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babunello [35]
2 years ago
12

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

r, a watermark will not resist tampering.
Physics
1 answer:
Marina86 [1]2 years ago
7 0

<em>The first blank is </em><em>robust watermark</em>; a robust watermark will not resist tampering.

<em>The second blank is </em><em>fragile watermark</em><em>;</em> 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:.

brainly.com/question/24206908

#SPJ1

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When steam condenses 1. All of these occur. 2. None of these occur. 3. molecules move closer together. 4. it changes from the ga
True [87]

2. None of these occur

5 0
3 years ago
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An atom of uranium 238 emits an alpha particle (an atom of He) and recoils with a velocity of 1.895 * 10^ 5 m/sec . With velocit
lora16 [44]

<u>Answer:</u> The velocity of released alpha particle is 1.127\times 10^7m/s

<u>Explanation:</u>

According to law of conservation of momentum, momentum can neither be created nor be destroyed until and unless, an external force is applied.

For a system:

m_1v_1=m_2v_2

where,

m_1\text{ and }v_1 = Initial mass and velocity

m_2\text{ and }v_2 = Final mass and velocity

We are given:

m_1=238u\\v_1=1.895\times 10^{5}m/s\\m_2=4u\text{ (Mass of }\alpha \text{ -particle)}\\v_2=?m/s

Putting values in above equation, we get:

238\times 1.895\times 10^5=4\times v_2\\\\v_2=\frac{238\times 1.895\times 10^5}{4}=1.127\times 10^7m/s

Hence, the velocity of released alpha particle is 1.127\times 10^7m/s

4 0
3 years ago
Which vector below goes from (0,0) to (1,-3)?
vitfil [10]

I think D. It starts at (0.0) and goes to the correct points so it makes sense

8 0
3 years ago
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24 POINTS!!!!!!!!!!!!!!!
TEA [102]
Potential Energy (Initial one) = m * g * h
P.E. = 60 * 9.8 * 10
P.E. = 5880

Kinetic Energy (Final One) = 1/2 mv²
K.E. = 1/2 * 60 * (10)²
K.E. = 6000/2
K.E. = 3000

Lost Energy = 5880 - 3000 = 2880 J

In short, Your Answer would be 2880 Joules

Hope this helps!
7 0
3 years ago
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The leg's force forward on the foot= 500N
Anna11 [10]

There's so much going on here, in a short period of time.

<u>Before the kick</u>, as the foot swings toward the ball . . .

-- The net force on the ball is zero.  That's why it just lays there and
does not accelerate in any direction.

-- The net force on the foot is 500N, originating in the leg, causing it to
accelerate toward the ball.


<u>During the kick</u> ... the 0.1 second or so that the foot is in contact with the ball ...

-- The net force on the ball is 500N.  That's what makes it accelerate from
just laying there to taking off on a high arc.

-- The net force on the foot is zero ... 500N from the leg, pointing forward,
and 500N as the reaction force from the ball, pointing backward. 

That's how the leg's speed remains constant ... creating a dent in the ball
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out of the dent, as the ball accelerates to exceed the speed of the foot and
draw away from it.


5 0
3 years ago
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