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3 years ago

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A 7.0-kg rock is subject to a variable force given by the equation f(x)=6.0n−(2.0n/m)x+(6.0n/m2)x2 if the rock initially is at r

est at the origin, find its speed when it has moved 9.0 m.

1 answer:

stealth61 [152]3 years ago

4 0

For Newton's second law, the force is equal to the product between the mass and the acceleration of the rocket:
$F=ma$
From which we can rewrite the acceleration as
$a(x)= \frac{F}{m} = \frac{1}{m} (6-2x+6x^2)$
where m=7.0 kg.

The velocity of the rocket is the derivative of the acceleration:
$v(x) = \frac{1}{m} (-2+12 x)$
and if we substitute x=9.0 m, we find the rocket velocity after 9.0 m:
$v(9)= \frac{1}{7}(-2+12\cdot 9)=15.1 m/s$

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The viewing screen in a double-slit experiment with monochromatic light. Fringe C is the central maximum. The fringe separation

makvit [3.9K]

Answer:

<em>Part A</em><em>:</em>

a) If the wavelength of the light is decreased the fringe spacing Δy will decrease.

<em>Part B</em><em>:</em>

b) If the spacing between the slits is decreased the fringe spacing Δy will increase.

<em>Part C</em><em>:</em>

a) If the distance to the screen is decreased the fringe spacing will decrease.

<em>Part D</em><em>:</em>

The dot in the center of fringe E is $920\ x\ 10^{-9} m$ farther from the left slit than from the right slit.

Explanation:

In the double-slit experiment there is a clear contrast between the dark and bright fringes, that indicate destructive and constructive interference respectively, in the central peak and then is less so at either side.

The position of bright fringes in the screen where the pattern is formed can be calculated with

$\vartriangle y =\frac{m \lambda L}{d}$

$m=0,\pm 1,\pm 2,\pm 3,.....$

m is the order number.
$\lambda$ is the wavelength of the monochromatic light.
L is the distance between the screen and the two slits.
d is the distance between the slits.

Part A: a) In the above equation for the position of bright fringes we can see that if the wavelength of the light $\lambda$ is decreased the overall effect will be that the fringes are going to be closer. That means that the fringe spacing Δy will decrease.

Part B: b) In the above equation for the position of bright fringes we can see that if the spacing between the slits d is decreased the fringes are going to be wider apart. That means the fringe spacing Δy will increase.
Part C: a) In the above equation we can see that if the distance to the screen L is decreased the fringes are going to be closer. That means the fringe spacing Δy will decrease.

Part D: We are told that the central maximum is the fringe C that corresponds with m=0. That means that fringe E corresponds with the order number m=2 if we consider it to be the second maximum at the rigth of the central one. To calculate how much farther from the left slit than from the right slit is a dot located at the center of the fringe E in the screen we use the condition for constructive interference. That says that the path length difference Δr between rays coming from the left and right slit must be $\vartriangle r=m \lambda$

We simply replace the values in that equation :

$\vartriangle r= m \lambda =2.\ 460\ nm$

$\vartriangle r= 920\ x\ 10^{-9} m$

The dot in the center of fringe E is $920\ x\ 10^{-9}m$ farther from the left slit than from the right slit.

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Astronomers have observed a small, massive object at the center of our Milky Way Galaxy. A ring of material orbits this massive

Setler [38]

Answer:

The mass of the massive object at the center of the Milky Way galaxy is $3.44\times10^{37}\ Kg$

Explanation:

Given that,

Diameter = 10 light year

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Suppose determine the mass of the massive object at the center of the Milky Way galaxy.

Take the distance of one light year to be 9.461×10¹⁵ m. I was able to get this it is 4.26×10³⁷ kg.

We need to calculate the radius of the orbit

Using formula of radius

$r=\dfrac{d}{2}$

$r=\dfrac{15\times9.461\times10^{15}}{2}$

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We need to calculate the mass of the massive object at the center of the Milky Way galaxy

Using formula of mass

$M=\dfrac{v^2r}{G}$

Put the value into the formula

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$M=3.44\times10^{37}\ Kg$

Hence, The mass of the massive object at the center of the Milky Way galaxy is $3.44\times10^{37}\ Kg$

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Yuliya22 [10]

<h2>Mercury, Neptune, and Jupiter </h2>

Explanation:

Mercury has the largest semimajor axis that is 5.791 x 107 in km.
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Neptune has the longest orbital speed around the sun of any planet in the Solar System which is equivalent to 164.8 years (or 60,182 Earth days)
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Hence, the answer is Mercury, Neptune, and Jupiter respectively.

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Answer:

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So, the mass of the bucket is 5.22 kg.

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B. is not a validated bu experimentation

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