2. Connect concepts on the left with a description on the right.

An object is 30 cm in front of a converging lens with a focal length of 10 cm. Use ray tracing to determine the location of the

Aleksandr-060686 [28]

Answer:

i). Inverted

ii). Magnification of the image = -0.5

iii). Real

Explanation:

As shown in the ray diagram attached,

An object AB has been placed in front of converging lens (convex lens) at u = 30 cm.

F (Focus) of the lens is at 10 cm. So F = 10 cm

By analyzing the ray diagram we can measure the distance of the image on the other side of the lens (By counting the small blocks of the graph)

V = 15 cm

Characteristics of the image is:

i) Inverted

ii) Magnification of the image = $-\frac{v}{u}=-\frac{15}{30}$

= -0.5

ii) Real

4 0

3 years ago

2. State Newton's third law of motion.

Grace [21]

Answer:

Action and reaction are equal but act in opposite directions

4 0

3 years ago

Read 2 more answers

Answers are - A 235 N B 376 N C 271 N D 188 N E 470 N

amm1812

Answer:

C

Explanation:

8 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

BARSIC [14]

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

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