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lidiya [134]
2 years ago
8

A cannonball is fired straight up at a speed of 25 m/s. What is the maximum altitude that it will reach?

Physics
1 answer:
valentinak56 [21]2 years ago
3 0

Answer:

When the projectile is launched straight up, there isn't a horizontal ... The initial acceleration was 9.8 m/s2 pointing up, so the acceleration at any other point should be the same.

Explanation:

Hope it helped =)

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The spherical annulus of cometary nuclei surrounding the solar system is called the ___.
Bingel [31]

Answer:

Oort Cloud

Explanation:

  • Hundreds of millions of comet nuclei forming a spherical region that surrounds the solar system is called Oort cloud.
  • The Oort cloud revolves at tens of thousands of the distance of the earth from the sun.
  • Comets are the condensed water and dust particles that orbit around the sun with inward helix leaving behind a tail of vapours and vaporize completely as they get closer to the sun during their revolution around it.
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3 0
3 years ago
When the mass is removed, the length of the cable is found to be l0 = 4.66 m. After the mass is added, the length is remeasured
zaharov [31]

Answer:

\gamma=6.07*10^5\frac{N}{m^2}

Explanation:

For a linear elastic material Young's modulus is a constant that is given by:

\gamma=\frac{F/A}{\Delta L/L_0}

Here, F is the force exerted on an object under tensio, A is the area of the cross-section perpendicular to the applied force, \Delta L is the amount by which the length of the object changes and  L_0 is the original length of the object. In this case the force is the weight of the mass:

F=mg\\F=55kg(9.8\frac{m}{s^2})\\F=539N

Replacing the given values in Young's modulus formula:

\gamma=\frac{F/\pi r^2}{(L_1-L_0)/L_0}\\\gamma=\frac{539N/\pi(0.045m)^2}{(5.31m-4.66m)/4.66m}\\\gamma=6.07*10^5\frac{N}{m^2}

6 0
2 years ago
Which one of the following does not accurately describe the universal gravitational law? Question 10 options: A) Gravitational f
just olya [345]

Answer:

Choice C seems to be the right answer.

Explanation:

5 0
3 years ago
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Identify the type of wave in the picture please asap
pychu [463]

Answer:

It is transverse wave

Explanation:

7 0
3 years ago
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A pebble is thrown into the air with a velocity of 19/m at an angle of 36 with respect to the horizontal.
kow [346]

Answer:

The maximum height the pebble reaches is approximately;

A. 6.4 m

Explanation:

The question is with regards to projectile motion of an object

The given parameters are;

The initial velocity of the pebble, u = 19 m/s

The angle the projectile path of the pebble makes with the horizontal, θ = 36°

The maximum height of a projectile, h_{max}, is given by the following equation;

h_{max} = \dfrac{\left (u \times sin(\theta) \right)^2}{2 \cdot g}

Therefore, substituting the known values for the pebble, we have;

h_{max} = \dfrac{\left (19 \times sin(36 ^{\circ}) \right)^2}{2 \times 9.8} = 6.3633894140470403035477570509439

Therefore, the maximum height of the pebble projectile, h_{max} ≈ 6.4 m.

3 0
3 years ago
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