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Pepsi [2]
3 years ago
5

The brake in most cars makes use of a hydraulic system. This system consists of a fluid filled tube connected at each end to a p

iston. Assume that the piston attached to the brake pedal has a cross sectional area of 3 cm2 and the piston attached to the brake pad has a 2 cross section area of 15 cm . When you apply a force of 50 Newton to the piston attached to the brake pedal, how much will be the force at the brake pad
Physics
1 answer:
Maksim231197 [3]3 years ago
7 0

Answer:

The force at the brake pad = 250 N

Explanation:

The hydraulic brake system works on the Pascal's Principle for pressure transmission in fluids; the pressure applied to a fluid is transmitted undiminished in all directions.

For hydraulic systems, the pressure applied to the brake pedal is transmitted undiminished through the fluid filled tube, connected at each end to a piston, to the brake pad.

Hence, mathematically,

P(brake pedal) = P(break pad)

Pressure is given as the force applied divided by the cross sectional Area perpendicular to the direction of applied force.

P(brake pedal) = (Force applied on the brake pedal) ÷ (Cross Sectional Area of the brake pedal)

Force applied on the brake pedal = 50 N

Cross Sectional Area of the brake pedal = 3 cm²

P(brake pedal) = (50/3) = 16.67 N/cm²

P(brake pad) = P(brake pedal) = 16.67 N/cm²

P(brake pad) = (Force applied on the brake pad) ÷ (Cross Sectional Area of the brake pad)

Force applied on the brake pad = F = ?

Cross Sectional Area of the brake pad = 15 cm²

16.67 = (F/15)

F = 16.67 × 15 = 250 N

Hence, the force at the brake pad = 250 N

Hope this Helps!!!

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PSYCHO15rus [73]

Hi there!

Recall the equation for electric potential of a point charge:

V = \frac{kQ}{r}

V = Electric potential (V)
k = Coulomb's Constant(Nm²/C²)

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r = distance (m)

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Upper right charge's potential:

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Lower left charge's potential:

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Add the two, and subtract from the total EP at the point:

3160 + 1123.75 - 2247.5 = 2036.25


The remaining charge must have a potential of 2036.25 V, so:

2036.25 = \frac{(8.99*10^9)(Q)}{\sqrt{0.02^2 + 0.04^2}}\\\\2036.25 = \frac{(8.99*10^9)Q}{0.0447} \\\\Q = 0.000000010127 = \boxed{10.13nC}


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2 years ago
A 1-kg iron frying pan is placed on a stove. The pan increases from 20°C to 250°C. If the same amount of heat is added to a pan
sergejj [24]

Here mass of the iron pan is given as 1 kg

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also its temperature rise is given from 20 degree C to 250 degree C

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<u>A) The second pan would reach a lower temperature.</u>

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