he read at a constant rate with

out stopping

1 answer:

atroni [7]4 years ago

6 0

Hello there! Your answer is (255/25=9) hope it helps and have a wonderful day!

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Vector A⃗ points in the positive y direction and has a magnitude of 12 m. Vector B⃗ has a magnitude of 33 m and points in the ne

gavmur [86]

vector A has magnitude 12 m and direction +y

so we can say

$\vec A = 12 \hat j$

vector B has magnitude 33 m and direction - x

$\vec B = -33 \hat i$

Now the resultant of vector A and B is given as

$\vec A + \vec B = 12 \hat j - 33 \hat i$

now for direction of the two vectors resultant will be given as

$\theta = tan^{-1}\frac{12}{-33}$

$\theta = 160 degree$

so it is inclined at 160 degree counterclockwise from + x axis

magnitude of A and B will be

$R = \sqrt{A^2 + B^2}$

$R = \sqrt{12^2 + 33^2} = 35.11 m$

so magnitude will be 35.11 m

6 0

3 years ago

A charge Q is distributed uniformly on a non-conducting ring of radius R and mass M. The ring is dropped from rest from a height

mihalych1998 [28]

Answer:

Check below for the explanation

Explanation:

Since it is stated that the ring is dropped from a height, h, through a non uniform magnetic field, two kinds of force will act on the ring, namely:

A magnetic force (that is non uniform since the field is non uniform)
Gravitational force

A certain amount of torque is provided by the non uniform magnetic force on the ring while the force gravity pulls it down. Due to the downward pull by the force of gravity on the ring and the torque acting on it as a result of the non uniform magnetic force, the ring begins to rotate.

5 0

3 years ago

A compass in a magnetic field will line up __________.

Anastaziya [24]

Line up in a direction parallel to the magnetic field lines<span />

5 0

3 years ago

Order the sounds made by these sources from highest to lowest pitch.

Trava [24]

Answer:

c

motorcycle, telephone, piano, lawn mower

4 0

3 years ago

7.22 Ignoring reflection at the air–water boundary, if the amplitude of a 1 GHz incident wave in air is 20 V/m at the water surf

Serga [27]

Answer:

z = 0.8 (approx)

Explanation:

given,

Amplitude of 1 GHz incident wave in air = 20 V/m

Water has,

μr = 1

at 1 GHz, r = 80 and σ = 1 S/m.

depth of water when amplitude is down to 1 μV/m

Intrinsic impedance of air = 120 π Ω

Intrinsic impedance of water = $\dfrac{120\pi}{\epsilon_r}$

Using equation to solve the problem

$E(z) = E_0 e^{-\alpha\ z}$

E(z) is the amplitude under water at z depth

E_o is the amplitude of wave on the surface of water

z is the depth under water

$\alpha = \dfrac{\sigma}{2}\sqrt{\dfrac{(120\pi)^2}{\Epsilon_r}}$

$\alpha = \dfrac{1}{2}\sqrt{\dfrac{(120\pi)^2}{80}}$

$\alpha =21.07\ Np/m$

now ,

$1 \times 10^{-6} = 20 e^{-21.07\times z}$

$e^{21.07\times z}= 20\times 10^{6}$

taking ln both side

21.07 x z = 16.81

z = 0.797

z = 0.8 (approx)

5 0

4 years ago