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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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So, using the formula: Circumference = 2πr = <span>πd

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A single loop of wire with an area of 0.0920 m2 is in a uniform magnetic field that has an initial value of 3.80 T, is perpendic
jekas [21]

Answer:

Induced emf in the loop is 0.02208 volt.

Induced current in the loop is 0.0368 A.

Explanation:

Given that,

Area of the single loop, A=0.092\ m^2

The initial value of uniform magnetic field, B = 3.8 T

The magnetic field is decreasing at a constant rate, \dfrac{dB}{dt}=0.24\ T/s

(a) The induced emf in the loop is given by the rate of change of magnetic flux.

\epsilon=\dfrac{-d\phi}{dt}\\\\\epsilon=A\times \dfrac{dB}{dt}\\\\\epsilon=0.092\times 0.24\\\\\epsilon=0.02208\ V

(b) Resistance of the loop is 0.6 ohms. Let I is the current induced in the loop. Using Ohm's law :

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Hence, this is the required solution.

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