Answer:
a. mass density
Explanation:
<em>Land and sea breeze that occur near the shore are due to the variation of mass density of air with change in temperature.</em>
- When the air gets heated it becomes rarer in density and thus rises up in the atmosphere and its space is occupied by a cooler and denser air that flows to the place.
<em>During the day the land is warmer than the sea so the sea breeze blows and during the night the water bodies are warmer than the land so the land breeze blows.</em>
Answer:
|v| = 8.7 cm/s
Explanation:
given:
mass m = 4 kg
spring constant k = 1 N/cm = 100 N/m
at time t = 0:
amplitude A = 0.02m
unknown: velocity v at position y = 0.01 m

1. Finding Ф from the initial conditions:

2. Finding time t at position y = 1 cm:

3. Find velocity v at time t from equation 2:

Answer:
Explanation:
total weight acting downwards
= 3g + 10g
13 g
volume of lead = 10 / 11.3 = .885 cm³
Let the volume of bobber submerged in water be v in floating position . buoyant force on bobber = v x 1 x g
Buoyant force on lead = .885 x 1 x g
total buoyant force = vg + .885 g
For floating
vg + .885 g = 13 g
v = 12.115 cm³
total volume of bobber
= 4/3 x 3.14 x 2³
= 33.5 cm³
fraction of volume submerged
= 12.115 / 33.5
= .36
= 36 %
Answer:
a)693.821N/m
b)17.5g
Explanation:
We the Period T we can find the constant k,
That is

squaring on both sides,

where,
M=hanging mass, m = spring mass,
k =spring constant
T =time period
a) So for the equation we can compare, that is,

the hanging mass M is x here, so comparing the equation we know that

b) In order to find the mass of the spring we make similar process, so comparing,

This question is incomplete, the complete question is;
The Figure shows a container that is sealed at the top by a moveable piston, Inside the container is an ideal gas at 1.00 atm. 20.0°C and 1.00 L.
"What will the pressure inside the container become if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant?"
Answer:
the pressure inside the container become 0.625 atm if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant
Explanation:
Given that;
P₁ = 1.00 atm
P₂ = ?
V₁ = 1 L
V₂ = 1.60 L
the temperature of the gas is kept constant
we know that;
P₁V₁ = P₂V₂
so we substitute
1 × 1 = P₂ × 1.60
P₂ = 1 / 1.60
P₂ = 0.625 atm
Therefore the pressure inside the container become 0.625 atm if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant