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natta225 [31]
2 years ago
9

Mass m1 on the frictionless table of the figure is connected by a string through a hole in the table to a hanging mass m2

Physics
1 answer:
Setler [38]2 years ago
8 0

For a mass m1 on the frictionless table, the speed m1 must rotate a with radius r if m2 is to remain hanging at rest is mathematically given as

v=\sqrt{m1/m2*xg}

<h3>What speed must m1 rotate in a circle of radius r if m2 is to remain hanging at rest?</h3>

Generally, the equation for the Force  is mathematically given as

Fc=Fw

Therefore

\frac{m1v1}{x}=m2g

v=\sqrt{m1/m2*xg}

In conclusion,  if m2 is to remain hanging at rest the speed of ratio of m1 is calcuylated using

v=\sqrt{m1/m2*xg}

Read more about Speed

brainly.com/question/4931057

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bagirrra123 [75]
Data:

The charge of a body depends on the amount of electrons it gains or loses. Q = n * e, where "Q" is charge, "n" is the number of plus or minus electrons, and "e" is the fundamental charge of an electron 1,6 * 10 ^{-19}C<span>. To know if the body has gained or lost, we look at the signal of its charge, remembering that the electron is negative. The charge of the body is 4 μC (positive), so there is a lack of electrons! 

Q = 4 </span>μC → Q = 4*10^{-6}
e = 1,6 * 10 ^{-19}C
n = ?<span>

We have:
</span>Q = n*e
n =  \frac{Q}{e}
n =  \frac{4*10^{-6}}{1,6 * 10 ^{-19}}
n = 2,5*10^{-6-(-19)}
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\boxed{n = 2,5*10^{13}electrons}
7 0
3 years ago
A toy car goes over a small ramp at a horizontal velocity of 1.21 m/s and decelerates at 0.131 m/s2 while in the air. The total
xz_007 [3.2K]
We need to considerate only the horizontal component of the motion of the toy car.

The formula for the distance in a decelerated motion is:
s = s₀ + v₀·t - 1/2·a·t²

where:
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t = time elapsed = 0.342 s
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Plugging in numbers:
s = 0 + 1.21×0.342 - 0.5×0.141×(0.342)²
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3 years ago
1. Si comprimes un globo hasta reducir su volumen a un tercio de su valor original, ¿cuánto aumenta la presión en su interior?
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Answer:

See step by step sexplanation

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3.-No se puede estar de acuerdo con el criterio del plomero. En su solución no plantea el aumento de la altura del tanque,  para el logro del aumento de la presión que es realmente lo que hay que hacer

4 0
3 years ago
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