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2 years ago

What happens to force when pressure is doubled?

Physics

2 answers:

san4es73 [151]2 years ago

7 0

When the force on some area is doubled and the area doesn't change,
then the pressure on that area has doubled.

Gelneren [198K]2 years ago

4 0

The force also doubbles

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A book on a 2-meter high shelf has a mass of 0.4 kg. What is its potential energy?

poizon [28]

Answer:

$\boxed {\boxed {\sf 7.84 \ Joules}}$

Explanation:

The formula for potential energy is:

$PE=m*g*h$

where m is the mass, g is the gravitational acceleration, and h is the height.

The mass of the book is 0.4 kilograms. The gravitational acceleration on Earth is 9.8 m/s². The height of the book is 2 meters.

$m=0.4 \ kg \\g=9.8 \ m/s^2 \\h=2\ m$

Substitute the values into the formula.

$PE=(0.4 \ kg)(9.8 \ m/s^2)(2 \ m)$

Multiply the first two numbers.

0.4 kg*9.8 m/s²= 3.92 kg*m/s²
If we convert the units now, the problem will be much easier later on.
1 kg*m/s² is equal to 1 Newton. So, our answer of 3.92 kg*m/s² is equal to 3.92 N

$PE=(3.92 \ N )(2 \ m)$

Multiply.

3.92 N* 2 m=7.84 N*m
1 Newton meter is equal to 1 Joule (this is why we converted the units).
Our answer is equal to 7.84 Joules.

$PE=7.84 \ J$

6 0

2 years ago

Read 2 more answers

How do kinetic and potential energy transfer to one throughout a roller coaster ride?

mojhsa [17]

Answer:

As the cars ascend the next hill, some kinetic energy is transformed back into potential energy. Then, when the cars descend this hill, potential energy is again changed to kinetic energy. This conversion between potential and kinetic energy continues throughout the ride.

Explanation:

hope it helps U

6 0

1 year ago

When you enter a toll road, your ticket is stamped 1:00 p.m. When you leave, after traveling 55 miles, your ticket is stamped 2:

Viefleur [7K]

27.5 because of you divide the 55miles with the time you get your velocity which is the speed.

4 0

2 years ago

Read 2 more answers

An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a

S_A_V [24]

Answer:

Given:

Thermal Kinetic Energy of an electron, $KE_{t} = \frac{3}{2}k_{b}T$

$k_{b} = 1.38\times 10^{- 23} J/k$ = Boltzmann's constant

Temperature, T = 1800 K

Solution:

Now, to calculate the de-Broglie wavelength of the electron, $\lambda_{e}$ :

$\lambda_{e} = \frac{h}{p_{e}}$

$\lambda_{e} = \frac{h}{m_{e}{v_{e}}$ (1)

where

h = Planck's constant = $6.626\times 10^{- 34}m^{2}kg/s$

$p_{e}$ = momentum of an electron

$v_{e}$ = velocity of an electron

$m_{e} = 9.1\times 10_{- 31} kg$ = mass of electon

Now,

Kinetic energy of an electron = thermal kinetic energy

$\frac{1}{2}m_{e}v_{e}^{2} = \frac{3}{2}k_{b}T$

$}v_{e} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{e}}}$

$}v_{e} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{9.1\times 10_{- 31}}}$

$v_{e} = 2.86\times 10^{5} m/s$ (2)

Using eqn (2) in (1):

$\lambda_{e} = \frac{6.626\times 10^{- 34}}{9.1\times 10_{- 31}\times 2.86\times 10^{5}} = 2.55 nm$

Now, to calculate the de-Broglie wavelength of proton, $\lambda_{e}$ :

$\lambda_{p} = \frac{h}{p_{p}}$

$\lambda_{p} = \frac{h}{m_{p}{v_{p}}$ (3)

where

$m_{p} = 1.6726\times 10_{- 27} kg$ = mass of proton

$v_{p}$ = velocity of an proton

Now,

Kinetic energy of a proton = thermal kinetic energy

$\frac{1}{2}m_{p}v_{p}^{2} = \frac{3}{2}k_{b}T$

$}v_{p} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{p}}}$

$}v_{p} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{1.6726\times 10_{- 27}}}$

$v_{p} = 6.674\times 10^{3} m/s$ (4)

Using eqn (4) in (3):

$\lambda_{p} = \frac{6.626\times 10^{- 34}}{1.6726\times 10_{- 27}\times 6.674\times 10^{3}} = 5.94\times 10^{- 11} m = 0.0594 nm$

7 0

2 years ago

Concrete colums are constructed with reinforcing steel in them to make them stronger and more ductile. The reinforcing bars are

Sergio039 [100]

Answer:

$21678.47223\ lbf-in^2$

$383.1109\ lbf-in^2$

Explanation:

d = Diameter of column = 0.5 inch

$A_c$ = Area of concrete = $119.4\ in^2$

The strain in the system is conserved

$\dfrac{F_sL}{A_sE_s}=\dfrac{F_cL}{A_cE_c}\\\Rightarrow F_c=\dfrac{F_sA_cE_c}{A_sE_s}\\\Rightarrow F_c=\dfrac{F_s \times 119.4\times 4.1\times 10^6}{8\times \dfrac{\pi \dfrac{1}{2^2}}{4}\times 29\times 10^6}\\\Rightarrow F_c=10.74658F_s$