We know V=IR (Ohm's law).
We are given R=180Ω and I=0.1A, then V=(0.1AΩ)(180Ω). Therefore
V=18V
There are none on the list you included with your question.
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.
Based on the equation KE = 1/2(m)(v^2), Kinetic Energy can be measured based on velocity. If an object has a large velocity, it have a larger kinetic energy than if the velocity is small.
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D is the point where the planet moves the fastest. This is because it is in the perihelion, where the planet is moving at it’s fastest pace
