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The watermark can be of different types depending on the application. A watermark will resist manipulations of the media. Howeve

Marina86 [1]

The first blank is robust watermark; a robust watermark will not resist tampering.

The second blank is fragile watermark; a fragile watermark will resist manipulations of the media.

<h3>What is a watermark?</h3>

A watermark is a faint design made in paper during manufacture that is visible when held against the light and clearly identifies the maker.

The watermark can be of different types depending on the application and they include:

A robust watermark will not resist tampering.
A fragile watermark will resist manipulations of the media.

Thus, The first blank is robust watermark; a robust watermark will not resist tampering.

The second blank is fragile watermark; a fragile watermark will resist manipulations of the media.

Learn more about watermark here:.

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7 0

1 year ago

Calculate the average orbital speed of Ceres in

marin [14]

] Ceres is composed of rock and ice and is estimated to comprise approximately one third of the mass of the entire asteroid belt. Ceres is the only object in the asteroid belt known to be rounded by its own gravity (though detailed analysis was required to exclude 4 Vesta). From Earth, the apparent magnitude of Ceres ranges from 6.7 to 9.3, peaking once every 15 to 16 months,[21]hence even at its brightest it is too dim to be seen with the naked eye except under extremely dark skies.

8 0

3 years ago

A physics book slides off a horizontal tabletop with a speed of 1.10 m/s. It strikes the floor in 0.350s. ignore air resistance.

Nookie1986 [14]

Answer:

a. 0.6 m b. 0.385 m c. 3.6 m/s at 287.78° to the horizontal

Explanation:

a. Using s = ut - 1/2gt² for motion under gravity where s = vertical distance = height of table, u = initial vertical velocity of book = 0 m/s, t = time of flight = 0.350 s and g = acceleration due to gravity = 9.8 m/s².

Substituting these these values into s and taking the top of the table as position 0 m, we have.

0 - s = 0t - 1/2gt²

-s = -1/2gt²

s = 1/2gt²

s = 1/2 × 9.8 m/s² × (0.350 s)²

s = 0.6 m

b. Using d = v't where d = horizontal distance from table, v' = horizontal velocity of book = 1.10 m/s and t = time of flight = 0.350 s

d = v't = 1.10 m/s × 0.350 s = 0.385 m

c. Using v² = u² - 2gs where u = initial vertical velocity of book = 0 m/s and g = 9.8 m/s², s = -0.6 m (negative since we are at the bottom and 0 m is at the top)and v = final vertical velocity of book

v² = u² - 2gs

= 0 - 2 × 9.8 m/s² × (-0.6 m)

= 11.76 m²/s²

v = √11.76 m/s

= 3.43 m/s

So, the magnitude of the resultant velocity is V = √(v² + v'²)

= √((3.43 m/s)² + (1.10 m/s)'²)

= √(11.76 m²/s² + 1.21 m²/s²)

= √12.97 m²/s²

= 3.6 m/s

Its direction Ф = tan⁻¹(-v/v') since v is in the negative y direction

= tan⁻¹(-3.43 m/s/1.10 m/s)

= tan⁻¹(-3.1182)

= -72.22°

Ф = -72.22°+ 360 = 287.78° since it is in the third quadrant

7 0

3 years ago

Tammy leaves the office, drives 34 km due north, then turns onto a second highway and continues in a direction of 35◦ north of e

Natasha2012 [34]

Answer:

The total displacement is 102 km $51^o$ north of east.

Explanation:

We can treat this problem as a trigonometric one, so we need to calculate the total displacement on the north and east.

$d_n=34km+79km*sin(35^o)=79km$

and

$d_e=79*cos(35^o)=65km$

The total displacement is given by:

$D=\sqrt{(79km)^2+(65km)^2}=102km$

with an agle of:

$\alpha =arctg(\frac{79km}{65km})=51^o$

4 0

3 years ago

Two spheres have identical charges and are 75 cm apart. the force between them is +0.30 n. what is the magnitude of the charge o

ella [17]

Answer:

$4.33\cdot 10^{-6}C$ , charges are both positive or both negative

Explanation:

The electrostatic force between the two spheres is given by

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1 and q2 are the charges on the two spheres

r is the distance between the centres of the two spheres

In this problem, we have

$F=+0.30 N$ is the force

$r=75 cm=0.75 m$ is the distance between the spheres

$q_1 =q_2 =q$ because the two spheres have identical charge

Solving the formula for q, we find

$q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(+0.30 N)(0.75 m)^2}{9\cdot 10^9}}=4.33\cdot 10^{-6}C$

And the two charges have the same sign (so, both positive or both negative), since the sign of the force is positive (+0.30 N), so it is a repulsive force.

5 0

3 years ago