Answer:
The time of Mars is 1.65 times larger on Mars than on Earth
Explanation:
The equation that describes the system is the final speed is equal to the speed minos the speed lost by the collision with the porhole
Vf = Vo - V pothole
B) let's transform the weight of free groin system and N international system
1 N = 0.2248 lb
2.8 lbs (1N / 0.2248lbs) = 12.5 N
c) Kinematic equations are the same in all inertial systems, Mars and Earth, so we can use the height equation, with zero initial velocity
Y = Vo t - ½ g t²
Y = - ½ g t²
t = √ 2Y / g
Mars
gm = 0.37g
gm = 0.37 9.8
gm = 3,626 m / s²
t = √( 2 1.9 / 3.626
)
t = 1.02 s
Earth
t = √( 2 1.9 / 9.8)
t = 0.62 s
To make the comparison of time we are the relationship between the two
tm / te = 1.02 / 0.62
tm / te = 1.65
The time of Mars is 1.65 times larger on Mars than on Earth
The complex, highly technical formula for capacitors is
<em>Q = C V</em>
Charge = (capacitance) (voltage)
Charge = (3 F) (24 V)
<em>Charge = 72 Coulombs</em>
The positive plate of the capacitor is missing 72 coulombs worth of electrons. They were sucked into positive terminal of the battery stack.
The negative plate of the capacitor has 72 coulombs worth of extra electrons. They came from the negative terminal of the battery stack.
You should be aware that this is a humongous amount of charge ! An average <u><em>lightning bolt</em></u>, where electrons flow between a cloud and the ground for a short time, is estimated to transfer around <u><em>15 coulombs</em></u> of charge !
The scenario in the question involves a "supercapacitor". 3 F is is no ordinary component ... One distributor I checked lists one of these that's able to stand 24 volts on it, but that product costs $35 apiece, you have to order at least 100 of them at a time, and they take 2 weeks to get.
Also, IF you can charge this animal to 24 volts, it will hold 864J of energy. You'd probably have a hard time accomplishing this task with a bag of leftover AA batteries.
Answer:
Explanation:
<h2><u>Given</u> :-</h2>
<h2><u>To Find</u> :-</h2>
<h2><u>Formula to be used</u> :-</h2>
Where,
- K.E. = Kinetic energy possessed by the body
- M = Mass of the body
- V = Velocity of the body
<h2><u>Solution</u> :-</h2>








- Velocity of the vehicle at the instant is

Answer:
(2x + 4)(x - 4)=2x^2-4x-16