Why did the Framers establish the Legislative Branch in Article I of the Constitution?

Physics

1 answer:

shepuryov [24]3 years ago

7 0

Answer:

I think it will be A.

Explanation:

they felt people would more easily accept a strong military if it was controlled by the legislative branch.

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The peak of the trajectory occurs at time t1. This is the point where the ball reaches its maximum height ymax. At the peak the

Digiron [165]

Part B
What are the values of the intial velocity vector components v0,x and v0,y (both in m/s) as well as the acceleration vector components a0,x and a0,y (both in m/s2)? Here the subscript 0 means "at time t0."
15.0, 26.0, 0, -9.80

Part C
What are the values of the velocity vector components v1,x and v1,y (both in m/s) as well as the acceleration vector components a1,x and a1,y (both in m/s2)? Here the subscript 1 means that these are all at time t1.
15, 26, 0, -9.81

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3 years ago

What's the difference in a balanced force in an unbalanced force

AlekseyPX

There's no such thing as one balanced force or one unbalanced force.

If ALL of the forces in a GROUP of forces acting on the same object
all add up to zero, then we say that the GROUP of forces is balanced.

If they don't, then the GROUP of forces is unbalanced.

Two or more forces can be balanced or unbalanced.
One force can't.

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3 years ago

Ah electron is a negatively charged particle that has a charge of magnitude, e - 1.60 x 10-19 C. Which one of the following stat

Advocard [28]

Answer:

The correct statement is "The electric field is directed toward the electron and has a magnitude of $\rm \dfrac{ke}{r^2}$ ".

Explanation:

According to Coulomb's law, the magnitude of the electric field due to a static point charge q at a point r distance away from it is given by

$\rm E = \dfrac{k|q|}{r^2}.$

k is the Coulmob's constant.

The direction of the electric field along the line joining the charge and the point where electric field is to be found and it is directed from positive charge to negative charge.

Conventionally, we assume a positive test charge placed at the point where electric field is to be found, the test charge has very small charge such that its charge does not affect the electric field due to the given charge.

The charge on the electron = -e.

The electric field due to an electron is given by

$\rm E = \dfrac{k|-e|}{r^2}=\dfrac{ke}{r^2}.$

The direction of this electric field is from positive test charge, placed at the point where electric field is to be found, towards the electron along the line joining the two.

Thus, the correct statement is "The electric field is directed toward the electron and has a magnitude of $\rm \dfrac{ke}{r^2}$ ".

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3 years ago

The initial and final velocities of two blocks experiencing constant acceleration are respectively −7.45 m/s and 14.9 m/s. (a) T

ikadub [295]

Answer:

a) $a_{1}=3.7 m/s^{2}$