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3 years ago

Explain the conditions through which friction can be increased.

Physics

2 answers:

Rufina [12.5K]3 years ago

8 0

Answer:

Friction can be increased the longer two objects are rubbed together, since the longer they are rubbed is the more heat produced.

Explanation:

Thepotemich [5.8K]3 years ago

6 0

Answer:

walking, car moving

Explanation:

basically anything with two surfaces pressing together ig

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3 years ago

Two circular disks spaced 0.50 mm apart form a parallel-plate capacitor. transferring 1.50 ×109 electrons from one disk to the o

blondinia [14]

The diameter of the disks is 1.32 cm.

If n electrons each of charge e are transferred from one disc to another, calculate the total charge Q transferred from one disc to the other using the expression,

$Q=ne$

Substitute 1.50×10⁹ for n and 1.6×10⁻¹⁹C for e.

$Q=ne\\ =(1.50*10^9)(1.6*10^-^1^9C)\\ =2.4*10^-^1^0C$

The potential difference V between the disks separated by a distance d is given by,

$V=Ed$

here, E is the electric field.

Substitute 2.00×10⁵N/C for E and 0.50×10⁻³m for d.

$V=Ed\\ =(2.00*10^5N/C)(0.50*10^-^3m)\\ =100V$

The capacitance C of the capacitor is given by,

$C=\frac{Q}{V} \\ =\frac{(2.4*10^-^1^0C)}{100V} \\ =2.4*10^-^1^2F$

The capacitance of a parallel plate capacitor is given by,

$C=\frac{\epsilon_0A}{d}$

Here, ε₀ is the permittivity of free space and A is the area of the disks.

Rewrite the expression for A.

$C=\frac{\epsilon_0A}{d}\\ A=\frac{Cd}{\epsilon_0} \\ =\frac{(2.4*10^-^1^2F)(0.50*10^-^3m)}{(8.85*10^-^1^2C^2/Nm^2)} \\ =1.36*10^-^4m^3$

the area A of the disks is given by,

$A=\frac{\pi D^2}{4} \\ D=\sqrt{\frac{4A}{\pi } }$

Here, D is the diameter of the disk.

$D=\sqrt{\frac{4A}{\pi } }\\ =\sqrt{\frac{4(1.36*10^-^4m^2)}{3.14} } \\ =0.01316m\\ =1.32cm$

The diameter of each disc is found to be 1.32 cm.

7 0

3 years ago

A 500-N weight sits on the small piston of a hydraulic machine. The small piston has area 2.0 cm2. If the large piston has area

weqwewe [10]

Answer:

W₂= 10000 N

Explanation:

Pascal´s Principle can be applied in the hydraulic press:

If we apply a small force (F1) on a small area piston A1, then, a pressure (P) is generated that is transmitted equally to all the particles of the liquid until it reaches a larger area piston and therefore a force (F2) can be exerted that is proportional to the area (A2) of the piston:

Pressure is defined as the force (F) applied per unit area (A)

P=F/A (N/m²)

P1=P2

$\frac{F_{1} }{A_{1} } = \frac{F_{2} }{A_{2} }$

$F_{2} = \frac{F_{1}*A_{2} }{A_{1}}$ Equation (1)

Data

W₁ = weight sits on the small piston

F₁ = W₁= 500 N

A₁ = 2.0 cm²

A₂ = 40 cm²

Calculation of the weight (W₂) can the large piston support

We replace data in the equation (1)

$F_{2} = \frac{(500)*(40) }{2}$

F₂ = 10000 N

W₂= F₂= 10000 N

6 0

3 years ago

Explain the conditions through which friction can be increased.​

Explain the conditions through which friction can be increased.