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3 years ago

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

1 answer:

Pachacha [2.7K]3 years ago

4 0

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.

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An electric motor rotating a workshop grinding wheel at a rate of 151 rev/min is switched off. Assume constant angular decelerat

tamaranim1 [39]

Answer:

Explanation:

Initial angular velocity ω₀ = 151 x 2π / 60

= 15.8 rad /s

final velocity = 0

Angular deceleration α = 2.23 rad / s

ω² = ω₀² - 2 α θ

0 = 15.8² - 2 x 2.23 θ

= 55.99 rad

one revolution = 2π radian

55.99 radian = 55.99 / 2 π no of terns

= 9 approx .

8 0

3 years ago

A block (mass = 61.2 kg) is hanging from a massless cord that is wrapped around a pulley (moment of inertia = 1/2MR2 kg · m2, wh

kolezko [41]

Answer:

The angular velocity is $w = 53.35 \ rounds /minute$

Explanation:

From the question we are told that

The mass of the block is $m = 61.2kg$

The of the pulley is $M = 14.2 kg$

The radius of the pulley is $R = 1.5m$

The radius of the cord around the pulley is $r = 1.5 m$

The distance of the block to the floor is $d = 8.0 m$

From the question we are told that the moment of inertia of the pulley is

$I = \frac{1}{2} MR^2 kg \cdot m^2$

Substituting value

$I = \frac{1}{2} * 14.2 * (1.5)^2$

$I = 15.975 kg \cdot m^2$

Using the Newtons law we can express the force acting on the vertical axis as

$ma = mg -T$

=> $T = mg -ma$

Now when the pulley is rotated that torque generated on the massless cord as a r result of the tension T and the radius of the cord around the pulley is mathematically represented as

$\tau = I \alpha$

Here $\alpha$ is the angular acceleration

Here $\tau$ is the torque which can be equivalent to

$\tau = T r$

Substituting this above

$Tr = I \alpha$

Substituting for T

$(mg - ma ) r = I\ r \alpha$

Here $a$ is the linear acceleration which is mathematically represented as

$a = r\alpha$

$(mg - m(r\alpha ) ) r = I\ r \alpha$

$mgr = I\alpha + m(r\alpha ) r$

$mgr = \alpha [ I + mr^2]$

making $\alpha$ the subject

$\alpha = \frac{mgr}{I -mr ^2}$

Substituting values

$\alpha = \frac{61.2 * 1.5 * 9.8}{15.975 + (61.2 ) * (1.5)^2}$

$\alpha =5.854 rad /s^2$

Now substituting into the equation above to obtain the acceleration

$a = 5.854 * 1.5$

$a=8.78 m/s^2$

This acceleration is $a = \frac{v}{t}$

and v is the linear velocity with is mathematically represented as

$v = \frac{d}{t}$

Substituting this into the formula acceleration

$a = \frac{d}{t^2}$

making t the subject

$t = \sqrt{\frac{d}{a} }$

substituting value

$t = \sqrt{\frac{8}{8.78}}$

$t = 0.9545 \ s$

Now the linear velocity is

$v = \frac{8}{0.9545}$

$v = 8.38 m/s$

The angular velocity is

$w = \frac{v}{r}$

So

$w = \frac{8.38}{1.5}$

$w = 5.59 rad/s$

Generally 1 radian is equal to 0.159155 rounds or turns

So 5.59 radian is equal to x

Now x is mathematically obtained as

$x = \frac{5.59 * 0.159155}{1}$

$= 0.8892 \ rounds$

Also

60 second = 1 minute

So 1 second = z

Now z is mathematically obtained as

$z = \frac{ 1}{60}$

z $= 0.01667 \ minute$

Therefore

$w = \frac{0.8892}{0.01667}$

$w = 53.35 \ rounds /minute$

8 0

3 years ago

PLEASE HELP ME SOMEONE

sasho [114]

I think its B or D, most likely D.

7 0

3 years ago

PLEASE HELP, NO ONE IS HELPING MEEEE :(

Blizzard [7]

Answer:

probably the trip where it took u 5 seconds

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What is the formular for lmpulse

Brums [2.3K]

Answer:

the change in momentum = Force x change in time

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2 years ago

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