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netineya [11]
3 years ago
10

If any state tribunal decides a federal question and the litigant has no further remedy within the state court system, from whic

h of the following remedies could the litigant possibly benefit?
A.
a hearing by a US district court
B.
a hearing by a federal court of appeals
C.
a hearing by the US Supreme Court
D.
a Congressional vote
Physics
1 answer:
ANEK [815]3 years ago
6 0

Answer:    B. A hearing by a federal court of appeals

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Which factor/s affect the amount of energy stored in a magnetic field?
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Answer:

The answer is I

Explanation:

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A gas sample is heated from -20.0°C to 57.0°C and the volume is increased from 2.00 L to 4.50 L. If the initial pressure is 0.14
Svetach [21]

Answer:

The answer is e.

Explanation:

We take:

T_{1}=253K

V_{1}=2.00 l

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V_{2}=4.50 l

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Taking the gas as an ideal gas, we can use the ideal gas law:

\frac{PV}{nRT}

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List 5 occasions in real life when you want friction (motion to stop)
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An object executing simple harmonic motion has a maximum speed of 4.3 m/s and a maximum acceleration of 0.65 m/s2 . find (a) the
Alexxx [7]
(1) The position around equilibrium of an object in simple harmonic motion is described by
x(t) = A \cos (\omega t)
where
A is the amplitude of the motion
\omega is the angular frequency.

The velocity is the derivative of the position:
v(t)=-\omega A \sin(\omega t) = -v_0 \sin (\omega t)
where 
v_0 = \omega A is the maximum velocity of the object.

The acceleration is the derivative of the velocity:
a(t)=- \omega^2 A \cos (\omega t) = -a_0 \cos (\omega t)
where
a_0=\omega^2 A is the maximum acceleration of the object.

We know from the problem both maximum velocity and maximum acceleration:
v_0 = \omega A = 4.3 m/s
a_0 = \omega^2 A = 0.65 m/s^2
From the first equation, we get
A= \frac{4.3 }{\omega} (1)
and if we substitute this into the second equation, we find the angular fequency
\omega=0.15 rad/s
while the amplitude is  (using (1)):
A=28.7 m


(b) We found in the previous step that the angular frequency of the motion is
\omega=0.15 rad/s
But the angular frequency is related to the period by
T= \frac{2 \pi}{\omega}
and so, the period is
T= \frac{2 \pi}{\omega}= \frac{2 \pi}{0.15 rad/s}=41.9 s
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