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andrew11 [14]
2 years ago
5

A projectile is launched at an angle of 29 degrees above the horizontal with an initial velocity of 36.6 at an unknown height.

Physics
1 answer:
alex41 [277]2 years ago
3 0

The magnitude of the unknown height of the projectile is determined as 16.1 m.

<h3>Magnitude of the height</h3>

The magnitude of the height of the projectile is calculated as follows;

H = u²sin²θ/2g

H = (36.6² x (sin 29)²)/(2 x 9.8)

H = 16.1 m

Thus, the magnitude of the unknown height of the projectile is determined as 16.1 m.

Learn more about height here: brainly.com/question/1739912

#SPJ1

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You cross two pure tall (TT) plants with two pure short (tt) plants and produce four offspring. Which of the following statement
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All will have a dominant trait I can't see the following statments

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5. A single slit illuminated with a 500 nm light gives a diffraction pattern on a far screen. The 5th minimum occurs at 7.00° aw
abruzzese [7]

For a single slit illuminated with a 500 nm light gives a diffraction pattern on a far screen,the angle  is mathematically given as

theta=25.3

Option A is correct

<h3> What angle does the 18th minimum occur?</h3>

Generally, the equation for the the angle   is mathematically given as

\theta=n(\lambda/d)

Therefore

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2 years ago
A 4.87-kg ball of clay is thrown downward from a height of 3.21 m with a speed of 5.21 m/s onto a spring with k = 1570 N/m. The
Yuki888 [10]

Answer:

Approximately 0.560\; {\rm m}, assuming that:

  • the height of 3.21\; {\rm m} refers to the distance between the clay and the top of the uncompressed spring.
  • air resistance on the clay sphere is negligible,
  • the gravitational field strength is g = 9.81\; {\rm m\cdot s^{-2}}, and
  • the clay sphere did not deform.

Explanation:

Notations:

  • Let k denote the spring constant of the spring.
  • Let m denote the mass of the clay sphere.
  • Let v denote the initial speed of the spring.
  • Let g denote the gravitational field strength.
  • Let h denote the initial vertical distance between the clay and the top of the uncompressed spring.

Let x denote the maximum compression of the spring- the only unknown quantity in this question.

After being compressed by a displacement of x, the elastic potential energy \text{PE}_{\text{spring}} in this spring would be:

\displaystyle \text{PE}_{\text{spring}} = \frac{1}{2}\, k\, x^{2}.

The initial kinetic energy \text{KE} of the clay sphere was:

\displaystyle \text{KE} = \frac{1}{2}\, m \, v^{2}.

When the spring is at the maximum compression:

  • The clay sphere would be right on top of the spring.
  • The top of the spring would be below the original position (when the spring was uncompressed) by x.
  • The initial position of the clay sphere, however, is above the original position of the top of the spring by h = 3.21\; {\rm m}.

Thus, the initial position of the clay sphere (h = 3.21\; {\rm m} above the top of the uncompressed spring) would be above the max-compression position of the clay sphere by (h + x).

The gravitational potential energy involved would be:

\text{GPE} = m\, g\, (h + x).

No mechanical energy would be lost under the assumptions listed above. Thus:

\text{PE}_\text{spring} = \text{KE} + \text{GPE}.

\displaystyle \frac{1}{2}\, k\, x^{2} = \frac{1}{2}\, m\, v^{2} + m\, g\, (h + x).

Rearrange this equation to obtain a quadratic equation about the only unknown, x:

\displaystyle \frac{1}{2}\, k\, x^{2} - m\, g\, x - \left[\left(\frac{1}{2}\, m\, v^{2}\right)+ (m\, g\, h)\right] = 0.

Substitute in k = 1570\; {\rm N \cdot m^{-1}}, m = 4.87\; {\rm kg}, v = 5.21\; {\rm m\cdot s^{-1}}, g = 9.81\; {\rm m \cdot s^{-2}}, and h = 3.21\; {\rm m}. Let the unit of x be meters.

785\, x^{2} - 47.775\, x - 219.453 \approx 0 (Rounded. The unit of both sides of this equation is joules.)

Solve using the quadratic formula given that x \ge 0:

\begin{aligned}x &\approx \frac{-(-47.775) + \sqrt{(-47.775)^{2} - 4 \times 785 \times (-219.453)}}{2 \times 785} \\ &\approx 0.560\; {\rm m}\end{aligned}.

(The other root is negative and is thus invalid.)

Hence, the maximum compression of this spring would be approximately 0.560\; {\rm m}.

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Explain what the core muscles do and why it's important to have a strong set of core muscles.
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Answer:

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A pharmacist attempts to weigh 100 milligrams of codeine sulfate on a balance with a sensitivity requirement of 4 milligrams. Ca
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Answer:

4 %

2 ) 3.42 %

Explanation:

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Given , 4 milligram is the maximum error possible .

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2 )

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6 milliliters of water should measure 6 grams

Deviation = 6 - 5.795 = - 0.205 gram.

Percentage of error =(.205 / 6 )x 100

= 3.42 %

3 0
3 years ago
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