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Kruka [31]
3 years ago
9

PLEASE HELP. THIS IS DUE TODAY AT 11:59PM ​

Physics
1 answer:
OLga [1]3 years ago
4 0

Answer:

ggn

Explanation:

rt

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Find the range of a projectile launched at an angle of 30° with an initial velocity of 20m/s.​
Tems11 [23]

Answer:

<em>The range is 35.35 m</em>

Explanation:

<u>Projectile Motion</u>

It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.

Being vo the initial speed of the object, θ the initial launch angle, and g=9.8m/s^2 the acceleration of gravity, then the maximum horizontal distance traveled by the object (also called Range) is:

\displaystyle d={\frac  {v_o^{2}\sin(2\theta )}{g}}

The projectile was launched at an angle of θ=30° with an initial speed vo=20 m/s. Calculating the range:

\displaystyle d={\frac  {20^{2}\sin(2\cdot 30^\circ )}{9.8}}

\displaystyle d={\frac  {400\sin(60^\circ )}{9.8}}

d=35.35\ m

The range is 35.35 m

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The neutron can be found in the nucleus of the atom with the proton.

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As the sum of the two right directed forces match exactly the left directed force, the only unbalanced force, and thus the net force, is the upward 25 N force.

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