Answer:
0.426 L
Explanation:
Boyles law is expressed as p1v1=p2v2 where
P1 is first pressure, v1 is first volume
P2 is second pressure, v2 is second volume.
Given information
P1=96 kPa, v1=0.45 l
P2=101.3 kpa
Unknown is v2
Making v2 the subject from Boyle's law

Substituting the given values then

Therefore, the volume is approximately 0.426 L
Answers:
a) -171.402 m/s
b) 17.49 s
c) 1700.99 m
Explanation:
We can solve this problem with the following equations:
(1)
(2)
(3)
Where:
is the bomb's final height
is the bomb's initial height
is the bomb's initial vertical velocity, since the airplane was moving horizontally
is the time
is the acceleration due gravity
is the bomb's range
is the bomb's initial horizontal velocity
is the bomb's final velocity
Knowing this, let's begin with the answers:
<h3>b) Time
</h3>
With the conditions given above, equation (1) is now written as:
(4)
Isolating
:
(5)
(6)
(7)
<h3>a) Final velocity
</h3>
Since
, equation (3) is written as:
(8)
(9)
(10) The negative sign only indicates the direction is downwards
<h3>c) Range
</h3>
Substituting (7) in (2):
(11)
(12)
Answer:
1020g
Explanation:
Volume of can=

Mass of can=80g=
1Kg=1000g
Density of lead=
By using 
We have to find the mass of lead which shot can it carry without sinking in water.
Before sinking the can and lead inside it they are floating in the water.
Buoyancy force =

Where
Density of water
Mass of can
Mass of lead
Volume of can
Substitute the values then we get




Hence, 1020 grams of lead shot can it carry without sinking water.
Apply Gay-Lussac's law:
P/T = const.
P = pressure, T = temperature, the quotient of P/T must stay constant.
Initial P and T values:
P = 180kPa, T = -8.0°C = 265.15K
Final P and T values:
P = 245kPa, T = ?
Set the initial and final P/T values equal to each other and solve for the final T:
180/265.15 = 245/T
T = 361K