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mart [117]
3 years ago
10

The net force acting on an object equals the applied force plus the force of friction.

Physics
1 answer:
Georgia [21]3 years ago
3 0

Answer:

False

Explanation:

The net force is equal to the applied force minus the force of friction. It is possible for friction to act in the same direction as an applied force, but that would mean there would have to be more than two forces acting on the object.

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Alfredo makes a diagram to organize his notes about charging objects.
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The region marked X in the diagram shows that the objects have the same charge.

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Read 2 more answers
A copper wire and a tungsten wire of the same length have the same resistance. What is the ratio of the diameter of the copper w
spayn [35]

Answer:

Therefore the ratio of diameter of the copper to that of the tungsten is

\sqrt{3} :\sqrt{10}

Explanation:

Resistance: Resistance is defined to the ratio of voltage to the electricity.

The resistance of a wire is

  1. directly proportional to its length i.eR\propto l
  2. inversely proportional to its cross section area i.eR\propto \frac{1}{A}

Therefore

R=\rho\frac{l}{A}

ρ is the resistivity.

The unit of resistance is ohm (Ω).

The resistivity of copper(ρ₁) is 1.68×10⁻⁸ ohm-m

The resistivity of tungsten(ρ₂) is 5.6×10⁻⁸ ohm-m

For copper:

A=\pi r_1^2 =\pi (\frac{d_1}{2} )^2

R_1=\rho_1\frac{l_1}{\pi(\frac{d_1}{2})^2 }

\Rightarrow (\frac{d_1}{2})^2=\rho_1\frac{l_1}{\pi R_1 }......(1)

Again for tungsten:

R_2=\rho_2\frac{l_2}{\pi(\frac{d_2}{2})^2 }

\Rightarrow (\frac{d_2}{2})^2=\rho_2\frac{l_2}{\pi R_2 }........(2)

Given that R_1=R_2   and    l_1=l_2

Dividing the equation (1) and (2)

\Rightarrow\frac{ (\frac{d_1}{2})^2}{ (\frac{d_2}{2})^2}=\frac{\rho_1\frac{l_1}{\pi R_1 }}{\rho_2\frac{l_2}{\pi R_2 }}

\Rightarrow( \frac{d_1}{d_2} )^2=\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}   [since R_1=R_2   and    l_1=l_2]

\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}}

\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{3}{10}}

\Rightarrow d_1:d_2=\sqrt{3} :\sqrt{10}

Therefore the ratio of diameter of the copper to that of the tungsten is

\sqrt{3} :\sqrt{10}

8 0
3 years ago
What are the 4 energy transformations of a dash toy
Hunter-Best [27]

Answer:

Thermal/Heat energy, kinetic energy, light energy, & Electromagnetic  energy

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