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3 years ago

14

when an object slides over a smooth horizontal surface, how does the force of friction depend on the surface area of blocks that

's are in contact with the table?

1 answer:

Marina CMI [18]3 years ago

6 0

Answer: with a greater surface area, there will be a greater force of friction

Explanation:

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1. A SUV along with 5 passengers has a mass of 3500 kg. It has a driving force of 2500 N directed along west on a perfectly hori

Dovator [93]

Answer:

The net acceleration of the SUV is 0.429 meters per square second due west.

Explanation:

Statement is incomplete. Description is presented below:

<em>A SUV along with 5 passengers has a mass of 3500 kg. It has a driving force of 2500 N directed along west on a perfectly horizontal road. The surface of the road exerts a resistance force of 500 N due east. At the same time a high wind is blowing a force of 500 N due east in the opposite direction of the car's drive force. Does the car has any acceleration? If yes, then what are the magnitude and direction of the car's acceleration?</em>

According to Newton's Laws of Motion, the SUV will accelerate if and only if net acceleration is different of zero. Let suppose as positive the direction of driving force ( $F$ ), measured in newtons:

$\Sigma F = F - R -f = F_{net}$ (1)

Where:

$R$ - Resistance force, measured in newtons.

$f$ - Wind force, measured in newtons.

$F_{net}$ - SUV net force, measured in newtons.

If we know that $F = 2500\,N$ , $R = 500\,N$ and $f = 500\,N$ , then net force experimented by the SUV is:

$F_{net} = 2500\,N-500\,N-500\,N$

$F_{net} = 1500\,N$

The car has acceleration.

By definition of force for systems with constant mass, we calculate the acceleration of the vehicle below:

$a_{net} = \frac{F_{net}}{m}$ (2)

Where $m$ is the mass of the SUV, measured in kilograms.

If we know that $F_{net} = 1500\,N$ and $m = 3500\,kg$ , then the net acceleration of the car is:

$a_{net} = \frac{1500\,N}{3500\,kg}$

$a_{net} = 0.429\,\frac{m}{s^{2}}$

The net acceleration of the SUV is 0.429 meters per square second due west.

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Read 2 more answers

Rank the wavelengths of the following quantum particles from the largest to the smallest. If any have equal wavelengths, display

goldfiish [28.3K]

The wavelengths of the following quantum particles from the largest to the smallest is (d) > (a) = (e) > (b) > (c)

De Broglie proposed that because light has both wave and particle properties, matter exhibits both wave and particle properties. This property has been explained as the dual behavior of matter.
From his observations, de Broglie derived the relationship between the wavelength and momentum of matter. This relationship is known as de Broglie's relationship.

De Broglie's relationship is given by $\lambda=\frac{h}{mv}$ .

This can be written as $\lambda=\frac{h}{p}$ .....(1) where λ is known as de Broglie wavelength and p is momentum , h = Plank’s constant .

As we know that mass of proton is greater than electron and photon .

(a)For photon , the momentum is given by $p=\frac{E}{c}$ ...(2) where c is the speed is the speed of light .

Putting E = 3eV in equation (2) , we get

$p=\frac{3\times 1.6\times10^-^1^9}{3\times 10^8} \\p=1.6\times10^-^2^7Js/m$

Putting this value of p in equation (1) , we get

$\lambda=\frac{6.62\times10^{-34}}{1.6\times10^{-27}}\\\lambda=4.13\times10^{-7}$

(b) As we know that kinetic energy is given by

$K.E=\frac{1}{2} mv^2\\\\2K.E = mv^2\\2K.E\times m = m^2v^2\\2K.E\times m = (mv)^2\\2K.E\times m = p^2\\\\\sqrt{2K.E\times m }=p$ ...(3)

Where mass of electron is $9.1\times10^-^3^1 kg$ .

Putting K.E = 3eV in equation (3) , we get

$\sqrt{2K.E\times m }=p\\p=\sqrt{2\times3\times1.6\times10^{-19} \times 9.31\times10^{-31} }\\p=\sqrt{89.376\times10^{-50}} \\p=9.45\times10^{-25}Js/m$

Putting this value of p in equation (1) , we get

$\lambda=\frac{6.62\times10^{-34}}{9.45\times10^{-25}}\\\lambda=0.7005\times10^{-9}$

(c) Putting $m=1.67\times10^{-27}kg$ and K.E = 3eV in equation (3) , we get

$\sqrt{2K.E\times m }=p\\p=\sqrt{2\times3\times1.6\times10^{-19} \times 1.67\times10^{-27} }\\p=\sqrt{16.032\times10^{-46}} \\p=4.003\times10^{-23}Js/m$

Putting this value of p in equation (1) , we get

$\lambda=\frac{6.62\times10^{-34}}{4.003\times10^{-23}}\\\lambda=1.65\times10^{-11}$

(d) Putting E = 0.3eV in equation (2) , we get

$p=\frac{0.3\times 1.6\times10^-^1^9}{3\times 10^8} \\p=1.6\times10^-^2^8Js/m$

Putting this value of p in equation (1) , we get

$\lambda=\frac{6.62\times10^{-34}}{1.6\times10^{-28}}\\\lambda=4.13\times10^{-6}$

(e) Putting $p=3eV/c$ in equation (1) , we get

$\lambda=\frac{6.62\times10^{-34}\\\times3\times10^8}{3\times1.6\times10^{-19}}\\\lambda=4.13\times10^{-7}$

On comparing the wavelength order should be (d) > (a) = (e) > (b) > (c) .

Learn more about de brogile here :

brainly.com/question/28165547

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6 0

2 years ago