Answer:
= 85.7 ° C
Explanation:
For this exercise we will use the calorimetry heat ratios, let's start with the heat lost by the evaporation of coffee, since it changes from liquid to vapor state
Q₁ = m L
Where m is the evaporated mass (m = 2.00 103-3kg) and L is 2.26 106 J / kg, where we use the latent heat of the water
Q₁ = 2.00 10⁻³ 2.26 10⁶
Q1 = 4.52 10³ J
Now the heat of coffee in the cup, which does not change state is
Q coffee = M 
( -
)
Since the only form of energy transfer is terminated, the heat transferred is equal to the evaporated heat
Qc = - Q₁
M ce ( -) = - Q₁
The coffee dough left in the cup after evaporation is
M = 250 -2 = 248 g = 0.248 kg
-Ti = -Q1 / M
= Ti - Q1 / M
Since coffee is essentially water, let's use the specific heat of water,
= 4186 J / kg ºC
Let's calculate
= 90.0 - 4.52 103 / (0.248 4.186 103)
= 90- 4.35
= 85.65 ° C
= 85.7 ° C
Answer:
fundamental frequency of pipe will be equal to 74 Hz
Explanation:
We have given for a particular organ pipe two adjacent frequency are 296 Hz and 370 Hz
Speed of the sound in air is 343 m/sec
We have to find the fundamental frequency for the pipe
Fundamental frequency will be equal to difference of the two adjacent frequency
So fundamental frequency = 370 - 296 = 74 Hz
So fundamental frequency of pipe will be equal to 74 Hz
Answer:
28,400 N
Explanation:
Let's start by calculating the pressure that acts on the upper surface of the hatch. It is given by the sum of the atmospheric pressure and the pressure due to the columb of water, which is given by Stevin's law:
On the lower part of the hatch, there is a pressure equal to
So, the net pressure acting on the hatch is
which acts from above.
The area of the hatch is given by:
So, the force needed to open the hatch from the inside is equal to the pressure multiplied by the area of the hatch:
Answer:
a) t=1s
y = 10.1m
v=5.2m/s
b) t=1.5s
y =11.475 m
v=0.3m/s
c) t=2s
y =10.4 m
v=-4.6m/s (The minus sign (-) indicates that the ball is already going down)
Explanation:
Conceptual analysis
We apply the free fall formula for position (y) and speed (v) at any time (t).
As gravity opposes movement the sign in the equations is negative.:
y = vi*t - ½ g*t2 Equation 1
v=vit-g*t Equation 2
y: The vertical distance the ball moves at time t
vi: Initial speed
g= acceleration due to gravity
v= Speed the ball moves at time t
Known information
We know the following data:
Vi=15 m / s
t=1s ,1.5s,2s
Development of problem
We replace t in the equations (1) and (2)
a) t=1s
=15-4.9=10.1m
v=15-9.8*1 =15-9.8 =5.2m/s
b) t=1.5s
=22.5-11.025=11.475 m
v=15-9.8*1.5 =15-14.7=0.3m/s
c) t=2s
= 30-19.6=10.4 m
v=15-9.8*2 =15-19.6=-4.6m/s (The minus sign (-) indicates that the ball is already going down)
Answer:
I just toughed it out and talked with friends
Explanation:
