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2 years ago

11

A coil with 55 loops has its flux change from 0 Wb to 0.0266 Wb in a certain amount of time, generating 5.22 V of EMF. How much

time elapsed?

1 answer:

Eduardwww [97]2 years ago

4 0

Answer:

0.280 s

Explanation:

I set it up as 5.22=(55)(0.0266)/x and then solved for x to be 2.80.

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dangina [55]

Answer:

C 2000v its obviously ans because if o is 1000 2 vo is 2000v

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2 years ago

A disk of mass M and radius R rotates at angular velocity ω0. Another disk of mass M and radius r is dropped on top of the rotat

AleksandrR [38]

Answer:

$\omega = \frac{(R^2\omega_o}{(R^2 + r^2)}$

Explanation:

As we know that there is no external torque on the system of two disc

then the angular momentum of the system will remains conserved

So we will have

$L_i = L_f$

now we have

$L_i = (\frac{1}{2}MR^2)\omega_o$

also we have

$L_f = (\frac{1}{2}MR^2 + \frac{1}{2}Mr^2)\omega$

now from above equation we have

$(\frac{1}{2}MR^2)\omega_o = (\frac{1}{2}MR^2 + \frac{1}{2}Mr^2)\omega$

now we have

$\omega = \frac{MR^2\omega_o}{(MR^2 + Mr^2)}$

$\omega = \frac{(R^2\omega_o}{(R^2 + r^2)}$

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3 years ago

An object is placed in front of a diverging lens, such that the object-to-image distance is 71 cm.

Pachacha [2.7K]

Explanation:

Given that,

Object-to-image distance d= 71 cm

Image distance = 26 cm

We need to calculate the object distance

$u -v= d$

$u=71+26=97\ cm$

We need to calculate the focal length

Using formula of lens

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

put the value into the formula

$\dfrac{1}{f}=\dfrac{1}{-26}+\dfrac{1}{97}$

$\dfrac{1}{f}=-\dfrac{71}{2522}$

$f=-35.52\ cm$

The focal length of the lens is 35.52.

(B). Given that,

Object distance = 95 cm

Focal length = 29 cm

We need to calculate the distance of the image

Using formula of lens

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

Put the value in to the formula

$\dfrac{1}{-29}=\dfrac{1}{v}-\dfrac{1}{95}$

$\dfrac{1}{v}=\dfrac{1}{-29}-\dfrac{1}{95}$

$\dfrac{1}{v}=-\dfrac{124}{2755}$

$v=-22.21\ cm$

We need to calculate the magnification

Using formula of magnification

$m=\dfrac{v}{u}$

$m=\dfrac{22.21}{95}$

$m=0.233$

The magnification is 0.233.

The image is virtual.

Hence, This is the required solution.

4 0

3 years ago

The height of a circular tower is 179 meters and its diameter is 48 meters. (Assume that the building is a perfect cylinder). Wh

EastWind [94]

The definition of a scale of 1: 166 will mean that the scale of 1 in the model will be equivalent to 166 times the measurement in the real model, therefore we will have that the height would be 166 times smaller than the 179m given:

$h = \frac{179m}{166} = 1.078m = 107.8cm$

The same for the diameter,

$\phi = \frac{48m}{166}= 0.2891m = 28.91cm$

The volume of a cylinder is given as

$V = (\pi r^2)(h)$

$V = (\pi (\frac{d}{2})^2)(h)$

$V = (\pi (\frac{ 28.91}{2})^2)(107.8)$

$V = 70762.8cm^3$

Therefore the volume would be $V = 70762.8cm^3$

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3 years ago

An 800 N man stands on a scale in a motionless elevator. When the elevator begins to move, the scale

DENIUS [597]

Answer:

Net force = 800 – 650

= 150 N

150 = (800 ÷ 9.8) a

a = 1470 ÷ 800

= 1.8375 m/s^2, downwards

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3 years ago