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Veseljchak [2.6K]
2 years ago
7

HELP

Physics
1 answer:
olganol [36]2 years ago
8 0
All the 4 processes are correct answer to change the state of matter. Hope it helps.
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The scientific inquiry process
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What clues are useful in reconstructing pangaea
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You can use map and notice one thinh. If you flipp over the edges of continents and put them together, you will get a big single continent that is called pangaea. Practically it's impossible but it could be imagined.
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A team of engineering students is designing a catapult to launch a small ball at A so that it lands in the box. If it is known t
Murljashka [212]

Answer:

Explanation:

If the initial velocity is U

Then the horizontal component of the velocity is

Ux= Ucosθ

Then the range for a projectile is give as

R=Ux.t

Where t is the time of flight

The time of flight is given as

t=2USinθ/g

Therefore,

R=Ux.t

R=UCosθ.2USinθ/g

R=U^2×2SinθCosθ/g

Then, from trigonometric ratio

2SinθCosθ= Sin2θ

R=U^2Sin2θ/g

Given that θ=32° and g=9.81m/s^2

Then

R=U^2Sin2×32/9.81

R=U^2Sin64/9.81

R=0.0916U^2

Then, range is given by R=0.0916U^2

A=0.0916U^2.

T

The box is at a distance A from the point of projection. Then the range R=A

R=0.0916U^2

A=0.0916U^2

Then,

U^2=A/0.0916

U^2=10.915A

Then the initial velocity should be

U=√10.915A

U=3.3√A

8 0
3 years ago
What is the range of motion of the elbow if extension is 0° and flexion is 145°?
inna [77]

Answer:

0 to 145 degrees

Explanation:

The normal range of flexion and extension is from 0 to 145 degrees.

6 0
2 years ago
A person stands at the base of a hill that is a straight incline making an angle φ with the horizontal. For a given initial spee
Brums [2.3K]

Answer:

θ = sin⁻¹\sqrt{2gd}

Explanation:

From one of the equations of motion, v² = u² + 2as.......... equation 1

Since the object thrown was moving against gravity, then the acceleration, a would change to -g and the initial velocity u would change to V₀ sin θ because the object is travelling at angle of θ to the horizontal. By inputting all these parameter into equation 1, we would arrive at:

v² = (u sin θ)² - 2gd

(u sin θ)² = 2gd

d = (u sin θ)²/2g

sin² θ = 2gd

sin θ = \sqrt{2gd}

θ = sin⁻¹ \sqrt{2gd}

4 0
3 years ago
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