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4 years ago

7

Which of the following is not true of a form?

1 answer:

Ludmilka [50]4 years ago

4 0

Answer:

It can only display one record at a time

Explanation:

Form ;

1. This is a document with spaces (also called placeholders or fields ) in which a series of documents with similar content can be written or selected.

2.This is the most popular method of data entry

3.It may contain images in the background.

4.This can be sorted data regardless of its source of information.

Only option C is wrong.

Therefore the answer C is correct.

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Can someone please help me

Zarrin [17]

They’re at a velocity of 2.0 m/s

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3 years ago

WILL MARK BRAINLIEST <br> How do you convert Farenheit to Celcius<br> and Celcius to Farenheit

olya-2409 [2.1K]

Answer:

F= $\frac{9}{5}C+32$

C= $\frac{5}{9}(F-32)$

F stands for farenheit

C stands for celcius

Explanation:

3 0

4 years ago

Read 2 more answers

A rotating space station is said to create "artificial gravity" - a loosely-defined term used for an acceleration that would be

FrozenT [24]

Answer:

$\omega=0.31\frac{rad}{s}$

Explanation:

The artificial gravity generated by the rotating space station is the same centripetal acceleration due to the rotational motion of the station, which is given by:

$a_c=\frac{v^2}{r}(1)$

Here, r is the radius and v is the tangential speed, which is given by:

$v=\omega r(2)$

Here $\omega$ is the angular velocity, we replace (2) in (1):

$a_c=\frac{(\omega r)^2}{r}\\\\a_c=\omega^2r$

Recall that $r=\frac{d}{2}=\frac{200m}{2}=100m$ .

Solving for $\omega$ :

$\omega=\sqrt{\frac{a_c}{r}}\\\omega=\sqrt{\frac{9.8\frac{m}{s^2}}{100m}}\\\omega=0.31\frac{rad}{s}$

3 0

4 years ago

A block of mass m slides down a frictionless track, then around the inside of a circular loop-the-loop of radius R. From what mi

Oliga [24]

Answer:

$H = \frac{5}{2}R$

Explanation:

As we know that if the block will complete the circular motion of the path then the speed at the bottom most part of the path must be equal to

$v = \sqrt{5Rg}$

now we know that

velocity at the bottom of the path is due to conversion of potential energy to kinetic energy

so we can say it is given as

$U = KE$

$mgH = \frac{1}{2}mv^2$

now we have

$mgH = \frac{1}{2}m(5Rg)$

$H = \frac{5}{2}R$

3 0

3 years ago

A wildlife photographer uses a moderate telephoto lens of focal length 135 mm and maximum aperture f/4.00 to photograph a bear t

labwork [276]

Answer:

<h3>(A) The width $x = 0.24 \times 10^{-3}$ m</h3><h3>(B) The new width is $1.32 \times 10^{-3}$ m</h3>

Explanation:

Given :

Focal length $f = 135 \times 10^{-3} m$

Maximum aperture $D = \frac{f}{4}$

Wavelength $\lambda = 550 \times 10^{-9}$ m

(A)

From rayleigh criterion,

$\theta = \frac{1.22 \lambda }{D}$

$\theta =\frac{ 1.22 \times 550 \times 10^{-9} }{33.75 \times 10^{-3} }$

$\theta = 1.98 \times 10^{-5}$ rad

From angle formula,

$x = R\theta$

Where $R =$ 12 m ( given in example )

$x = 12 \times 1.98 \times 10^{-5}$ m

$x = 23.76 \times 10^{-5}$

$x = 0.24 \times 10^{-3}$ m

(B)

We know that $\theta$ is proportional to the $x$ and inversely proportional to the $D$

so we write the new width, here $x$ is 5.5 times larger than above case

$x = 0.24 \times 10^{-3} \times \frac{22}{4}$

$x = 1.32 \times 10^{-3}$ m

6 0

3 years ago