A В C D At which point will an image be formed? ОА Ов Ос OD

17- Find the magnitude of vector product \BxĀ| for A=– 23 +3Â and B = 2î – 3+ Å vectors.

aksik [14]

Answer:

alam ko sagot pero mataas

8 0

3 years ago

A 1.0-cm-tall object is 13 cm in front of a converging lens that has a 40 cm focal length.

kicyunya [14]

A) Image position: -19.3 cm

B) Image height: 1.5 cm, upright

Explanation:

A)

In order to calculate the image position, we can use the lens equation:

$\frac{1}{p}+\frac{1}{q}=\frac{1}{f}$

where

p is the distance of the object from the lens

q is the distance of the image from the lens

f is the focal length

In this problem, we have:

p = 13 cm (object distance)

f = 40 cm (focal length, positive for a converging lens)

So the image distance is

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{40}-\frac{1}{13}=-0.0519\\q=\frac{1}{-0.0519}=-19.3 cm$

The negative sign means that the image is virtual.

B)

In order to calculate the image height, we use the magnification equation:

$\frac{y'}{y}=-\frac{q}{p}$

where

y' is the image height

y is the object height

In this problem, we have:

y = 1.0 cm (object height)

p = 13 cm

q = -19.3 cm

Therefore, the image heigth is

$y'=-\frac{qy}{p}=-\frac{(-19.3)(1.0)}{13}=1.5 cm$

And the positive sign means the image is upright.

6 0

3 years ago

which words are readable red word on white paper in red light or blue words on white paper in blue light

Angelina_Jolie [31]

Answer:

second

Explanation:

the first let's everything appear red, the send would produce black(-ish) words on paper that's appealing blue

6 0

3 years ago

Read 2 more answers

Distance Between Two Cars A car leaves an intersection traveling west. Its position 4 sec later is 19 ft from the intersection.

faltersainse [42]

Answer:

The answer to the question is;

The rate at which the distance between the two cars is changing is equal to 14.4 ft/sec.

Explanation:

We note that the distance traveled by each car after 4 seconds is

Car A = 19 ft in the west direction.

Car B = 26 ft in the north direction

The distance between the two cars is given by the length of the hypotenuse side of a right angled triangle with the north being the y coordinate and the west being the x coordinate.

Therefore, let the distance between the two cars be s

we have

s² = x² + y²

= (19 ft)² + (26 ft)² = 1037 ft²

s = $\sqrt{1037 ft^2}$ = 32.202 ft.

The rate of change of the distance from their location 4 seconds after they commenced their journeys is given by;

Since s² = x² + y² we have

$\frac{ds^{2} }{dt} = \frac{dx^{2} }{dt} + \frac{dy^{2} }{dt}$

→ $2s\frac{ds }{dt} = 2x\frac{dx}{dt} + 2y\frac{dy }{dt}$ which gives

$s\frac{ds }{dt} = x\frac{dx}{dt} + y\frac{dy }{dt}$

We note that the speeds of the cars were given as

Car B moving north = 12 ft/sec, which is the y direction and

Car A moving west = 8 ft/sec which is the x direction.

Therefore

$\frac{dy }{dt}$ = 12 ft/sec and

$\frac{dx}{dt}$ = 8 ft/sec

$s\frac{ds }{dt} = x\frac{dx}{dt} + y\frac{dy }{dt}$ becomes

$32.202 ft.\frac{ds }{dt} = 19 ft \times 8 \frac{ft}{sec} + 26ft\times 12\frac{ft}{sec}$ = 464 ft²/sec

$\frac{ds }{dt} = \frac{464\frac{ft^{2} }{sec} }{32.202 ft.}$ = 14.409 ft/sec ≈ 14.4 ft/sec to one place of decimal.

3 0

3 years ago

An office window has dimensions 3.5 m by 2.1 m. As a result of the passage of a storm, the outside air pressure drops to 0.96 at

Nina [5.8K]

Answer:

Force=29789.55 N

Explanation:

Given data

dimension=3.5m by 2.1m

P(outside pressure)=0.96 atm

P(inside pressure)=1.00 atm

Force=?

Solution

Area=L×W

A=(3.5)×(2.1)

$A=7.35m^{2}$

As we know

F=PA

For P(pressure)

P=P(inside)-P(outside)

P=1.00-0.96

P=0.04 atm

convert atm to pascal (pa) we get

P=4053 pa

F=PA

F=(4053)×(7.35)

F=29789.55 N

8 0

3 years ago