Answer:
T=183.21K
Explanation:
We have to take into account that the system is a ideal gas. Hence, we have the expression

where P is the pressure, V is the volume, n is the number of moles, T is the temperature and R is the ideal gas constant.
Thus, it is necessary to calculate n and V
V is the volume of a sphere

V=8.86*10^{50}L
and for n

Hence, we have (1 Pa = 9.85*10^{-9}atm)

hope this helps!!
Answer:
a) a= 8.33 m/s², T = 12.495 N
, b) a = 2.45 m / s²
Explanation:
a) this is an exercise of Newton's second law. As the upper load is secured by a cable, it cannot be moved, so the lower load is determined by the maximum acceleration.
We apply Newton's second law to the lower charge
fr₁ + fr₂ = ma
The equation for the force of friction is
fr = μ N
Y Axis
N - W₁ –W₂ = 0
N = W₁ + W₂
N = (m₁ + m₂) g
Since the beams are the same, it has the same mass
N = 2 m g
We replace
μ₁ 2mg + μ₂ mg = m a
a = (2μ₁ + μ₂) g
a = (2 0.30 + 0.25) 9.8
a= 8.33 m/s²
Let's look for cable tension with beam 2
T = m₂ a
T = 1500 8.33
T = 12.495 N
b) For maximum deceleration the cable loses tension (T = 0 N), so as this beam has less friction is the one that will move first, we are assuming that the rope is horizontal
fr = m₂ a₂
N- w₂ = 0
N = W₂ = mg
μ₂ mg = m a₂
a = μ₂ g
a = 0.25 9.8
a = 2.45 m / s²
The acceleration produced in a body is always in the direction of the resultant force acting on the body. Therefore, we may determine the horizontal acceleration using the horizontal force applied. To do this, we may apply the mathematical form of Newton's second law:
Force = mass * acceleration
acceleration = force / mass
Substituting the values,
a = 100 / 0.15
a = 666.7 m/s²
The acceleration of the hockey puck is 670 m/s²
The displacement is the shortest distance between two points, which is 546.41. The displacement for both is 546.41 meters
Average velocity of X = (200 + 200 + 200) / 30
Average velocity of X = 20 m/s
Average velocity of Y = 546.41 / 30 = 18.2 m/s
Answer: 13.1° from the horizontal
Explanation: For projectile
Horizontal distance R of a projectile is:
R= U×Usin2x/g
Where U is the speed of projectile, x is angle of projectile and g= acceleration due to gravity=9.8m/s
R= 15×15sin(2×31)/9.8= 225sin(62)/9.8=20.27m
Therefore if R is halved.
R/2 = 20.27/2=10.14m
Hence the angle for this case is.
Making sin(2x) the subject of formula
Sin2x= Rg/U×U
R is now 10.14 in this case.
Sin2x= 10.14×9.8/15×15=0.4415
2x=arcsin0.4415=26.2
x= 26.2/2
x= 13.1°
Good luck...