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yawa3891 [41]
3 years ago
11

Physics problem: piston? The piston of a hydraulic automobile lift is 0.300 m in diameter. 1. What gauge pressure, in pascals, i

s required to lift a car with a mass of 1200 kg?
Physics
2 answers:
Blababa [14]3 years ago
6 0
Given:
d = 0.3 m, piston diameter
m = 1200 kg, mass to be lifted

The piston area is
A = (n/4)*(0.3 m)² = 0.0707 m².

Let p (Pa) be the required piston pressure.
Then (p N/m²)*(0.0707 m²) = (1200 kg)*(9.8 m/s²)
p = 1.6637 x 10⁵ Pa = 166.4 kPa

Answer: 1.664 x 10⁵ Pa or 166.4 kPa
k0ka [10]3 years ago
3 0
Pressure is given by:
P=F/A
force=1200*10=12000 N
Area=pi*r^2=3.14*0.15^2
=0.07065 m^2
Thus;
P=12000/0.7065
P=169,851.38 N/m^2
but
1 N/m^2=1 Pa
thus our pressure is:
169,851.38 Pa
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"How do changes in voltage or resistance affect current in an electric circuit?"
kicyunya [14]
We can answer the question by looking at the Ohm's law, which gives us the relationship between voltage (V), current (I) and resistance (R) of a circuit:
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I= \frac{V}{R}

by looking at the equation, we can make the following observations:
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5 0
3 years ago
In the Rutherford model of hydrogen the electron (mass=9.11x10-31 kg) is in a planetary type orbit around the proton (mass=1.67x
Andrei [34K]

Answer:

(a) F_g=1.62*10^{-48}N

(b) F_e=3.68*10^{-9}N

Explanation:

(a) We use Newton's law of universal gravitation, in order to calculate the gravitational force between electron and proton:

F_g=-G\frac{m_1m_2}{r^2}

Where G is the Cavendish gravitational constant, m_1 and m_2 are the masses of the electron and the proton respectively and r is the distance between them:

F_g=-6.67*10^{-11}\frac{N\cdot m^2}{kg^2}\frac{(9.11*10^{-31}kg)(1.67*10^{-27}kg)}{(2.5*10^{-10}m)^2}\\F_g=-1.62*10^{-48}N

The minus sing indicates that the force is repulsive. Thus, its magnitude is:

F_g=1.62*10^{-48}N

(b) We use Coulomb's law, in order to calculate the electric force between electron and proton, here k is the Coulomb constant and e is the elementary charge:

F_e=-k\frac{e^2}{r^2}\\F_e=-8.99*10^{9}\frac{N\cdot m^2}{C^2}\frac{(1.6*10^{-19}C)^2}{(2.5*10^{-10}m)^2}\\F_e=-3.68*10^{-9}N

Its magnitude is:

F_e=3.68*10^{-9}N

6 0
3 years ago
A particle moving at speed 0. 32 c has momentum p0. the speed of the particle is increased to 0. 72 c. what is its momentum now?
Semenov [28]

A particle moving at speed 0.32 c has momentum P₀. the speed of the particle is increased to 0.72 c then its momentum would be 2.25P₀.

<h3>What is momentum?</h3>

It can be defined as the product of the mass and the speed of the particle, it represents the combined effect of mass and the speed of any particle, and the momentum of any particle is expressed in Kg m/s unit.

Mathematically the formula of the momentum is

P = mv

where P is the momentum of the particle

m is the mass of the particle

v is the velocity by which the particle is moving

As given in the question a particle moving at speed 0.32 c has momentum P₀. the speed of the particle is increased to 0.72 c

by using the formula of the momentum and substituting the values of the velocity

P =mv

As mentioned in the question when the particle is moving with 0.32c velocity it has a momentum of P₀

P₀ = m*(0.32c)

If the speed of the particle is increased to 0.72 c the momentum would be

P=m*(0.72c)

by dividing the second equation from the first one

P₀/P = 0.32c/0.72c

P₀/P = 0.44

P =2.25P₀

Thus, the increased momentum of the particle would be 2.25P₀

Learn more about momentum from here

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6 0
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