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4 years ago

What are three inventions used to increase friction?

Physics

1 answer:

Sophie [7]4 years ago

4 0

The answer I believe is shoes, tires and gloves.

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Air at 3 104 kg/s and 27 C enters a rectangular duct that is 1m long and 4mm 16 mm on a side. A uniform heat flux of 600 W/m2 is

ad-work [718]

Answer:

$T_{out}=27.0000077$ ºC

Explanation:

First, let's write the energy balance over the duct:

$H_{out}=H_{in}+Q$

It says that the energy that goes out from the duct (which is in enthalpy of the mass flow) must be equals to the energy that enters in the same way plus the heat that is added to the air. Decompose the enthalpies to the mass flow and specific enthalpies:

$m*h_{out}=m*h_{in}+Q\\m*(h_{out}-h_{in})=Q$

The enthalpy change can be calculated as Cp multiplied by the difference of temperature because it is supposed that the pressure drop is not significant.

$m*Cp(T_{out}-T_{in})=Q$

So, let's isolate $T_{out}$ :

$T_{out}-T_{in}=\frac{Q}{m*Cp}\\T_{out}=T_{in}+\frac{Q}{m*Cp}$

The Cp of the air at 27ºC is 1007 $\frac{J}{kgK}$ (Taken from Keenan, Chao, Keyes, “Gas Tables”, Wiley, 1985.); and the only two unknown are $T_{out}$ and Q.

Q can be found knowing that the heat flux is 600W/m2, which is a rate of heat to transfer area; so if we know the transfer area, we could know the heat added.

The heat transfer area is the inner surface area of the duct, which can be found as the perimeter of the cross section multiplied by the length of the duct:

Perimeter:

$P=2*H+2*A=2*0.004m+2*0.016m=0.04m$

Surface area:

$A=P*L=0.04m*1m=0.04m^2$

Then, the heat Q is:

$600\frac{W}{m^2} *0.04m^2=24W$

Finally, find the exit temperature:

$T_{out}=T_{in}+\frac{Q}{m*Cp}\\T_{out}=27+\frac{24W}{3104\frac{kg}{s} *1007\frac{J}{kgK} }\\T_{out}=27.0000077$

$T_{out}$ =27.0000077 ºC

The temperature change so little because:

The mass flow is so big compared to the heat flux.
The transfer area is so little, a bigger length would be required.

3 0

3 years ago

A student is experimenting with some insulated copper wire and a power supply. She winds a single layer of the wire on a tube wi

OverLord2011 [107]

Answer:

$P=214.7187\,W$

Explanation:

Given that:

Diameter of the solenoid, $D=10\,cm=0.1\,m$

length of the solenoid, $L=90\,cm=0.9\,m$

diameter of the wire, $d=0.1\,cm=10^{-3}\,m$

magnetic field at the center of the solenoid, $B=7.4\times 10^{-3}\,T$

Now we need the no. of turns incorporated in the length of 90 cm:

$N=\frac{Length\,\,of\,\,solenoid}{diameter\,\,of\,\, wire}$

$N=\frac{L}{d}$

$N=\frac{0.9}{10^{-3}}$

$N=900\,\,turns$

For solenoids we have:

$B=\mu.n.I$ ...............................(1)

where:

$\mu=$ permeability of the medium

n = no. of turns per unit length

I = current in the coil

So,

$n=\frac{900}{0.9}$

$n=1000\,turns\,.\,m^{-1}$

Now putting the respective values in the eq. (1)

$7.4\times 10^{-3}=4\pi\times10^{-7}\times 1000\times I$

$I=5.8887\,A$

For copper we have resistivity:

$\rho=1.72\times 10^{-8}\, \Omega.m$

We know that resistance is given by:

$R=\rho.\frac{l}{a}$ .....................................(2)

where:

l = length of the conducting wire

a = cross sectional area of the conducting wire

Now we need the length (l) of the wire:

Circumference of the solenoid,

$C=\pi.D$

$C=0.1\pi\,m$

$\therefore l=C\times N$

$l=90\pi\,m$

Cross-sectional area of wire:

$a=\pi.\frac{d^2}{4}$

$a=\pi. \frac{(10^{-3})^2}{4}\,m^2$

Resistance from eq. (2):

$R=1.72\times 10^{-8}\times \frac{90\pi}{\pi. \frac{(10^{-3})^2}{4}}$

$R=6.192 \,\Omega$

For power we have:

$P=I^2.R$

$P=5.8887^2 \times 6.192$

$P=214.7187\,W$

6 0

3 years ago

Increasing the mass attached to a spring will decrease the angular frequency of its vibrations. True or False

Scrat [10]

Answer:

decreases

Explanation:

The angular frequency of the mass is given by

$\omega =\sqrt{\frac{k}{m}}$

where, k is the spring constant and m be the mass of the body which is doing oscillations.

if the mass of the body increases, so the value of angular frequency decreases, as the angular frequency is inversely proportional to the square root of the mass of the body.

4 0

3 years ago

Help please, I don't get it

Sergio [31]

Answers:

a) $\hat F=(0.83,-0.55) N$

b) $\hat D=(-0.44,-0.89) m$

c) $\hat V=(-0.47,0.88) m/s$

Explanation:

A unit vector is a vector whose magnitude (length) is equal to 1. This kind of vector is identified as $\hat v$ and the way to calculate is as follows: