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stiks02 [169]
3 years ago
14

A larger truck takes more force to move What law of motion is it?

Physics
1 answer:
Y_Kistochka [10]3 years ago
6 0
Newton's Second Law would probably best describe this. 
F = ma 
Where F = force
m = mass
a = acceleration 

The force required is dependant on the mass, and where the mass is greater, the force required will be greater. 
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4 years ago
Determine the farthest away a car can be at night so that your eyes can resolve the two headlights. Assume that the the diameter
Zigmanuir [339]

Answer:

The farthest distance is L_{max}=11.48 km

Explanation:

The angular resolution is mathematically represented as

              \theta = \frac{Z}{L} = \frac{1.22 \lambda}{d}

Where Z is the distance between the car head lamp with a value Z = 1.4 m

L is the distance between the eye and the car

\lambda is the wavelength  with a value 500 nm = = 500 *10^{-9}m

d is the diameter of the eye with a value  d = 0.50 cm = = \frac{0.50}{100} = 0.005m

   The maximum distance L_{max} is given as

              L_{max} = \frac{(1.4) *0.005}{1.22 * (500 *10^{-9})}

                       = 11.48*10^{3}m

                        =11.48 km

             

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8 0
4 years ago
Consider an electron with charge −e and mass m orbiting in a circle around a hydrogen nucleus (a single proton) with charge +e.
alexandr1967 [171]

Answer:

v=\sqrt{k\frac{e^2}{m_e r}}, 2.18\cdot 10^6 m/s

Explanation:

The magnitude of the electromagnetic force between the electron and the proton in the nucleus is equal to the centripetal force:

k\frac{(e)(e)}{r^2}=m_e \frac{v^2}{r}

where

k is the Coulomb constant

e is the magnitude of the charge of the electron

e is the magnitude of the charge of the proton in the nucleus

r is the distance between the electron and the nucleus

v is the speed of the electron

m_e is the mass of the electron

Solving for v, we find

v=\sqrt{k\frac{e^2}{m_e r}}

Inside an atom of hydrogen, the distance between the electron and the nucleus is approximately

r=5.3\cdot 10^{-11}m

while the electron mass is

m_e = 9.11\cdot 10^{-31}kg

and the charge is

e=1.6\cdot 10^{-19} C

Substituting into the formula, we find

v=\sqrt{(9\cdot 10^9 m/s) \frac{(1.6\cdot 10^{-19} C)^2}{(9.11\cdot 10^{-31} kg)(5.3\cdot 10^{-11} m)}}=2.18\cdot 10^6 m/s

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4 years ago
Explain how you would draw magnetic field lines in the vicinity of this magnet (it is a normal north and south magnet)
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D. All of the above

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