Answer: 100cm
Explanation:
The force of friction on a surface normal to gravity where µ is the coefficient of friction is
F = µmg
Where
F = the friction force
µ = coefficient of friction
m = mass of the object
g = acceleration due to gravity
Also, the Kinetic Energy of the object, E = Fs, where
E = Kinetic Energy
s = stopping distance. So that,
E = µmgs
40 J = 0.4 * 10 kg * 10 m/s² * s
40 J = 40 kgm/s² * s
s = 40 J / 40 kgm/s²
s = 1 m or 100 cm
The added weight of the sand puts more downward pressure on the wheels contacting the rails, which would cause the trains speed to decrease.
PV = 400 x 0.08 = 32 J
Hope this helps
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:
a= 4.4×10 m/s^2
Explanation:
pressure P = E/c
Where, E = 100 W/m^2 intensity of light
c= speed of light = 3×10^8 m/s
P = 1000/ 3×10^8
P = 3.33×10^(-6) Pa
Force F = P×A
- P is the pressure and c= speed of light
F = 3.33×10^{-6}×6.65×10(-29)
= 2.22×10^{-6}
acceleration a = F/m = 2.22×10^{-6}/ 5.10×10^{-27}
a= 4.4×10 m/s^2
