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2 years ago

Which best describes the law of conservation of mass?

Physics

1 answer:

Sergio039 [100]2 years ago

7 0

Explanation:

The law of conservation of mass states that mass in an isolated system is neither created nor destroyed by chemical reactions or physical transformations. According to the law of conservation of mass, the mass of the products in a chemical reaction must equal the mass of the reactants.

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It is known that the gravitational force of attraction between two alpha particles is much weaker than the electrical repulsion.

natali 33 [55]

Answer:

The ratio of gravitational force to electrical force is 3.19 x 10^-36

Explanation:

mass of an alpha particle = 6.64 x $10^{-27}$ kg

charge on an alpha particle = +2e = +2(1.6 x $10^{-19}$ C) = 3.2 x $10^{-19}$ C

distance between particles = d

For gravitational attraction:

The force of gravitational attraction F = $\frac{Gm^{2} }{r^{2} }$

where G = gravitational constant = 6.67 x $10^{-11}$ m^3 kg^-1 s^-2

r = the distance between the particles = d

m = the mass of each particle

therefore, gravitational force = $\frac{6.67*10^{-11}*(6.64*10^{-27} )^{2} }{d^{2} }$ = $\frac{2.94*10^{-63} }{d^{2} }$ Newton

For electrical repulsion:

Electrical force between the particles = $\frac{-kQ^{2} }{r^{2} }$

where k is the Coulomb's constant = 9.0 x $10^{9}$ N•m^2/C^2

r = distance between the particles = d

Q = charge on each particle

therefore, electrical force = $\frac{-9*10^{9}*(3.2*10^{-19} )^{2} }{d^{2} }$ = $\frac{-9.216*10^{-28} }{d^{2} }$ Newton

the negative sign implies that there is a repulsion on the particles due to their like charges.

Ratio of the magnitude of gravitation to electrical force = $\frac{2.94*10^{-63} }{9.216*10^{-28} }$

==> 3.19 x 10^-36

8 0

2 years ago

A proton, mass 1.67 × 10−27 kg and charge +1.6 × 10−19 c, moves in a circular orbit perpendicular to a uniform magnetic field of

stellarik [79]

Time taken by proton to complete one complete circular orbit= 7.28 x 10⁻⁸ s

Explanation:

For proton, the centripetal force required for circular motion is provided by the magnetic force,

so Fm= Fc

q v B = m v²/r

m= mass of charged particle

v= velocity

B =magnetic field

q= charge

r= radius of circular path

v= q B r/m

now v= r ω

ω= angular velocity

ω r = q B r /m

ω=q B /m

now ω= 2π/T where T =time period

so 2π/T=q B/m

T= 2 πm/q B

T= 2π (1.67 x 10⁻²⁷)/ [( 1.6 x 10⁻¹⁹)* (0.9)]

T= 7.28 x 10⁻⁸ s

6 0

2 years ago

A friction force of 280 N exists between a cart and the path. If the force of reaction is 766 N, what is the minimum action forc

MatroZZZ [7]

force of friction on the cart is given as

$F_f = 280 N$

here we also know the reaction force due to surface

$R = 766 N$

so we can say reaction force is given as

$R = \sqrt{F_n^2 + F_f^2}$

$766 = \sqrt{F_n^2 + 280^2}$

$F_n^2 = 766^2 - 280^2$

$F_n = 713 N$

now by force balance we will say

$F_y + F_n = mg$

$F_y = mg - F_n$

$F_x = F_f$

also we know that

$F_f = \mu * F_n$

$280 = \mu * 713$

$\mu = 0.39$

now minimum force required to set this into motion

$F_{min} = \frac{\mu mg}{\sqrt{1 + \mu^2}}$

here we know that

$mg = F_n = 713 N$

$F_{min} = \frac{0.39* 713}{\sqrt{1 + 0.39^2}}$

$F_{min} = 259 N$

So it will require 259 N minimum force to move it

3 0

2 years ago

Pain in the pleura

kow [346]

Answer:

b . peuritis .

hope it helped

8 0

2 years ago

Read 2 more answers

A Ferris wheel starts at rest and builds up to a final angular speed of 0.70 rad/s while rotating through an angular displacemen

PilotLPTM [1.2K]

Answer:

The average angular acceleration is 0.05 radians per square second.

Explanation:

Let suppose that Ferris wheel accelerates at constant rate, the angular acceleration as a function of change in angular position and the squared final and initial angular velocities can be clear from the following expression:

$\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha\cdot (\theta-\theta_{o})$

Where:

$\omega_{o}$ , $\omega$ - Initial and final angular velocities, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$\theta_{o}$ , $\theta$ - Initial and final angular position, measured in radians.

Then,

$\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}$

Given that $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $\theta-\theta_{o} = 4.9\,rad$ , the angular acceleration is:

$\alpha = \frac{\left(0.70\,\frac{rad}{s} \right)^{2}-\left(0\,\frac{rad}{s} \right)^{2}}{2\cdot \left(4.9\,rad\right)}$

$\alpha = 0.05\,\frac{rad}{s^{2}}$

Now, the time needed to accelerate the Ferris wheel uniformly is described by this kinematic equation:

$\omega = \omega_{o} + \alpha \cdot t$

Where $t$ is the time measured in seconds.

The time is cleared and obtain after replacing every value:

$t = \frac{\omega-\omega_{o}}{\alpha}$

If $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $\alpha = 0.05\,\frac{rad}{s^{2}}$ , the required time is:

$t = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{0.05\,\frac{rad}{s^{2}} }$

$t = 14\,s$

Average angular acceleration is obtained by dividing the difference between final and initial angular velocities by the time found in the previous step. That is:

$\bar \alpha = \frac{\omega-\omega_{o}}{t}$

If $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $t = 14\,s$ , the average angular acceleration is:

$\bar \alpha = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{14\,s}$

$\bar \alpha = 0.05\,\frac{rad}{s^{2}}$

The average angular acceleration is 0.05 radians per square second.

4 0

2 years ago