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Kryger [21]
2 years ago
8

What are the types of federal court cases

Physics
2 answers:
Advocard [28]2 years ago
8 0

Answer:

More specifically, federal courts hear criminal, civil, and bankruptcy cases. And once a case is decided, it can often be appealed.

Explanation:

I hope this helps and pls mark me brainliest! Thank you :)

Flura [38]2 years ago
8 0

Answer:

There are many types of federal court cases. Examples are civil, criminal, bankruptcy. There are also special "military courts" for cases related to military crimes like desertion, etc.

Explanation:

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The accepted value for the acceleration due to gravity is 9.8 m/s2. What factors do you think are responsible for the difference
kakasveta [241]
It’s hard to perfectly measure the distance something travels, as well as the exact time it takes, making the results have some variation.
3 0
3 years ago
For a given initial projectile speed Vo, calculate what launch angle A gives the longest range R. Show your work, don't just quo
pickupchik [31]
The optimal angle of 45° for maximum horizontal range is only valid when initial height is the same as final height. 

<span>In that particular situation, you can prove it like this: </span>

<span>initial velocity is Vo </span>
<span>launch angle is α </span>

<span>initial vertical velocity is </span>
<span>Vv = Vo×sin(α) </span>

<span>horizontal velocity is </span>
<span>Vh = Vo×cos(α) </span>

<span>total time in the air is the the time it needs to fall back to a height of 0 m, so </span>
<span>d = v×t + a×t²/2 </span>
<span>where </span>
<span>d = distance = 0 m </span>
<span>v = initial vertical velocity = Vv = Vo×sin(α) </span>
<span>t = time = ? </span>
<span>a = acceleration by gravity = g (= -9.8 m/s²) </span>
<span>so </span>
<span>0 = Vo×sin(α)×t + g×t²/2 </span>
<span>0 = (Vo×sin(α) + g×t/2)×t </span>
<span>t = 0 (obviously, the projectile is at height 0 m at time = 0s) </span>
<span>or </span>
<span>Vo×sin(α) + g×t/2 = 0 </span>
<span>t = -2×Vo×sin(α)/g </span>

<span>Now look at the horizontal range. </span>
<span>r = v × t </span>
<span>where </span>
<span>r = horizontal range = ? </span>
<span>v = horizontal velocity = Vh = Vo×cos(α) </span>
<span>t = time = -2×Vo×sin(α)/g </span>
<span>so </span>
<span>r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) </span>
<span>r = -(Vo)²×sin(2α)/g </span>

<span>To find the extreme values of r (minimum or maximum) with variable α, you must find the first derivative of r with respect to α, and set it equal to 0. </span>

<span>dr/dα = d[-(Vo)²×sin(2α)/g] / dα </span>
<span>dr/dα = -(Vo)²/g × d[sin(2α)] / dα </span>
<span>dr/dα = -(Vo)²/g × cos(2α) × d(2α) / dα </span>
<span>dr/dα = -2 × (Vo)² × cos(2α) / g </span>

<span>Vo and g are constants ≠ 0, so the only way for dr/dα to become 0 is when </span>
<span>cos(2α) = 0 </span>
<span>2α = 90° </span>
<span>α = 45° </span>
4 0
3 years ago
A race car accelerates on a straight road from rest to speed of 180 km/hr in 25s . Assuming uniform acceleration of the car thro
adelina 88 [10]
X-x0=0.5(v0+v)t ergo x=0.5(50)25=625m
3 0
3 years ago
A balloon is ascending at a rate of +4.00 m/s to a height of 11.0 m above the ground when a package is dropped. In the absence o
o-na [289]

Answer:

Vf = 14.7 m/s

Explanation:

Vf² = Vi² + 2 * a * Δy

given:

a = 9.81 m/s²

Δy = 11m

Vi = 0 when upon release

Vf² = 0 + 2 (9.81) 11

Vf = 14.7 m/s

6 0
3 years ago
PLEASE HELP also i know this isnt physics but there was no HOPE/Gym subject for me to choose.
Andreyy89

Answer:

Plan B.

Because flexibility is best improved by stretching.

Explanation:

Improving and increasing flexibility is done by having stretching sessions daily which maintains and widens the range of motion in the joints and stretches muscles.

4 0
2 years ago
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