Answer:
This has all the answer (Data sheet, graph, answers to the questions, and the summary) :)
What is a gas-like mixture that is made of charged particles? (Physical Science)
1 answer:
You might be interested in
From ideal gas law, PV=nRT
where P is the pressure, V is the volume of the container, n is number of moles, R is the gas constant and T is the temperature.
Hence,
T= 110.65 k
Kinetic Energy =
K.E=
The energy an object has as a result of motion is known as kinetic energy.
A force must be applied to an object in order to accelerate it. We must put in effort in order to apply a force. After the work is finished, energy is transferred to the item, which then moves at a new, constant speed. Kinetic energy is the type of energy that is transferred and is dependent on the mass and speed attained.
Kinetic energy can be converted into other types of energy and transported between objects. A flying squirrel may run into a chipmunk that is standing still, for instance. Some of the squirrel's initial kinetic energy may have been transferred to the chipmunk or changed into another kind of energy after the collision.
To know more about kinetic energy, visit:
#SPJ4
Answer:
54.9 m/s at 44.9 degrees
Explanation:
If the ball has a total velocity of 46.2 m/s, at an angle of -32.7 degrees, we can decompose its speed into its horizontal and vertical components.
Vx = V * cos(a) = 46.2 * cos(-32.7) = 38.9 m/s
Vy = V * sin(a) = 46.2 * sin(-32.7) = -25 m/s
SInce there is no force on the horizontal direction (omitting air drag), we can assume constant horizontal speed.
Since a ball thrown is at free fall, only affected by gravity (omitting air drag), we can say it is affected by constant acceleration, therefore we can use
Y(t) = Y0 + Vy0 *t + 1/2 * a * t^2
We consider t=0 as the moment when the ball was hit, so in this case Y0 = 1 m
If we take the first derivative of the equation of position, we get the equation for speed
V(t) = Vy0 + a * t
We know that being t2 the moment the ball goes over the wall
V(t2) = -25 m/s
Y(t2) = 45.4 m
So:
45.4 = 1 + Vy0 * t2 + 1/2 * a * t2^2
-25 = Vy0 + a * t2
Then:
Vy0 = -25 - a * t2
So:
45.4 = 1 + (-25 - a * t2) * t2 + 1/2 * a * t2^2
0 = -44.4 - 25 * t2 - 1/2 * a * t2^2
a = -9.81 m/s^2
0 = -44.4 - 25 * t2 + 4.9 * t2^2
Solving this quadratic equation we get:
t1 = -1.39 s
t2 = 6.5 s
Since we are looking for a positive value we disregard t1.
Now we can obtain Vy0:
Vy0 = -25 + 9.81 * 6.5 = 38.76 m/s
Since horizontal speed is constant Vx0 = 38.9 m/s
By Pythagoras theorem we obtain the value of the initial speed:
The angle is in the the first quadrant because both comonents ate positive, so: 0 < a < 90
a = atan(Vy0/Vx0) = 44.9 degrees