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Romashka [77]
4 years ago
13

an object 8.25 cm from a lens creates a virtual image of magnification 2.40 what is the focal length of the lens (mind your minu

s sign) (unit=cm)
Physics
1 answer:
murzikaleks [220]4 years ago
5 0

Answer:

Focal length of the lens is 14.14 cm

Explanation:

As we know that the image formed due to lens is virtual image

so here we have

M = 2.40

now we have

\frac{d_i}{d_o} = 2.40

now we have

distance of object is 8.25 cm

so we have

d_i = 8.25(2.40) = 19.8 cm

now by lens formula

\frac{1}{d_i} - \frac{1}{d_o} = \frac{1}{f}

\frac{1}{-19.8} - \frac{1}{-8.25} = \frac{1}{f}

f = 14.14 cm

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Psychology:
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Bosons &amp; Fermions A new type of quantum object has been discovered. When a large collection of these are cooled to almost ab
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Answer:

Only 2,3,4 are true

Explanation:

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5 0
3 years ago
The angle between the two force of magnitude 20N and 15N is 60 degrees (20N force being horizontal) determine the resultant in m
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A) The resultant force is 30.4 N at 25.3^{\circ}

B) The resultant force is 18.7 N at 43.9^{\circ}

Explanation:

A)

In order to find the resultant of the two forces, we must resolve each force along the x- and y- direction, and then add the components along each direction to find the components of the resultant.

The two forces are:

F_1 = 20 N at 0^{\circ} above x-axis

F_2 = 15 N at 60^{\circ} above y-axis

Resolving each force:

F_{1x}=F_1 cos \theta = (20)(cos 0)=20 N\\F_{1y}=F_1 sin \theta =(20)(sin 0)=0 N

F_{2x}=F_2 cos \theta = (15)(cos 60)=7.5 N\\F_{2y}=F_2 sin \theta =(15)(sin 60)=13.0 N

So, the components of the resultant are:

F_x = F_{1x}+F_{2x}=20+7.5 = 27.5 N\\F_y = F_{1y}+F_{2y}=0+13.0=13.0 N

And the magnitude of the resultant is:

F=\sqrt{F_x^2+F_y^2}=\sqrt{27.5^2+13.0^2}=30.4 N

And the direction is:

\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{13.0}{27.5})=25.3^{\circ}

B)

In this case, the 15 N is applied in the opposite direction to the 20 N force. Therefore we need to re-calculate its components, keeping in mind that the angle of the 15 N force this time is

\theta=180^{\circ}-60^{\circ}=120^{\circ}

So we have:

F_{2x}=F_2 cos \theta = (15)(cos 120)=-7.5 N\\F_{2y}=F_2 sin \theta =(15)(sin 120)=13.0 N

So, the components of the resultant this time are:

F_x = F_{1x}+F_{2x}=20-7.5 = 12.5 N\\F_y = F_{1y}+F_{2y}=0+13.0=13.0 N

And the magnitude is:

F=\sqrt{F_x^2+F_y^2}=\sqrt{13.5^2+13.0^2}=18.7 N

And the direction is:

\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{13.0}{13.5})=43.9^{\circ}

Learn more about vector addition:

brainly.com/question/4945130

brainly.com/question/5892298

#LearnwithBrainly

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