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ratelena [41]
2 years ago
9

What do executive departments do?​

Physics
1 answer:
enot [183]2 years ago
8 0

Answer:

Image result for What do executive departments do?​

Under Article II of the Constitution, the President is responsible for the execution and enforcement of laws created by Congress. Fifteen executive departments—each led by an appointed member of the President's Cabinet—carry out the day-to-day administration of the Federal Government.

Explanation:

The Cabinet and independent federal agencies are responsible for the day-to-day enforcement and administration of federal laws. ... Fifteen executive departments — each led by an appointed member of the President's Cabinet — carry out the day-to-day administration of the federal government.

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On the earth, when an astronaut throws a 0.250-kg stone vertically upward, it returns to his hand a time T later. On planet X he
Pachacha [2.7K]

Answer:

d) g/2

Explanation:

We need to use one of Newton's equations of motion to find the position of the stone at any time t.

x(t) = x₀(t) + ut - ¹/₂at²

Where

x₀(t) = initial position of the stone.

x(t) - x₀(t) = distance traveled by the stone at any time.

u = initial velocity of the stone

a = acceleration of the stone

t = time taken

On both planets, before the stone was thrown by the astronaut, x = 0 and t = 0.

=> 0 = x₀(t)

=> x₀(t) = 0

On earth, when the stone returns into the hand of the astronaut at time T on earth, x = 0.

=> 0 = 0 + uT - ¹/₂gT² (a = g)

=> uT = ¹/₂gT²

=> g = 2u/T

On planet X, when the stone returns into the hand of the astronaut, time = 2T , x = 0.

=> 0 = 0 + u(2T) - ¹/₂a(2T)²

=> 2uT = 2aT²

=> a = u/T

By comparing we see that a = g/2.

5 0
3 years ago
The resistance of the coil of a
ololo11 [35]

Answer:

54 is the correct answer to this question

7 0
3 years ago
Is there gravitational force between two students sitting in a classroom?
drek231 [11]

Answer:

Yes.

Explanation: the magnitude of the force is extremely small because the masses of the students are small relative to Earth's mass.

8 0
3 years ago
How many revolutions per minute would a 23 m -diameter Ferris wheel need to make for the passengers to feel "weightless" at the
kirza4 [7]

Answer:

Approximately 6.2\; {\rm rpm}, assuming that the gravitational field strength is g = 9.81\; {\rm m\cdot s^{-2}}.

Explanation:

Let \omega denote the required angular velocity of this Ferris wheel. Let m denote the mass of a particular passenger on this Ferris wheel.

At the topmost point of the Ferris wheel, there would be at most two forces acting on this passenger:

  • Weight of the passenger (downwards), m\, g, and possibly
  • Normal force F_\text{normal} that the Ferris wheel exerts on this passenger (upwards.)

This passenger would feel "weightless" if the normal force on them is 0- that is, F_\text{normal} = 0.

The net force on this passenger is (m\, g - F_\text{normal}). Hence, when F_\text{normal} = 0, the net force on this passenger would be equal to m\, g.

Passengers on this Ferris wheel are in a centripetal motion of angular velocity \omega around a circle of radius r. Thus, the centripetal acceleration of these passengers would be a = \omega^{2}\, r. The net force on a passenger of mass m would be m\, a = m\, \omega^{2}\, r.

Notice that m\, \omega^{2} \, r = (\text{Net Force}) = m\, g. Solve this equation for \omega, the angular speed of this Ferris wheel. Since g = 9.81\; {\rm m\cdot s^{-2}} and r = 23\; {\rm m}:

\begin{aligned} \omega^{2} = \frac{g}{r}\end{aligned}.

\begin{aligned} \omega &= \sqrt{\frac{g}{r}} \\ &= \sqrt{\frac{9.81\; {\rm m \cdot s^{-2}}}{23\; {\rm m}}} \\ &\approx 0.653\; {\rm rad \cdot s^{-1}} \end{aligned}.

The question is asking for the angular velocity of this Ferris wheel in the unit {\rm rpm}, where 1\; {\rm rpm} = (2\, \pi\; {\rm rad}) / (60\; {\rm s}). Apply unit conversion:

\begin{aligned} \omega &\approx 0.653\; {\rm rad \cdot s^{-1}} \\ &= 0.653\; {\rm rad \cdot s^{-1}} \times \frac{1\; {\rm rpm}}{(2\, \pi\; {\rm rad}) / (60\; {\rm s})} \\ &= 0.653\; {\rm rad \cdot s^{-1} \times \frac{60\; {\rm s}}{2\, \pi\; {\rm rad}} \times 1\; {\rm rpm} \\ &\approx 6.2\; {\rm rpm} \end{aligned}.

3 0
1 year ago
In astronomy, distances are often expressed in light-years. One light-year is the distance traveled by light in one year. If the
amid [387]

Answer:

dd=4.3 ly=4.3 •9.46•10°15 m=4.1 10°16m.

Explanation:

5 0
2 years ago
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