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1 year ago

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A rocket is fired in deep space, where gravity is negligible. If the rocket has a mass of 6000 kg and ejects gas at a relative v

elocity of 2000 m/s, how much gas must it eject in the first second to have an initial acceleration of 25.0 m/s?

1 answer:

shutvik [7]1 year ago

6 0

25 miles per second worth of gas

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The speed of light in benzene is 2.00×108 m/s. what is the index of refraction of benzene?

Klio2033 [76]

The index of refraction of a material is the ratio between the speed of light in vacuum, c, and the speed of light in that material, v:
$n= \frac{c}{v}$
where the speed of light in vacuum is $c=3 \cdot 10^8 m/s$ . The speed of light in benzene is $v=2.00 \cdot 10^8 m/s$ , so we can use the previous relationship to find the refractive index of benzene:
$n= \frac{3 \cdot 10^8 m/s}{2.00 \cdot 10^8 m/s}=1.5$

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1 year ago

Medical cyclotrons need efficient sources of protons to inject into their center. In one kind of ion source, hydrogen atoms (i.e

Arlecino [84]

Answer:

The radius is $r = 4.434 *10^{-5} \ m$

Explanation:

From the question we are told that

The magnetic field is $B = 90 mT = 90*10^{-3} \ T$

The electron kinetic energy is $KE = 1.4 eV = 1.4 * (1.60*10^{-19}) =2.24*10^{-19} \ J$

Generally for the collision to occur the centripetal force of the electron in it orbit is equal to the magnetic force applied

This is mathematically represented as

$\frac{mv^2}{r} = qvB$

=> $r = \frac{m* v}{q * B}$

Where m is the mass of electron with values $m = 9.1 *10^{-31} \ kg$

v is the escape velocity which is mathematically represented as

$v = \sqrt{\frac{2 * KE}{m} }$

So

$r = \frac{m}{qB} * \sqrt{\frac{2 * KE}{m} }$

apply indices

$r = \frac{\sqrt{2 * KE * m} }{qB}$

substituting values

$r = \frac{\sqrt{2 * 2.24*10^{-19}* 9.1 *10^{-31}} }{ 1.60 *10^{-19}* 90*10^{-3}}$

$r = 4.434 *10^{-5} \ m$

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1 year ago

A damped harmonic oscillator consists of a block of mass 2.5 kg attached to a spring with spring constant 10 N/m to which is app

Cerrena [4.2K]

Answer:

0.5% per oscillation

Explanation:

The term 'damped oscillation' means an oscillation that fades away with time. For Example; a swinging pendulum.

Kinetic energy, KE= 1/2×mv^2-------------------------------------------------------------------------------------------------------------(1).

Where m= Mass, v= velocity.

Also, Elastic potential energy,PE=1/2×kX^2----------------------------------------------------------------------------------------------------------------------(2).

Where k= force constant, X= displacement.

Mechanical energy= potential energy (when a damped oscillator reaches maximum displacement).

Therefore, we use equation (3) to get the resonance frequency,

W^2= k/m--------------------------------------------------------------------------------------(3)

Slotting values into equation (3).

= 10/2.5.

= ✓4.

= 2 s^-1.

Recall that, F= -kX

F^2= (-0.1)^2

Potential energy,PE= 1/2 ×0.01

Potential energy= 0.05 ×100

= 0.5% per oscillation.

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1 year ago

An 1876 N crate is being pushed across a level force at a constant speed by a force of 747 N. What is the coefficient of kinetic

nekit [7.7K]

The crate only moves horizontally, so its net vertical force is 0. The only forces acting in the vertical direction are the crate's weight (pointing downward) and the normal force of the surface on the crate (pointing upward). By Newton's second law, we have

∑ <em>F</em> (vertical) = <em>n</em> - <em>mg</em> = 0 → <em>n</em> = <em>mg</em> = 1876 N

where <em>n</em> is the magnitude of the normal force.

In the horizontal direction, the crate is moving at a constant speed and thus with no acceleration, so it's completely in equilibrium and the net horizontal force is also 0. The only forces acting on it in this direction are the 747 N push (pointing in the direction of the crate's motion) and the kinetic friction opposing it (pointing in the opposite direction). By Newton's second law,

∑ <em>F</em> (horizontal) = 747 N - <em>f</em> = 0 → <em>f</em> = 747 N

The frictional force is proportional to the normal force by a factor of the coefficient of kinetic friction, <em>µ</em>, such that

<em>f</em> = <em>µn</em> → <em>µ</em> = <em>f</em> / <em>n</em> = (747 N) / (1876 N) ≈ 0.398188 ≈ 0.40

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2 years ago

You throw a ball of mass 1 kg straight up. You observe that it takes 2.2 s to go up and down, returning to your hand. Assuming w

Elina [12.6K]

Answer:

10.791 m/s

5.93505 m

Explanation:

m = Mass of ball

$v_f$ = Final velocity

$v_i$ = Initial velocity

$t_f$ = Final time

$t_i$ = Initial time

g = Acceleration due to gravity = 9.81 m/s²

From the momentum principle we have

$\Delta P=F\Delta t$

Force

$F=mg$

So,

$m(v_f-v_i)=mg(t_f-t_i)\\\Rightarrow v_i=v_f-g(t_f-t_i)\\\Rightarrow v_i=0-(-9.81)(1.1-0)\\\Rightarrow v_i=10.791\ m/s$

The speed that the ball had just after it left the hand is 10.791 m/s

As the energy of the system is conserved

$K_i=U\\\Rightarrow \dfrac{1}{2}mv_i^2=mgh\\\Rightarrow h=\dfrac{v_i^2}{2g}\\\Rightarrow h=\dfrac{10.791^2}{2\times 9.81}\\\Rightarrow h=5.93505\ m$

The maximum height above your hand reached by the ball is 5.93505 m

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2 years ago