2 years ago

Where did the missouri compromise imaginary line run

1 answer:

7 0

In the Missouri Compromise, the slavery line for future US states ran along the southern border of Missouri at 36 degrees north 30 minutes

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10 points and brailiest to correct answer plz

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Mars is the planet that's closest in size to Earth.

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An object has a position given by r = [2.0 m + (5.00 m/s)t] i^ + [3.0m−(2.00 m/s2)t2] j^, where all quantities are in SI units.

makkiz [27]

Answer:

Acceleration of the object is $4\ m/s^2$ .

Explanation:

It is given that, the position of the object is given by :

$r=[2\ m+(5\ m/s)t]i+[3\ m-(2\ m/s^2)t^2]j$

Velocity of the object, $v=\dfrac{dr}{dt}$

Acceleration of the object is given by :

$a=\dfrac{d^2r}{dt^2}$

$a=\dfrac{d^2}{dt^2}([2\ m+(5\ m/s)t]i+[3\ m-(2\ m/s^2)t^2]j)$

Using the property of differentiation, we get :

$a=\dfrac{d^2r}{dt^2}=-4\ m/s^2$

So, the magnitude of the acceleration of the object at time t = 2.00 s is $4\ m/s^2$ . Hence, this is the required solution.

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2 years ago

If 27 J of work are needed to stretch a spring from 15 cm to 21 cm and 45 J are needed to stretch it from 21 cm to 27 cm, what i

kupik [55]

Answer:

9 cm.

Explanation:

The energy used for stretch the spring from $15 cm$ to $21 cm$ will be , $E_{1}=27J$

The energy used for stretch the spring from $21 cm$ to $27 cm$ will be , $E_{2}=45J$

using the energy of spring formula ,we find that

$27 = \frac{1}{2}K((21-L^{2})-(15-L^{2}))$

$45 = \frac{1}{2}K((27-L^{2})-(21-L^{2}))$

Dividing both the equation will get,

$\frac{3}{5}=\frac{(21-L)^{2}-(15-L)^{2}}{(27-L)^{2}-(21-L)^{2}}\\5((21-L)^{2}-(15-L)^{2})=3((27-L)^{2}-(21-L)^{2})\\3(729 - 54L + L^{2}- 441 + 42L - L^{2} ) = 5(441 - 42L + L^{2} - 225 + 30L - L^{2} )\\3(288 - 12L) = 5(216 - 12L)\\24L = 216\\L = 9 cm$

Therefore, the natural length of the spring is, 9 cm.

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eimsori [14]

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What is the process called when a solid changes to a<br> liquid?<br><br> ???20 points?ASAP please

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Answer:

melting

Explanation:

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