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2 years ago

Desde que altura debes de lanzar una canica de 50g para que adquiera una energia de 100j

Physics

1 answer:

AnnZ [28]2 years ago

7 0

Answer:

204 m

Explanation:

When the marble is dropped from a certain height, its gravitational potential energy converts into kinetic energy. So the kinetic energy gained is equal to the variation of gravitational potential energy:

$\Delta U=mg\Delta h$

where

m is the mass of the marble

g = 9.8 m/s^2 is the acceleration of gravity

$\Delta h$ is the change in height

In this problem, we have

m = 50 g = 0.05 kg

$\Delta U = 100 J$

Solving the formula for $\Delta h$ , we find the necessary height from which the marble should be dropped:

$\Delta h = \frac{\Delta U}{mg}=\frac{100}{(0.05)(9.8)}=204 m$

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In Niels Bohr's 1913 model of the hydrogen atom, the single electron is in a circular orbit of radius 5.29×10⁻¹¹m and its speed

Svet_ta [14]

The magnitude of the magnetic moment due to the electron's motion is $87.87 * 10^{-37}$ .

<h3>What is magnetic moment?</h3>

The magnetic pull and direction of a magnet or other object that produces a magnetic field are referred to as the magnetic moment in electromagnetism. Things that have magnetic moments include electromagnets, permanent magnets, various compounds, elementary particles like electrons, and a number of celestial objects (such as many planets, some moons, stars, etc).

The term "magnetic moment" really refers to the magnetic dipole moment of a system, which is the portion of the magnetic moment that can be represented by an equivalent magnetic dipole or a pair of magnetic north and south poles that are only very slightly apart. The magnetic dipole component is adequate for sufficiently small magnets or over sufficiently large distances.

Calculations:

radius= $5.29 * 10^{-11} m\\$

velocity= $2.9* 10^{6} m/s$

Working formula, M=N/A

$I=\frac{charge flow }{time taken} =\frac{e}{time taken\\}$

$T= \frac{2xr}{v} =\frac{2xx * 5.29 * 10^{-11} }{2.9* 10^{6} }$

= $15.16 * 10^{-5} s$

$I= \frac{1.6 * 10^{-19} }{15.16 * 10^{-5} }$ $= 0.10 * 10^{-14}$

= $1 * 10^{-15} C$

M= $1x (1* 10^{-15} * (5.29 * 10^{-11} )^{2}$

= $87.87 * 10^{-37}$

To learn more about magnetic moment ,visit:

brainly.com/question/14298729

#SPJ4

4 0

9 months ago

A 1,508 kg car rolling on a horizontal surface has a speed of 20.8 km/hr when it strikes a horizontal coiled spring and is broug

natali 33 [55]

Answer:

Approximately $1.79 \times 10^{5}\; {\rm N \cdot m^{-1}}$ , assuming friction between the vehicle and the ground is negligible.

Explanation:

Let $m$ denote the mass of the vehicle. Let $v$ denote the initial velocity of the vehicle. Let $k$ denote the spring constant (needs to be found.) Let $x$ denote the maximum displacement of the spring.

Convert velocity of the vehicle to standard units (meters per second):

$\begin{aligned}v &= 20.8\; {\rm km \cdot h^{-1}} \times \frac{1000\; {\rm m}}{1\; {\rm km}} \times \frac{1\; {\rm h}}{3600\; {\rm s}} \\ &\approx 1.908\; {\rm m \cdot s^{-1}}\end{aligned}$ .

Initial kinetic energy ( ${\rm KE}$ ) of the vehicle:

$\begin{aligned}\frac{1}{2}\, m \, v^{2}\end{aligned}$ .

When the vehicle is brought to a rest, the elastic potential energy ( $\text{EPE}$ ) stored in the spring would be:

$\displaystyle \frac{1}{2}\, k\, x^{2}$ .

By the conservation of energy, if the friction between the vehicle and the ground is negligible, the initial $\text{KE}$ of the vehicle should be equal to the ${\rm EPE}$ of the vehicle. In other words:

$\begin{aligned}\frac{1}{2}\, m \, v^{2} &= \frac{1}{2}\, k\, x^{2}\end{aligned}$ .

Rearrange this equation to find an expression for $k$ , the spring constant:

$\begin{aligned}k &= \frac{m\, v }{x^{2}}\end{aligned}$ .

Substitute in the given values $m = 1508\; {\rm kg}$ , $v \approx 1.908\; {\rm m\cdot s^{-1}}$ , and $x = 6.87\; {\rm m}$ :

$\begin{aligned}k &= \frac{m\, v }{x^{2}} \\ &\approx \frac{1508\; {\rm kg} \times 1.908\; {\rm m\cdot s^{-1}}}{(6.87\; {\rm m})^{2}} \\ &\approx 1.79 \times 10^{5}\; {\rm kg \cdot m \cdot s^{-2} \cdot m^{-3}}\\ &\approx 1.79 \times 10^{5}\; {\rm N \cdot m^{-1}}\end{aligned}$

8 0

1 year ago

What is the difference between fertilization and pollination

lana [24]

<span>Pollination is when the male gamete or pollen reaches the female flower through different modes...wind,air,animals,...
Fertilization </span><span>is when the male gamete "join" with the female gamete to become an embryo.
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4 0

3 years ago

Two stationary positive point charges, charge 1 of magnitude 3.90 nC and charge 2 of magnitude 1.80 nC, are separated by a dista

soldi70 [24.7K]

Answer:

$v = 7793150 m/s$

Explanation:

First, we are going to calculate the electrical potential in the point middle between the two charges

Remember that the electrical potential can be calculated as:

$v = \frac{kQ}{r}$

Where $k = 8.9874 x 10^{9} \frac{Nm^{2} }{C^{2} }$

and it is satisfy the superposition principle, thus

$v = \frac{8.9874x10^{9}(3.90x10^{-9} ) }{0.23} + \frac{8.9874x10^{9}(1.80x10^{-9} ) }{0.23}$

$v = 222.73v$

The electrical potential at 10 cm from charge 1 is:

$v = \frac{8.9874x10^{9}(3.90x10^{-9} ) }{0.1} + \frac{8.9874x10^{9}(1.80x10^{-9} ) }{0.36}$

$v = 395.44 v$

Since the work - energy theorem, we have:

$q\Delta v = \frac{mv^{2} }{2}$

where q is the electron's charge and m is the electron's mass

Therefore:

$v = \sqrt{\frac{2q\Delta v}{m} }$

$v = 7793150 m/s$

6 0

2 years ago

What is the period and frequency of a water wave if 4.0 complete waves pass a fixed point in 10 seconds

olga2289 [7]

The correct answer for this question is this one: C) 2.5s. T<span>he period and frequency of a water wave if 4.0 complete waves pass a fixed point in 10 seconds is that 2.5 s
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Here are the following choices:
<span>A) 0.25s
B) 0.40s
C) 2.5s
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7 0

2 years ago