(amount of heat)Q = ? , (Mass) m= 4 g , ΔT = T f - T i = 180 c° - 20 °c = 160 °c ,
Ce = 0.093 cal/g. °c
Q = m C ΔT
Q = 4 g × 0.093 cal/g.c° × ( 180 °c- 20 °c )
Q= 4×0.093 × 160
Q = 59.52 cal
I hope I helped you^_^
Answer:
Laws are statements about something that's been observed and stated while a theory is an explanation of what's been observed. This connection between them forms a main idea that many people regulate as "what's normal."
Explanation:
The initial velocity of the ball is 1.01 m/s
Explanation:
The motion of the ball rolling off the desk is a projectile motion, which consists of two independent motions:
- A uniform horizontal motion with constant horizontal velocity
- A vertical accelerated motion with constant acceleration (
, acceleration due to gravity)
We start by analyzing the vertical motion: we can find the time of flight of the ball by using the following suvat equation
where
s = 1.20 m is the vertical displacement (the height of the desk)
u = 0 is the initial vertical velocity
t is the time of flight
Solving for t,
Now we analyze the horizontal motion. We know that the ball covers a horizontal distance of
d = 0.50 m
in a time
t = 0.495 s
Therefore, since the horizontal velocity is constant, we can calculate it as
So, the ball rolls off the table at 1.01 m/s.
Learn more about projectile motion:
brainly.com/question/8751410
#LearnwithBrainly
Answer:
m = 35.98 Kg ≈ 36 Kg
Explanation:
I₀ = 125 kg·m²
R₁ = 1.50 m
ωi = 0.600 rad/s
R₂ = 0.905 m
ωf = 0.800 rad/s
m = ?
We can apply The law of conservation of angular momentum as follows:
Linitial = Lfinal
⇒ Ii*ωi = If*ωf <em>(I)</em>
where
Ii = I₀ + m*R₁² = 125 + m*(1.50)² = 125 + 2.25*m
If = I₀ + m*R₂² = 125 + m*(0.905)² = 125 + 0.819025*m
Now, we using the equation <em>(I) </em>we have
(125 + 2.25*m)*0.600 = (125 + 0.819025*m)*0.800
⇒ m = 35.98 Kg ≈ 36 Kg
Answer:
D
Explanation:
The greater the mass, the greater the inertia, and vice versa.
Remark: This means that a more massive object has a greater tendency to resist a change in its state of rest or motion.
