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3 years ago

Which one of the following is a correct statement......

Physics

1 answer:

IceJOKER [234]3 years ago

5 0

Answer:

Just 3

Explanation:

I believe the other two are incorrect

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We want to construct a solenoid with a resistance of 4.30 Ω and generate a magnetic field of 3.70 × 10−2 T at its center when ap

marshall27 [118]

Answer with Explanation:

We are given that

Resistance of solenoid,R=4.3 ohm

Magnetic field,B= $3.7\times 10^{-2} T$

Current,I=4.6 A

Diameter of wire,d=0.5 mm= $0.5\times 10^{-3} m$

Radius of wire,r= $\frac{d}{2}=\frac{0.5\times 10^{-3}}{2}=0.25\times 10^{-3} m$

$1mm=10^{-3} m$

Radius of solenoid,r'=1 cm= $1\times 10^{-2} m$

$1 cm=10^{-2} m$

Resistivity of copper, $\rho=1.68\times 10^{-8}\Omega m$

We know that

$R=\frac{\rho l}{A}$

Where $A=\pi r^2$

Using the formula

$4.3=\frac{1.68\times 10^{-8}\times l}{\pi(0.25\times 10^{-3})^2}$

$l=\frac{4.3\times \pi(0.25\times 10^{-3})^2}{1.68\times 10^{-8}}=50.23 m$

Number of turns of wire= $\frac{l}{2\pi r'}$

Number of turns of wire= $\frac{50.26}{2\pi(1\times 10^{-2}}=800$

Hence, the number of turns of the solenoid,N=799

Magnetic field in solenoid,B= $\mu_0 nI$

$3.7\times 10^{-2}=4\pi\times 10^{-7} n\times 4.6$

$n=\frac{3.7\times 10^{-2}}{4\times 3.14\times 10^{-7}\times 4.6}$

$n=6404 turns/m$

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

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

$L=\frac{799}{6404}$

$L=0.125 m=0.125\times 100=12.5 cm$

Length of solenoid=12.5 cm

1m=100 cm

8 0

3 years ago

A woman can row her canoe at 2.5 m/s. If she faces an opposing current of 3.0 m/s, how fast will she go forward?

taurus [48]

Answer:

V = - 0.5 [m/s]

Explanation:

In order to solve this problem, we must use the principle of relative speeds. This is for an observer who is on the edge of the river he can see how the river moves to the left and the woman tries to move to the right but can not since:

$V_{total}=-3+2.5\\V_{total}=-0.5 [m/s]$

That is, the person sees how the woman moves to the left but with avelocity of 0.5 [m/s] to the left

8 0

2 years ago

A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is r

aivan3 [116]

Answer:

Explanation:

Given that,

Initial angular velocity is 0

ωo=0rad/s

It has angular velocity of 11rev/sec

ωi=11rev/sec

1rev=2πrad

Then, wi=11rev/sec ×2πrad

wi=22πrad/sec

And after 30 revolution

θ=30revolution

θ=30×2πrad

θ=60πrad

Final angular velocity is

ωf=18rev/sec

ωf=18×2πrad/sec

ωf=36πrad/sec

a. Angular acceleration(α)

Then, angular acceleration is given as

wf²=wi²+2αθ

(36π)²=(22π)²+2α×60π

(36π)²-(22π)²=120πα

Then, 120πα = 8014.119

α=8014.119/120π

α=21.26 rad/s²

Let. convert to revolution /sec²

α=21.26/2π

α=3.38rev/sec

b. Time Taken to complete 30revolution

θ=60πrad

∆θ= ½(wf+wi)•t

60π=½(36π+22π)t

60π×2=58πt

Then, t=120π/58π

t=2.07seconds

c. Time to reach 11rev/sec

wf=wo+αt

22π=0+21.26t

22π=21.26t

Then, t=22π/21.26

t=3.251seconds

d. Number of revolution to get to 11rev/s

∆θ= ½(wf+wo)•t

∆θ= ½(0+11)•3.251

∆θ= ½(11)•3.251

∆θ= 17.88rev.

5 0

3 years ago

Read 2 more answers

Write a simple rule that will tell a person how many water molecule will be lost while putting monosaccharides together to form

belka [17]

<span>For hydrolysis to monosaccharides, one molecule of a disaccharide needs only one molecule of water. C12H22O11 (sucrose) + H2O = C6H12O6 (glucose) + C6H12O6 (fructose) Structurally, a disaccharide molecule may be viewed as a product formed by the condensation of two molecules of monosaccharides with the elimination of a water molecule. So, only one H2O molecule is needed for the reverse process.</span>

3 0

2 years ago

A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 11.0 cm ,

viktelen [127]

Part A)
First of all, let's convert the radii of the inner and the outer sphere:
$r_A = 11.0 cm = 0.110 m$
$r_B = 16.5 cm=0.165 m$
The capacitance of a spherical capacitor which consist of two shells with radius rA and rB is
$C=4 \pi \epsilon _0 \frac{r_A r_B}{r_B- r_A}=4\pi(8.85 \cdot 10^{-12}C^2m^{-2}N^{-1}) \frac{(0.110m)(0.165m)}{0.165m-0.110m}=$
$=3.67\cdot 10^{-11}F$

Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
$Q=CV=(3.67\cdot 10^{-11}F)(100 V)=3.67\cdot 10^{-9}C$

Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
$E\cdot (4 \pi r^2) = \frac{Q}{\epsilon _0}$
from which
$E(r) = \frac{Q}{4 \pi \epsilon_0 r^2}$
In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get
$E(0.111m)=2680 N/C$

And so, the energy density at r=0.111 m is
$U= \frac{1}{2} \epsilon _0 E^2 = \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(2680 N/C)^2=3.17 \cdot 10^{-5}J/m^3$

Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor: $Q=3.67 \cdot 10^{-9}C$ . We just need to calculate the electric field E at a different value of r: r=16.4 cm=0.164 m, so
$E(0.164 m)= \frac{Q}{4 \pi \epsilon_0 r^2}=1228 N/C$

And therefore, the energy density at this distance from the center is
$U= \frac{1}{2}\epsilon_0 E^2 = \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(1228 N/C)^2=6.68 \cdot 10^{-6}J/m^3$

8 0

3 years ago