Given:
Gasoline pumping rate, R = 5.64 x 10⁻² kg/s
Density of gasoline, D = 735 kg/m³
Radius of fuel line, r = 3.43 x 10⁻³ m
Calculate the cross sectional area of the fuel line.
A = πr² = π(3.43 x 10⁻³ m)² = 3.6961 x 10⁻⁵ m²
Let v = speed of pumping the gasoline, m/s
Then the mass flow rate is
M = AvD = (3.6961 x 10⁻⁵ m²)*(v m/s)*(735 kg/m³) = 0.027166v kg/s
The gasoline pumping rate is given as 5.64 x 10⁻² kg/s, therefore
0.027166v = 0.0564
v = 2.076 m/s
Answer: 2.076 m/s
The gasoline moves through the fuel line at 2.076 m/s.
It takes the shape of the cup and it can be sucked through a straw
Magnitude of acceleration = (change in speed) / (time for the change).
Change in speed = (27 - 0) = 27 m/s
Time for the change = 10 s
Magnitude of acceleration = (27 m/s) / (10 s) = 2.7 m/s² .
Answer:
Explanation:
Given data
Electric potential at point a is Ua=5.4×10⁻⁸J
q₂ moves to point b where a negative work done on it 
Required
Electric potential energy Ub
Solution
When a particle moves from a point where the potential is Ua to a point where it is Ub the change in potential energy is equal to work done where the force exerted on the charge is conservative and work done is given by:
Now substitute the given values
So
Answer:
Explanation:
a )
The stored elastic energy of compressed spring
= 1 / 2 k X²
= .5 x 19.6 x (.20)²
= .392 J
b ) The stored potential energy will be converted into gravitational potential energy of the block earth system when the block will ascend along the incline . So change in the gravitational potential energy will be same as stored elastic potential energy of the spring that is .392 J .
c ) Let h be the distance along the incline which the block ascends.
vertical height attained ( H ) =h sin30
= .5 h
elastic potential energy = gravitational energy
.392 = mg H
.392 = 2 x 9.8 x .5 h
h = .04 m
4 cm .
=
