Answer:
Da=(1/4)Db
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s²
When s = Da, t = t

When s = Db, t = 2t

Dividing the two equations

Hence, Da=(1/4)Db
Answer:
Explanation:
The mass of the block is 0.5kg
m = 0.5kg.
The spring constant is 50N/m
k =50N/m.
When the spring is stretch to 0.3m
e=0.3m
The spring oscillates from -0.3 to 0.3m
Therefore, amplitude is A=0.3m
Magnitude of acceleration and the direction of the force
The angular frequency (ω) is given as
ω = √(k/m)
ω = √(50/0.5)
ω = √100
ω = 10rad/s
The acceleration of a SHM is given as
a = -ω²A
a = -10²×0.3
a = -30m/s²
Since we need the magnitude of the acceleration,
Then, a = 30m/s²
To know the direction of net force let apply newtons second law
ΣFnet = ma
Fnet = 0.5 × -30
Fnet = -15N
Fnet = -15•i N
The net force is directed to the negative direction of the x -axis
When wool is rubbed with a balloon, the wool is left with a positive charge as electrons have travelled from the wool to the balloon which means the balloon now has a negative charge.
Now that the balloon has a negative charge, you need to know:
The tissue paper originally contains electrons and protons
The fact that the balloon has a negative charge, it will ATTRACT protons because protons are POSITIVE and electrons are NEGATIVE.
So once they are attracted, they will move closer to one another.
Answer:

Explanation:
first write the newtons second law:
F
=δma
Applying bernoulli,s equation as follows:
∑
Where,
is the pressure change across the streamline and
is the fluid particle velocity
substitute
for {tex]γ[/tex] and
for 

integrating the above equation using limits 1 and 2.

there the bernoulli equation for this flow is 
note:
=density(ρ) in some parts and change(δ) in other parts of this equation. it just doesn't show up as that in formular
Answer:
assume nitrogen is an ideal gas with cv=5R/2
assume argon is an ideal gas with cv=3R/2
n1=4moles
n2=2.5 moles
t1=75°C <em>in kelvin</em> t1=75+273
t1=348K
T2=130°C <em>in kelvin</em> t2=130+273
t2=403K
u=пCVΔT
U(N₂)+U(Argon)=0
<em>putting values:</em>
=>4x(5R/2)x(Tfinal-348)=2.5x(3R/2)x(T final-403)
<em>by simplifying:</em>
Tfinal=363K