Which of these would have the lowest albedo?

A square loop of side 7 cm is placed with the nearest side 2 cm from a long wire carrying a current that varies with time at a c

ioda

Answer:

Explanation:

side of the square loop, a = 7 cm

distance of the nearest side from long wire, r = 2 cm = 0.02 m

di/dt = 9 A/s

Integrate on both the sides

$\int _{0}^{i}di =9\int _{0}^{t}dt$

i = 9t

(a) The magnetic field due to the current carrying wire at a distance r is given by

$B = \frac{\mu_{0}i}{2\pi r}$

$B = \frac{\mu_{0}\times 9t}{2\pi r}$

(b)

Magnetic flux,

$\phi=\int B\times a dr$

$\phi=\int \frac{\mu_{0}\times 9t}{2\pi r}\times a dr$

$\phi=\frac{\mu_{0}\times 9t\times a}{2\pi}\times ln\left ( \frac{2 + 7}{2} \right )$

$\phi=\frac{\mu_{0}\times 9t\times 0.07}{2\pi}\times ln(4.5)$

$\phi = 1.89 \times 10^{-7}t$

(c)

R = 3 ohm

$e = -\frac{d\phi}{dt}$

magnitude of voltage is

e = 1.89 x 10^-7 V

induced current, i = e / R = (1.89 x 10^-7) / 3

i = 6.3 x 10^-8 A

8 0

3 years ago

An air-filled pipe is found to have successive harmonics at 700 Hz , 900 Hz , and 1100 Hz . It is unknown whether harmonics belo

Artyom0805 [142]

Answer:

the length of the pipe is 0.85 m or 85 cm

Explanation:

Given the data in the question;

The successive harmonics are; 700 Hz , 900 Hz , and 1100 H

Now, for a closed pipe,

length of pipe (L) = λ/4

Harmonics; 1x, 3x, 5x, 7x, 9x, 11x

1100Hz - 900Hz = 200Hz

⇒ 2x = 200Hz

x = 100Hz ( fundamental frequency )

λ = V/f = 340 /100 = 3.4 m

Now

Length L = λ / 4

L = 3.4 / 4

L = 0.85 m or 85 cm

Therefore, the length of the pipe is 0.85 m or 85 cm

7 0

3 years ago

A man is dragging a trunk up the loading ramp of a mover’s truck. The ramp has a slope angle of 20.0°, and the man pulls upward

AleksandrR [38]

Answer:

(a) 104 N

(b) 52 N

Explanation:

Given Data

Angle of inclination of the ramp: 20°

F makes an angle of 30° with the ramp

The component of F parallel to the ramp is Fx = 90 N.

The component of F perpendicular to the ramp is Fy.

(a)

Let the +x-direction be up the incline and the +y-direction by the perpendicular to the surface of the incline.

Resolve F into its x-component from Pythagorean theorem:

Fx=Fcos30°

Solve for F:

F= Fx/cos30°

Substitute for Fx from given data:

Fx=90 N/cos30°

=104 N

(b) Resolve r into its y-component from Pythagorean theorem:

Fy = Fsin 30°

Substitute for F from part (a):

Fy = (104 N) (sin 30°)

= 52 N

5 0

3 years ago

Add me and ill add you back!! you have to bc i just gave you free coins :)

Rainbow [258]

Answer:

yoooooooooooooooooooo

Explanation:

6 0

3 years ago

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Determine the acceleration due to gravity for low Earth orbit (LEO) given: MEarth = 6.00 x 1024 kg, rEarth = 6.40 x 106 m, G = 6

Nana76 [90]

Answer:

The answer to the question is as follows

The acceleration due to gravity for low for orbit is 9.231 m/s²

Explanation:

The gravitational force is given as

$F_{G}= \frac{Gm_{1} m_{2}}{r^{2} }$

Where $F_{G}$ = Gravitational force

G = Gravitational constant = 6.67×10⁻¹¹ $\frac{Nm^{2} }{kg^{2} }$

m₁ = mEarth = mass of Earth = 6×10²⁴ kg

m₂ = The other mass which is acted upon by $F_{G}$ and = 1 kg

rEarth = The distance between the two masses = 6.40 x 10⁶ m

therefore at a height of 400 km above the erth we have

r = 400 + rEarth = 400 + 6.40 x 10⁶ m = 6.80 x 10⁶ m

and $F_{G}$ = $\frac{6.67*10^{-11} *6.40*10^{24} *1}{(6.8*10^{6})^{2} }$ = 9.231 N

Therefore the acceleration due to gravity = $F_{G}$ /mass

9.231/1 or 9.231 m/s²

Therefore the acceleration due to gravity at 400 kn above the Earth's surface is 9.231 m/s²

4 0

3 years ago

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