Answer:
Classification of the Elements. These three groups are: metals, nonmetals, and inert gases.
Explanation:
Answer:
I= 3.5 amps
Explanation:
Step one:
given data
rating of resistor R= 8 ohms
power P= 100W
Required
The current I
Step two
Yet this power is also given by

make I subject of the formula we have

substitute

Slightly raising your body temperature while increase oxygen and blood circulation throughout your body
To explain, I will use the equations for kinetic and potential energy:

<h3>Potential energy </h3>
Potential energy is the potential an object has to move due to gravity. An object can only have potential energy if 1) <u>gravity is present</u> and 2) <u>it is above the ground at height h</u>. If gravity = 0 or height = 0, there is no potential energy. Example:
An object of 5 kg is sitting on a table 5 meters above the ground on earth (g = 9.8 m/s^2). What is the object's gravitational potential energy? <u>(answer: 5*5*9.8 = 245 J</u>)
(gravitational potential energy is potential energy)
<h3>Kinetic energy</h3>
Kinetic energy is the energy of an object has while in motion. An object can only have kinetic energy if the object has a non-zero velocity (it is moving and not stationary). An example:
An object of 5 kg is moving at 5 m/s. What is the object's kinetic energy? (<u>answer: 5*5 = 25 J</u>)
<h3>Kinetic and Potential Energy</h3>
Sometimes, an object can have both kinetic and potential energy. If an object is moving (kinetic energy) and is above the ground (potential), it will have both. To find the total (mechanical) energy, you can add the kinetic and potential energies together. An example:
An object of 5 kg is moving on a 5 meter table at 10 m/s. What is the objects mechanical (total) energy? (<u>answer: KE = .5(5)(10^2) = 250 J; PE = (5)(9.8)(5) = 245 J; total: 245 + 250 = 495 J</u>)
Answer:
v_{f} = 115.95 m / s
Explanation:
This is an exercise of a variable mass system, let's form a system formed by the masses of the rocket, the mass of the engines and the masses of the injected gases, in this case the system has a constant mass and can be solved using the conservation the amount of movement. Which can be described by the expressions
Thrust =
-v₀ = v_{e} 
where v_{e} is the velocity of the gases relative to the rocket
let's apply these expressions to our case
the initial mass is the mass of the engines plus the mass of the fuel plus the kill of the rocket, let's work the system in SI units
M₀ = 25.5 +12.7 + 54.5 = 92.7 g = 0.0927 kg
The final mass is the mass of the engines + the mass of the rocket
M_{f} = 25.5 +54.5 = 80 g = 0.080 kg
thrust and duration of ignition are given
thrust = 5.26 N
t = 1.90 s
Let's start by calculating the velocity of the gases relative to the rocket, where we assume that the rate of consumption is linear
thrust = v_{e}
v_{e} = thrust
v_{e} = 5.26
v_{e} = - 786.93 m / s
the negative sign indicates that the direction of the gases is opposite to the direction of the rocket
now we look for the final speed of the rocket, which as part of rest its initial speed is zero
v_{f}-0 = v_{e} 
we calculate
v_{f} = 786.93 ln (0.0927 / 0.080)
v_{f} = 115.95 m / s