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Allushta [10]
3 years ago
14

When a court has "___ ___" they can hear a case once a lower court has ruled on it.

Physics
1 answer:
notka56 [123]3 years ago
3 0
Your answer is going to be Appellate jurisdiction.
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In an electric vehicle, each wheel is powered by its own motor. The vehicle weight is 4,000 lbs. By regenerative braking, its sp
Slav-nsk [51]

Answer:

the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh

Explanation:

Given that;

weight of vehicle = 4000 lbs

we know that 1 kg = 2.20462

so

m = 4000 / 2.20462 =  1814.37 kg

Initial velocity V_{i} = 60 mph = 26.8224 m/s

Final velocity V_{f} = 30 mph = 13.4112 m/s

now we determine change in kinetic energy

Δk = \frac{1}{2}m(  V_{i}² - V_{f}² )

we substitute

Δk = \frac{1}{2}×1814.37( (26.8224)² - (13.4112)² )

Δk = \frac{1}{2} × 1814.37 × 539.5808

Δk = 489500 Joules

we know that; 1 kilowatt hour = 3.6 × 10⁶ Joule

so

Δk = 489500 / 3.6 × 10⁶

Δk = 0.13597 ≈ 0.136 kWh

Therefore, the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh

4 0
3 years ago
A 7.00 kg bowling ball is held 2.00 m above the ground. Using g- 9.8 m/s^2, how much energy does the bowling ball have due to it
Alexeev081 [22]
It should be B.137 just took the test.
3 0
3 years ago
1. Aunt Minnie gives you $10 per second for 4 seconds. How much money do you have after 4 se
9966 [12]

Answer:

$40

Explanation:

10 × 4 = 40

or

10 + 10 + 10 + 10 = 40

4 0
3 years ago
Read 2 more answers
The function of a capacitor in an electric circuit is to
jok3333 [9.3K]

Answer: B. store electric charges.

Explanation: I JUST TOOK THE PF EXAM AND I GOT IT CORRECT!!!!

7 0
3 years ago
A charging RC circuit controls the intermittent windshield wipers in a car. The emf is 12.0 V. The wipers are triggered when the
lilavasa [31]

Answer:

R=803k\Omega

Explanation:

We have the following information,

V_0 = 12V\\V=10V\\c= 1.25*10^{-6}F\\t=1.8s

We apply the equation for capacitor charging the voltage across it,

V=V_0 (1-e^{-t/x})\\e^{-t/x}=1-(\frac{V}{V_0})\\-\frac{t}{Rc}=ln(\frac{V}{V_0})\\R=-\frac{t}{ln(\frac{V}{V_0})*c}

Replacing values,

R=-\frac{1.8}{ln(10/12)*1.25*10^{-6}}

R=803k\Omega

3 0
3 years ago
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