Answer:
c) F = 16000 N
Explanation:
For this exercise we use Newton's second law
F = ma
they tell us that adding the other wagons the acceleration of the locomotive must be maintained
F = m a
by adding the other four wagons
mass = 4 no
therefore to maintain the force you must also raise the same factor
Fe = 4Fo
Fe = 4 4000
F = 16000 N
y = 0m
y0 = 166m
v0y = 0 m/s
g = 9.8 m/s^2
t = ?
Solve for t:
y = y0 + v0y*t - (0.5)gt^2
0 = 166 - (0.5)(9.8)t^2
t = 5.82 s
Now, using time, we can solve for the range using the equation:
x = vx(t)
x = (40)(5.82)
x = 232.8 m
The impact horizontal component of velocity will be 40 m/s as velocity in terms of x is always constant. To find the impact vertical component of velocity, we use the equation:
v = v0y - gt
v = 0 - (9.8)(5.82)
v = -57.04 m/s
Given data:
- It is a graphical display where the data is grouped in to ranges
- A diagram consists rectangles, whose area is proportional to frequency of a variable and whose width is equal to the class interval.
- It is an accurate representation of the distribution of numerical data.
<em>From Figure:</em>
Each box in the graph (small rectangle box) is assumed to be one download. So, in the graph the time between 8 p.m to 9 p.m, the number of downloads are 8.75 approximately (because the last box is incomplete, therefore 8 complete boxes and 9th is more than half).
<em>So, We conclude that the total number of downloads are approximately 9 in the time span of 8 p.m. to 9 p.m.</em>
The temperature of both the halves of the body remain same and thus the body remain in thermal equilibrium.
Both the parts of the body even after being in contact with each other will not transfer heat and thus maintain equal temperature and the resultant of the two will be zero.
Answer:
6926.4J
Explanation:
Given parameters:
Mass of iron = 200g
Initial temperature = 100°C
Final temperature = 22°C
Unknown:
Amount of heat transferred to the water = ?
Solution:
The quantity of heat transferred to the water is a function of mass and temperature of the iron;
H = m c Ф
m is the mass of the iron
Ф is the change in temperature
C is the specific heat capacity of iron = 0.444 J/g°C
Now;
insert the parameters and solve;
H = 200 x 0.444 x (100-22)
H = 6926.4J