Answer:
the ratio of Hank's mass to Harry's mass is 0.7937 or [ 0.7937 : 1
Explanation:
Given the data in the question;
Hank and Harry are two ice skaters, since both are on top of ice, we assume that friction is negligible.
We know that from Newton's Second Law;
Force = mass × Acceleration
F = ma
Since they hold on to opposite ends of the same rope. They have the same magnitude of force |F|, which is the same as the tension in the rope.
Now,
Mass
× Acceleration = Mass
× Acceleration
so
Mass / Mass = Acceleration / Acceleration
given that; magnitude of Hank's acceleration is 1.26 times greater than the magnitude of Harry's acceleration,
Mass / Mass = 1 / 1.26
Mass / Mass = 0.7937 or [ 0.7937 : 1 ]
Therefore, the ratio of Hank's mass to Harry's mass is 0.7937 or [ 0.7937 : 1 ]
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.
Answer:
Explanation:
Given that
Mass , m = 25 kg
We know that when body is in rest condition then static friction force act on the body and when body is in motion the kinetic friction force act on the body .That is why these two forces are given as follows
Static friction force ,fs= 165 N
Kinetic friction force ,fk = 127 N
If the body is moving with constant velocity ,it means that acceleration of that body is zero and all the forces are balanced.
Lets take coefficient of kinetic friction = μk
The kinetic friction is given as follows
fk = μk m g
Now by putting the values
127 = μk x 25 x 9.81
Therefore the value of coefficient of kinetic friction will be 0.51
When a liquid changes to gas, this phase change is called vaporization or evaporization.
