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Nostrana [21]
3 years ago
10

In the simulation above, as the projectile travels upward, how does the vertical velocity change? Vertical velocity is zero. Ver

tical velocity decreases. Vertical velocity increases. Vertical velocity remains the same.
Physics
2 answers:
PolarNik [594]3 years ago
7 0

the vertical velocity decreases.

Butoxors [25]3 years ago
5 0

Answer:

Vertical velocity decreases.

Explanation:

The figure of the simulation is missing, however we can still answer. In fact, the problem tells us that the projectile is moving upward. We are only interested in the vertical velocity of the projectile, so we can consider only its vertical motion.

The vertical motion of the projectile is an accelerated motion, with constant acceleration g = 9.8 m/s^2, directed downward (acceleration due to gravity). The problem says that the projectile is still moving upward, so its velocity is directed upward. This means that the vertical acceleration is in the opposite direction of the vertical velocity: therefore, the vertical velocity must be decreasing, according to the equation

v(t)=v_0 -gt

where v0 is the initial vertical velocity of the projectile, and t the time. Due to the negative sign in the formula, we see that as t increases, v(t) decreases.

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Afina-wow [57]

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3 years ago
To measure the acceleration due to gravity on a distant planet, an astronaut hangs a 0.070-kg ball from the end of a wire. The w
mote1985 [20]

Answer:

g = 1.64m/s²

Explanation:

1.5m in 0.078s

V = 15 / 0.078

= 19.23m/s

Tension = mg

μ = 3.10 × 10⁻⁴

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g = ((19.23)²(3.10 × 10⁻⁴)) / (0.070)

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The amount of friction divided by the weight of an object forms a unit less number called the
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Answer:

Coefficient of friction.

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3 0
3 years ago
g An electron enters a region of space containing a uniform 1.63 × 10 − 5 T magnetic field. Its speed is 121 m/s and it enters p
kolbaska11 [484]

Answer:

i. The radius 'r' of the electron's path is 4.23 × 10^{-5} m.

ii. The frequency 'f' of the motion is 455.44 KHz.

Explanation:

The radius 'r' of the electron's path is called a gyroradius. Gyroradius is the radius of the circular motion of a charged particle in the presence of a uniform magnetic field.

                 r = \frac{mv}{qB}

Where: B is the strength magnetic field, q is the charge, v is its velocity and m is the mass of the particle.

From the question, B = 1.63 × 10^{-5}T, v = 121 m/s, Θ = 90^{0} (since it enters perpendicularly to the field), q = e  = 1.6 × 10^{-19}C and m = 9.11 × 10^{-31}Kg.

Thus,

         r = \frac{mv}{qB} ÷ sinΘ

But,  sinΘ =  sin 90^{0} = 1.

So that;

          r = \frac{mv}{qB}

            = (9.11 × 10^{-31} × 121) ÷ (1.6 × 10^{-19}  × 1.63 × 10^{-5})

            = 1.10231 × 10^{-28}   ÷ 2.608 × 10^{-24}

            = 4.2266 × 10^{-5}

            = 4.23 × 10^{-5} m

The radius 'r' of the electron's path is 4.23 × 10^{-5} m.

B. The frequency 'f' of the motion is called cyclotron frequency;

           f = \frac{qB}{2\pi m}

             =  (1.6 × 10^{-19}  × 1.63 × 10^{-5}) ÷ (2 ×\frac{22}{7} × 9.11 × 10^{-31})

             =  2.608 × 10^{-24} ÷  5.7263 × 10^{-30}

             = 455442.4323

          f  = 455.44 KHz

The frequency 'f' of the motion is 455.44 KHz.

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