a projectile is launched straight up at 141 m/s . How fast is it moving at the top of its trajectory? suppose it is launched upw
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As we know that reaction time will be
so the distance moved by car in reaction time
now the distance remain after that from intersection point is given by
So our distance from the intersection will be 100 m when we apply brakes
now this distance should be covered till the car will stop
so here we will have
now from kinematics equation we will have
so the acceleration required by brakes is -2 m/s/s
Now total time taken to stop the car after applying brakes will be given as
total time to stop the car is given as
The hoop is attached.
Consider that the friction force is given by:
F = μ·N
= μ·m·g·cosθ
We also know, considering the forces of the whole system, that:
F = -m·a + m·g·sinθ
and
a = (1/2)·<span>g·sinθ
Therefore:
</span>-(1/2)·m·g·sinθ + m·g·sinθ = <span>μ·m·g·cosθ
</span>(1/2)·m·g·sinθ = <span>μ·m·g·cosθ
</span>μ = (1/2)·m·g·sinθ / <span>m·g·cosθ
= </span>(1/2)·tanθ
Now, solve for θ:
θ = tan⁻¹(2·μ)
= tan⁻¹(2·0.9)
= 61°
Therefore, the maximum angle <span>you could ride down without worrying about skidding is 61°.</span>
Consider that the friction force is given by:
F = μ·N
= μ·m·g·cosθ
We also know, considering the forces of the whole system, that:
F = -m·a + m·g·sinθ
and
a = (1/2)·<span>g·sinθ
Therefore:
</span>-(1/2)·m·g·sinθ + m·g·sinθ = <span>μ·m·g·cosθ
</span>(1/2)·m·g·sinθ = <span>μ·m·g·cosθ
</span>μ = (1/2)·m·g·sinθ / <span>m·g·cosθ
= </span>(1/2)·tanθ
Now, solve for θ:
θ = tan⁻¹(2·μ)
= tan⁻¹(2·0.9)
= 61°
Therefore, the maximum angle <span>you could ride down without worrying about skidding is 61°.</span>
D. Atoms.
Explanation:
All the matter is made of elementary particles called "atoms".
Further, an atom is made of electrons, protons and neutrons. The electrons & protons are again made of the fundamental sub-particles, electrons (leptons) and the protons(quarks).
The classification of particles is shown in the figure attached
