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2 answers:
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Answer:
Final temperature, T2 = 314.9 Kelvin
Explanation:
Given the following data:
Mass = 0.9kg
Initial temperature, T1 = 10°C to Kelvin = 10 + 273 = 283K
Quantity of heat = 120,000 J
Specific heat capacity = 4182 j/kgK
To find the final temperature;
Heat capacity is given by the formula;
Where;
Q represents the heat capacity or quantity of heat.
m represents the mass of an object.
c represents the specific heat capacity of water.
dt represents the change in temperature.
Making dt the subject of formula, we have;
Substituting into the equation, we have;
dt = 31.9K
Now, the final temperature T2 is;
But, dt = T2 - T1
T2 = dt + T1
T2 = 31.9 + 283
T2 = 314.9 Kelvin
Answer: Got It!
<em>Explanation: </em>let s = speed at launch
v = 0 at top = s sin 63 - g t
so at top
t = s sin 63/g = .0909 s
h = 13.6 = s sin 63 t - 4.9 t^2
13.6 = .081s^2 - .0405 s^2
s^2 = 336
s = 18.3 m/s
0 0
Answer:
Explanation:
Given that, the range covered by the sphere, , when released by the robot from the height,
, with the horizontal speed
is
as shown in the figure.
The initial velocity in the vertical direction is .
Let be the acceleration due to gravity, which always acts vertically downwards, so, it will not change the horizontal direction of the speed, i.e.
will remain constant throughout the projectile motion.
So, if the time of flight is , then
Now, from the equation of motion
Where is the displacement is the direction of force,
is the initial velocity,
is the constant acceleration and
is time.
Here, and
(negative sign is for taking the sigh convention positive in
direction as shown in the figure.)
So, from equation (ii),
Similarly, for the launched height , the new time of flight,
, is
From equation (iii), we have
Now, the spheres may be launched at speed or
.
Let, the distance covered in the direction be
for
and
for
, we have
[from equation (iv)]
[from equation (i)]
(approximately)
This is in the points range as given in the figure.
Similarly,
[from equation (iv)]
[from equation (i)]
(approximately)
This is out of range, so there is no point for .
Hence, students must choose the speed to launch the sphere to get the maximum number of points.
