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maks197457 [2]
3 years ago
5

A(n) _________ is an object that is thrown.​

Physics
2 answers:
mart [117]3 years ago
7 0

Answer:

projectile

Explanation:

mars1129 [50]3 years ago
5 0

Answer:

the answer is projectile

Explanation:

this is because like take a nerf gun for example when u fire it PROJECTS the dart from the gun. may i be marked brainliest?

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How many neutrons does an element have of its atomic number is 50 and its mass number is 156?
pshichka [43]

Ok so I think you would first subtract


156 - 50 = 106


Since atomic number is based on the number of protons, you should subtract the 50 from 156


106 should be your answer


Hope this helps!

3 0
4 years ago
The first astronomer to show that spiral nebulae (today called spiral galaxies) have large Doppler shifts was
solong [7]

William Parsons was the first astronomer to show that spiral nebulae have large Doppler shifts.

<h3>What are Spiral nebulae?</h3>
  • In space, a nebula is a huge cloud of gas and dust. Some nebulae, including multiple nebulae, are made of gas and dust that have been released from the explosion of a dead star called a supernova. There are other nebulae that are star-forming regions.
  • Edwin Hubble began observing Cepheid variables in many spiral nebulae, including the alleged Andromeda Nebula, in 1923, demonstrating that they are, in fact, full-fledged galaxies outside our own. Since then, the phrase spiral nebula has lost favor.
  • One group of astronomers believed that spiral nebulae were parts of our Milky Way galaxy, whereas the other group believed that these objects were actual galaxies that were outside of the Milky Way galaxy.

To learn more about Nebulae visit:

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8 0
2 years ago
An object following a straight-line path at constant speed
Rom4ik [11]

Answer:

C

Explanation:

5 0
2 years ago
Which of the following is the volume of the shaded in Of the sphere?
marysya [2.9K]

Answer:

The correct option is;

πR³(2/3 + (1/3)cos³θ - cosθ)

Explanation:

The volume of a segment of a sphere is given by the relation;

V = \pi \cdot h^2 \cdot \left (R - \dfrac{h}{3} \right)

We note that h = R - R·cos(θ)

Therefore by substituting the value of h in the equation of a segment of a sphere, we have;

V = \pi \cdot \left (R - R\cdot cos(\theta) \right ) ^2 \cdot \left (R - \dfrac{\left (R - R\cdot cos(\theta) \right )}{3} \right)

Which gives;

\dfrac{R^3\cdot \pi \cdot cos^3 (\theta) -3 \cdot R^3 \cdot\pi \cdot cos (\theta) +  2 \cdot R^3 \cdot \pi}{3}

R^3\cdot \pi \cdot  \left (\dfrac{cos^3 (\theta) -3 \cdot cos (\theta) +  2 }{3} \right )

R^3\cdot \pi \cdot  \left (\dfrac{cos^3 (\theta) +  2 }{3} -  cos (\theta)  \right )

R^3\cdot \pi \cdot  \left (\dfrac{cos^3 (\theta) }{3} + \dfrac{2}{3} -  cos (\theta)  \right )

Therefore, the correct option is πR³(2/3 + (1/3)cos³θ - cosθ).

7 0
3 years ago
A projectile is launched at an angle of 30° and lands 20 s later at the same height as it was launched. (a) What is the initial
Elina [12.6K]

Answer:

a)Initial speed of the projectile = 196.2 m/s

b)Maximum altitude = 490.5 m

c) Range of projectile = 3398.28 m

d) Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m

Explanation:

Time of flight of a projectile is given by the expression,

               t=\frac{2usin\theta}{g}

           Here θ = 30° and t = 20 s

a) t=\frac{2usin\theta}{g}\\\\20=\frac{2\times usin30}{9.81}\\\\u=196.2m/s

  Initial speed of the projectile = 196.2 m/s

b) Maximum altitude is given by

                  H=\frac{u^2sin^2\theta}{2g}=\frac{196.2^2\times sin^230}{2\times 9.81}=490.5m

      Maximum altitude = 490.5 m

c) Range of projectile is given by

                              R=\frac{u^2sin2\theta}{g}=\frac{196.2^2\times sin(2\times 30)}{9.81}=3398.28m

    Range of projectile = 3398.28 m

d) Horizontal velocity = ucosθ = 196.2 x cos 30 = 169.91 m/s

   Vertical velocity = usinθ = 196.2 x sin 30 = 98.1 m/s

   We have equation of motion s = ut + 0.5 at²

   Horizontal motion

                         u = 169.91 m/s

                         a = 0 m/s²

                          t = 15 s

                Substituting

                          s = 169.91 x 15 + 0.5 x 0 x 15² = 2548.71 m

      Vertical motion

                         u = 98.1 m/s

                         a = -9.81 m/s²

                          t = 15 s

                Substituting

                          s = 98.1 x 15 + 0.5 x -9.81 x 15² = 367.88 m

   \texttt{Total displacement =}\sqrt{2548.71^2+367.88^2}=2575.12m

   Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m

7 0
4 years ago
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