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laila [671]
3 years ago
8

If the accuracy in measuring the position of a particle increases, what happens to the accuracy in measuring its velocity? - The

accuracy in measuring its velocity becomes uncertain.
- The accuracy in measuring its velocity remains the same.
- The accuracy in measuring its velocity also increases.
- The accuracy in measuring its velocity decreases.
Physics
1 answer:
timofeeve [1]3 years ago
5 0

Answer:

- The accuracy in measuring its velocity decreases.

Explanation:

This can be explained by Heisenberg's uncertainty principle which states that the position and velocity of a particle can be determined together exactly in reality.

This principle, unlike Newtonian mechanics deal with particles at microscopic level like that of an electron where if the accuracy in measurement of particle's position increases there will be  decreased accuracy in measurement of velocity of that particle.

There will be an uncertainty in accuracy in the measurement of particle's position and its velocity.

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Only kinetic.

Explanation:

Potential energy means it has the potential to move. Not something already in motion.

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In a physics lab, a 0.500-kg cart (Cart A) moving with a speed of 129 cm/s encounters a magnetic collision with a 1.50-kg cart (
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58 cm/s

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' A ' is the graph that shows it.

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On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a 6 iron. The acceleration due to gravity on t
kozerog [31]

Answer:

a) 6 times farther.  b) 6 times longer.

Explanation:

Once released, in the horizontal direction, no other forces act on the ball, so it continues moving at the same initial velocity, which is given by the projection of the velocity vector in the horizontal direction, as follows:

vₓ = v* cos (25º) = 23 m/s * 0.906 = 20.8 m/s

In the vertical direction, the initial velocity is the projection of the velocity vector along the vertical axis, as follows:

vy = v* sin (25º) = 23 m/s * 0.422 = 9.72 m/s

Assuming that the acceleration is constant, and equal to 1/6*g, we can calculate the total time of flight, with the following kinematic equation for the vertical displacement:

y = voy*t - (\frac{1}{2}*\frac{g}{6} * t^{2} )

If the total displacement in the vertical direction is 0 (which means  that the time if the total time of flight), we can solve for t, as follows:

t = \frac{voy*12}{g} = \frac{9.72 m/s*12}{9.8m/s2} = 11. 9 s

On earth, this time could be calculated in the same way:

t = \frac{voy*12}{g} = \frac{9.72 m/s*2}{9.8m/s2} = 1.98 s

As the time is defined by the vertical movement, we can find the horizontal distance travelled on the moon, as follows:

Δx = v₀ₓ * t = 20.8 m/s * 11. 9 s = 248.1 m

On earth, the distance travelled had been as follows:

Δx = v₀ₓ * t = 20.8 m/s * 1.98 s = 41.3 m

⇒ Δx(moon) / Δx(earth) = 248.1 / 41.3 = 6.00

b) As we have just found, the time of flight on the moon and on the earth are as follows:

tmoon = 11. 9 s

tearth = 1.98 s

⇒ t(moon) / t(earth) = 11.9 / 1.98 = 6.0

8 0
3 years ago
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V125BC [204]

Answer:

A

Explanation:

Kinetic energy is the energy of motion

KE=.5mv^2

>m= mass

>v= velocity (m/s)

PE=mgh

>m= mass

>g= acceleration due to graviry

>h= height

8 0
3 years ago
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