Answer:
The unknown substance is Aluminum.
Explanation:
We'll begin by calculating the change in the temperature of substance. This can be obtained as follow:
Initial temperature (T₁) = 25 ⁰C
Final temperature (T₂) = 100 ⁰C
Change in temperature (ΔT) =?
ΔT = T₂ – T₁
ΔT = 100 – 25
ΔT = 75 ⁰C
Finally, we shall determine the specific heat capacity of the substance. This can be obtained as follow:
Change in temperature (ΔT) = 75 ⁰C
Mass of the substance (M) = 135 g
Heat (Q) gained = 9133 J
Specific heat capacity (C) of substance =?
Q = MCΔT
9133 = 135 × C × 75
9133 = 10125 × C
Divide both side by 10125
C = 9133 / 10125
C = 0.902 J/gºC
Thus, the specific heat capacity of substance is 0.902 J/gºC
Comparing the specific heat capacity (i.e 0.902 J/gºC) of substance to those given in the table above, we can see clearly that the unknown substance is aluminum.
Answer: True
Explanation: Following are the properties that are used to identify minerals:
(1) Color
(2) Hardness
(3) Magnetism : It is the property of mineral to attract or repel with other magnetic materials.
(4) Luster : It is the property that shows the surface is reflecting light or not.
(5) Reactivity: Reactivity is also responsible to identify mineral. How the mineral is reacting with acids, bases etc helps to identify type of minerals.
Answer:
1.55 N
Explanation:
Density = mass / volume
0.789 g/mL = m / 200.0 mL
m = 157.8 g
Weight = mass × acceleration due to gravity
W = (0.1578 kg) (9.8 m/s²)
W = 1.55 N
a. The disk starts at rest, so its angular displacement at time
is

It rotates 44.5 rad in this time, so we have

b. Since acceleration is constant, the average angular velocity is

where
is the angular velocity achieved after 6.00 s. The velocity of the disk at time
is

so we have

making the average velocity

Another way to find the average velocity is to compute it directly via

c. We already found this using the first method in part (b),

d. We already know

so this is just a matter of plugging in
. We get

Or to make things slightly more interesting, we could have taken the end of the first 6.00 s interval to be the start of the next 6.00 s interval, so that

Then for
we would get the same
.