Answer:
Heat of vaporization will be 22.59 j
Explanation:
We have given mass m = 10 gram
And heat of vaporization L = 2.259 J/gram
We have to find the heat required to vaporize 10 gram mass
We know that heat of vaporization is given by
, here m is mass and L is latent heat of vaporization.
So heat of vaporization Q will be = 10×2.259 = 22.59 J
Answer:
T=0.372 s, f=2.7 Hz, w=16.9 rad/s, k=179.2 N/m, v= 8.78 m/s, F= 48.4 N
Explanation:
a.)
Period: It is already given in the question "oscillator repeats its motion every 0.372 s".
So T=0.372 s
b)
frequency= f = 1/ T
f = 1/ 0.372
f=2.7 Hz
c).
Angular frequency= w= 2πf
w= 2*π*2.7
w=16.9 rad/s
d)
Spring Constant:
As w=
⇒w²= k/m
⇒k= m*w²
⇒k= 0.628 * 16.9² N/m
⇒k=179.2 N/m
e)
The mass will have maximum speed when it passes through the mean position.
At mean position
Maximum elastic potential energy = Maximum kinetic energy
1/2 k A² = 1/2 m v² ( A is amplitude of oscillation)
⇒ v=
⇒ v=
\
⇒ v= 8.78 m/s
f)
Maximum force will be exerted on the block when it is at maximum distance.
F= k* A ( A is amplitude of oscillation)
F= 179.2 * 0.27 N
F= 48.4 N
Answer:
So the mass of the second object M will be 1.951 kg
Explanation:
We have given mass of the first object
and its velocity 
Mass of the second object
it is at rest so its velocity 
From conservation of momentum we know that
Initial momentum = final momentum
So 


M = 1.951 kg
The answer is B) evaporation,condensation, precipitation, runoff/storage