To solve this problem we will apply the theoretical concepts and definitions given for the specific weight, density and specific gravity. Consider also that a barrel contains 159 liters or
Consider the net weight of the oil which would be
The specific weight is defined as the proportion of weight by volume thereof, therefore
Through the information given we could find mass and density through the following relations:
Mass
Density
Specific gravity,
Answer:
The initial mass of oxygen is 0.38538 Kg.
Explanation:
Given that,
Volume of tank
Molar mass = 32.0 g/mol
Gauge pressure
Initial temperature = 35.4°C
Final temperature = 20.8°C
Gauge pressure
We need to calculate initial pressure
Using formula of pressure
Put the value into the formula
We need to calculate the number of moles
Using equation of ideal gas
Put the value into the formula
We need to calculate the initial mass of oxygen
Using formula of mass
Put the value into the formula
Hence, The initial mass of oxygen is 0.38538 Kg.
Answer:
(a) 0.833 j
(b) 2.497 j
(c) 4.1625 j
(d) 4.995 watt
Explanation:
We have given force F = 5 N
Mass of the body m = 15 kg
So acceleration
As the body starts from rest so initial velocity u = 0 m/sec
(a) From second equation of motion
For t = 1 sec
We know that work done W =force × distance = 5×0.1666 =0.833 j
(b) For t = 2 sec
We know that work done W =force × distance = 5×0.666 =3.33 j
So work done in second second = 3.33-0.833 = 2.497 j
(c) For t = 3 sec
We know that work done W =force × distance = 5×1.4985 =7.4925 j
So work done in third second = 7.4925 - 2.497 -0.833 = 4.1625 j
(d) Velocity at the end of third second v = u+at
So v = 0+0.333×3 = 0.999 m /sec
We know that power P = force × velocity
So power = 5× 0.999 = 4.995 watt
Answer:
option (c)
Explanation:
When an object thrown upwards, the value of acceleration acting on the object is acceleration due to gravity which is always acting towards the earth.
As it falls downwards, the acceleration is again equal to the acceleration due to gravity.
So, the ball's acceleration is constant.
Answer:
Explanation:
Let the velocity of projectile be v and angle of throw be θ.
The projectile takes 5 s to touch the ground during which period it falls vertically by 100 m
considering its vertical displacement
h = - ut +1/2 g t²
100 = - vsinθ x 5 + .5 x 9.8 x 5²
5vsinθ = 222.5
vsinθ = 44.5
It covers 160 horizontally in 5 s
vcosθ x 5 = 160
v cosθ = 32
squaring and adding
v²sin²θ +v² cos²θ = 44.4² + 32²
v² = 1971.36 + 1024
v = 54.73 m /s