Answer:
63.5 °C
Explanation:
The expression for the calculation of work done is shown below as:
Where, P is the pressure
is the change in volume
Also,
Considering the ideal gas equation as:-
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 8.314 J/ K mol
So,
Also, for change in volume at constant pressure, the above equation can be written as;-
So, putting in the expression of the work done, we get that:-
Given, initial temperature = 28.0 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (28.0 + 273.15) K = 301.15 K
W=1770 J
n = 6 moles
So,
Thus,
The temperature in Celsius = 336.63-273.15 °C = 63.5 °C
<u>The final temperature is:- 63.5 °C</u>
<span>Part B
What are the values of the intial velocity vector components v0,x and v0,y (both in m/s) as well as the acceleration vector components a0,x and a0,y (both in m/s2)? Here the subscript 0 means "at time t0."
15.0, 26.0, 0, -9.80
</span><span>Part C
What are the values of the velocity vector components v1,x and v1,y (both in m/s) as well as the acceleration vector components a1,x and a1,y (both in m/s2)? Here the subscript 1 means that these are all at time t1.
15, 26, 0, -9.81</span><span>
</span>
Answer:
f = 5.3 Hz
Explanation:
To solve this problem, let's find the equation that describes the process, using Newton's second law
∑ F = ma
where the acceleration is
a = 
B- W = m \frac{d^2 y}{dt^2 }
To solve this problem we create a change in the reference system, we place the zero at the equilibrium point
B = W
In this frame of reference, the variable y' when it is oscillating is positive and negative, therefore Newton's equation remains
B’= m 
the thrust is given by the Archimedes relation
B = ρ_liquid g V_liquid
the volume is
V = π r² y'
we substitute
- ρ_liquid g π r² y’ = m \frac{d^2 y'}{dt^2 }
this differential equation has a solution of type
y = A cos (wt + Ф)
where
w² = ρ_liquid g π r² /m
angular velocity and frequency are related
w = 2π f
we substitute
4π² f² = ρ_liquid g π r² / m
f = 
calculate
f = 
f = 5.3 Hz
Answer:
D) The ball exerts a force on the wall and the wall exerts a force back.
Explanation:
Newton's third law of motion states that:
"When an object A exerts a force on another object B, then object B exerts an equal and opposite force on object A"
In this problem, we can identify (for instance) object A with tha ball and object B with the wall. Therefore, if we apply Newton's third law, we get:
The ball (object A) exerts a force on the wall (object B), therefore the wall (object B) exerts an equal and opposite force on the ball (object A). So, option D is the correct one.
