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Agata [3.3K]
3 years ago
9

15. यदि दिइएको केन्द्र 0 भएको वृत्तको

Physics
1 answer:
miv72 [106K]3 years ago
8 0

Answer:

what language is that

Explanation:

i don't understand the languge u used please can you change it

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Two boxes (m1 = 43.0 kg and m2 = 39.5 kg) are connected by a light string that passes over a light, frictionless pulley. One box
wlad13 [49]
The net force on the system:
F = m₂g - m₁gsin(∅)
F = 39.5 x 9.81 - 43 x 9.81 x sin(30)
F = 176.58 N

Now, we use F = ma to find the acceleration on each mass.
F = m₁a₁
a₁ = 176.58 / 43
a₁ = 4.11 m/s²

F = m₂a₂
a₂ = 176.58 / 39.5
a₂ = 4.47 m/s²
5 0
3 years ago
Suppose that the speed of an electron traveling 2.0 km/s is known to an accuracy of 1 part in 105 (i.e., within 0.0010%). What i
OverLord2011 [107]

Answer: 2.89(10)^{-3} m

Explanation:

The <u>Heisenberg uncertainty principle</u> postulates that the fact each particle has a wave associated with it, imposes restrictions on the ability to determine its position and speed at the same time.  

In other words:  

It is impossible to measure simultaneously (according to quantum physics), and with absolute precision, the value of the position and the momentum (linear momentum) of a particle. Thus, in general, the greater the precision in the measurement of one of these magnitudes, the greater the uncertainty in the measure of the other complementary variable.

Mathematically this principle is written as:

\Delta x \geq \frac{h}{4 \pi m \Delta V} (1)

Where:

\Delta x is the uncertainty in the position of the electron

h=6.626(10)^{-34}J.s is the Planck constant

m=9.11(10)^{-31}kg is the mass of the electron

\Delta V is the uncertainty in the velocity of the electron.

If we know the accuracy of the velocity is 0.001\% of the velocity of the electron V=2 km/s=2000 m/s, then \Delta V is:

\Delta V=2000 m/s(0.001\%)

\Delta V=2000 m/s(\frac{0.001}{100})

\Delta V=2(10)^{-2} m/s (2)

Now, the least possible uncertainty in position \Delta x_{min} is:

\Delta x_{min}=\frac{h}{4 \pi m \Delta V} (3)

\Delta x_{min}=\frac{6.626(10)^{-34}J.s}{4 \pi (9.11(10)^{-31}kg) (2(10)^{-2} m/s)} (4)

Finally:

\Delta x_{min}=2.89(10)^{-3} m

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Is it a magnifying glass?
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