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3 years ago

How long would it take a 4,560 watt motor to raise a 166 kg piano to an apartment window

Physics

1 answer:

IrinaK [193]3 years ago

8 0

Answer:

Explanation:

We need the power equation here:

P = W/t where W is work and is defined as

W = F*displacement.

Force is a measure in Newtons, which is also weight. We have the mass of the piano, but we need to find the weight:

w = mg so

w = 166(9.8) so

w = 1600N, rounded to the correct number of sig dig. We use that now in the power equation:

$4560=\frac{(1600)(15)}{t}$ and isolating the unknown:

$t=\frac{(1600)(15)}{4560}$ so

t = 5.3 seconds

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A length of copper wire carries a current of 14 A, uniformly distributed through its cross section. The wire diameter is 2.5 mm,

gavmur [86]

Answer:

a. ρ $_\beta=1.996J/m^3$

b. $U_E=9.445x10^{-15} J/m^3$

Explanation:

a. To find the density of magnetic field given use the gauss law and the equation:

$i=14A$ , $d=2.5mm$ , $R=3.3$ Ω, $l=1 km$ , $E_o=8.85x10^{-12}F/m$ , $u_o=4*x10^{-7}H/m$

ρ $_\beta=\frac{\beta^2}{2*u_o}$

ρ $_\beta=\frac{1}{2*u_o}*(\frac{u_o*i^2}{2\pi *r})^2$

ρ $_\beta=\frac{u_o*i^2}{8\pi*r}=\frac{4\pi *10^{-7}H/m*(14A)^2}{8\pi*(1.25x10^{-3}m)^2}$

ρ $_\beta=1.996J/m^3$

b. The electric field can be find using the equation:

$U_E=\frac{1}{2}*E_o*E^2$

$E=(\frac{i*R}{l})^2$

$U_E=\frac{1}{2}*8.85x10^{-12}*(\frac{14A*3.3}{1000m})^2$

$U_E=9.445x10^{-15} J/m^3$

4 0

3 years ago

A solenoid 50-cm long with a radius of 5.0 cm has 800 turns. You find that it carries a current of 10 A. The magnetic flux throu

ioda

Answer:

126 mWb

Explanation:

Given that:

length (L) = 50 cm = 0.5 m, radius (r) = 5 cm = 0.05 m, current (I) = 10 A, number of turns (N) = 800 turns.

We assume that the magnetic field in the solenoid is constant.

The magnetic flux is given as:

$\phi_m=NBAcos(\theta)\\Where\ B\ is\ the\ magnetic\ field\ density,A\ is \ the\ area.\\But\ B =\mu_onI.\ n \ is\ the\ number\ of\ turns\ per\ unit \ length=N/L\\Therefore,B=\frac{\mu_oNI}{L} \\substituting\ the\ value\ of\ B\ in\ the\ equation: \\\phi_m=\frac{NAcos(\theta)*\mu_oNI}{L} .\ But \ \theta=0,cos(\theta)=1\ and\ A=\pi r^2\\ \phi_m=\frac{N^2\pi r^2\mu_oI}{L} \\Substituting\ values:\\\phi_m=\frac{800^2*(\pi*0.05^2)*(4\pi*10^{-7})*10}{0.5}=0.126\ Wb=126\ mWb$

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3 years ago