Answer:
true
Explanation:
The maximum height of a object in a projectile trajectory occurs when the vertical component of velocity, vy , equals zero. As the projectile moves upwards it goes against gravity, and therefore the velocity begins to decelerate. ... Once the projectile reaches its maximum height, it begins to accelerate downward.
Suppose you have a cylinder filled with diatomic oxygen (O2) and it is running low. The cylinder is shown above, is made of steel, and has a fixed volume of 10 L.
You are asked to determine the number of O2 molecules that are left in the cylinder, so you take a measurement of the temperature to be 20℃. You then note that the pressure gauge reads 100 psi, which you checked at sea level in Bellingham, where the local pressure is one atm (14.7 psi). Calculate the number of O2 molecules left in the container.
Answer:
The number of O₂ molecules that are left in the cylinder is 1.70x10²⁴.
Explanation:
The number of oxygen molecules can be found using the Ideal Gas law:
[tex] PV = nRT [/tex]
Where:
P: is the pressure = 100 psi
V: is the volume = 10 L
n: is the number of moles =?
T: is the temperature = 20 °C = 293 K
R: is the gas constant = 0.082 L*atm/(K*mol)
Hence, the number of moles is:
[tex]n = \frac{PV}{RT} = \frac{100 psi*\frac{1 atm}{14.7 psi}*10 L}{0.082 L*atm/(K*mol)*293 K} = 2.83 moles[/tex]
Now, the number of molecules can be found with Avogadro's number:
[tex]n_{m} = \frac{6.022 \cdot 10^{23}\: molecules}{1\: mol}*2.83 moles = 1.70 \cdot 10^{24} \: molecules[/tex]
Therefore, the number of O₂ molecules that are left in the cylinder is 1.70x10²⁴.
I hope it helps you!
Which factors affect the gravitational force between two objects?
-
distance and velocity
O mass and distance
O mass and weight
acceleration and weight
TELE
Answer:
mass and distance
Explanation:
mass, and distance. The force of gravity depends directly upon the masses of the two objects, and inversely on the square of the distance between them.
A person is standing on a level floor. His head, upper torso, arms, and hands together weigh 458 N and have a center of gravity that is 1.34 m above the floor. His upper legs weigh 120 N and have a center of gravity that is 0.766 m above the floor. Finally, his lower legs and feet together weigh 89.8 N and have a center of gravity that is 0.204 m above the floor. Relative to the floor, find the location of the center of gravity for the entire body.
Answer:
the location of the center of gravity for the entire body is 1.08 m
Explanation:
Given the data in the question;
w1 = 458 N, y1 = 1.34 m
w2 = 120 N, y2 = 0.766 m
w3 = 89.8 N, y2 = 0.204 m
The location arrangement of the body part is vertical, locate the overall centre of gravity by simply replacing the horizontal position x by the vertical position y as measured relative to the floor.
so,
[tex]Y_{centre of gravity}[/tex] = (w1y1 + w2y2 + w3y3 ) / ( w1 + w2 + w3 )
so we substitute in our values
[tex]Y_{centre of gravity}[/tex] = (458×1.34 + 120×0.766 + 89.8×0.204 ) / ( 458 + 120 + 89.8 )
[tex]Y_{centre of gravity}[/tex] = 723.9592 / 667.8
[tex]Y_{centre of gravity}[/tex] = 1.08 m
Therefore, the location of the center of gravity for the entire body is 1.08 m
I need help with this, I can't figure it out.
Here is the link:
https://platform.breakoutedu.com/game/play/fun-in-the-sun-140559-XMWCQF1LPL
Answer:
Star, Triangle, Circle, Rhombus, Square.Left, Down, Right, Down, Up.2,3,2,4.L,O,O,K,I,N,G,F,L,Y.Explanation: You're welcome ✓
An unfortunate astronaut loses his grip during a spacewalk and finds himself floating away from the space station, carrying only a rope and a bag of tools. First he tries to throw a rope to his fellow astronaut, but the rope is too short. In a last ditch effort, the astronaut throws his bag of tools in the direction of his motion, away from the space station. The astronaut has a mass of ma=102 kgma=102 kg and the bag of tools has a mass of mb=10.0 kg.mb=10.0 kg. If the astronaut is moving away from the space station at vi=2.10 m/svi=2.10 m/s initially, what is the minimum final speed vb,fvb,f of the bag of tools with respect to the space station that will keep the astronaut from drifting away forever?
Answer:
The answer is "[tex]2.352 \ \frac{m}{s}[/tex]"
Explanation:
[tex]\to mass(m_1)=102 \ kg\\\\\to mass(m_2)=10 \ kg \\\\\to v=2.10\ \frac{m}{s}\\\\[/tex]
momentum before:
[tex]\to p=(m_1+m_2)v[/tex]
[tex]=(102+10)2.10\\\\=(102\times 2.10 +10 \times 2.10)\\\\=214.2+21\\\\=235.2[/tex]
momentum After:
[tex]\to p=(m_1+m_2)v[/tex]
[tex]=(102\times 0 +10 \times v)\\\\ =(0 +10v)\\\\=10v\\[/tex]
Calculating the conservation of momentum:
[tex]\to \text{momentum before = momentum After}[/tex]
[tex]\to 235.2=10v\\\\\to v= \frac{235.2}{10}\\\\ \to v=2.352 \ \frac{m}{s}[/tex]