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3 years ago

7

Im just a little confused right now

1 answer:

defon3 years ago

8 0

Answer:

C

YExplanation:

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In each of two coils the rate of change of the magnetic flux in a single loop is the same. The emf induced in coil 1, which has

Nat2105 [25]

Answer:

303

Explanation:

We are given that

Emf in coil 1, $E_1=3.13 V$

Emf induced in coil 2, $E_2=4.16 V$

Number of loops in coil 1, $N_1=228$

We have to find the number of loops in coil 2.

Rat of change of magnetic flux in a single loop is same.

Let $\phi_1$ and $\phi_2$ be the magnetic flux in coil 1 and coil 2.

$\frac{d\phi_1}{dt}=\frac{d\phi_2}{dt}$

$\frac{V_2}{V_1}=\frac{N_2}{N_1}$

Using the formula

$\frac{E_2}{E_1}=\frac{N_2}{N_1}$

$\frac{4.16}{3.13}=\frac{N_2}{228}$

$N_2=\frac{4.16}{3.13}\times 228$

$N_2=303$

Hence, the number of loops in coil 2=303

5 0

3 years ago

A runner wants to run 12.0 km . Her running pace is 8.2 mi/hr . How many minutes must she run? Express your answer using two sig

Hunter-Best [27]

Answer:

53 minutes

Explanation:

To solve this problem, we need to use common proportions to convert.

= 11.7km * 1mi/1.61km * 1hr/8.2mi * 60min/hr

≈ 53 minutes

Best of LUck!

5 0

3 years ago

What’s the difference between applied and pure research

Ket [755]

Answer:

Pure research is conducted without any specific goal in mind. Applied research is conducted in order to solve a specific and practical problem, unlike pure research. Hope this helped.

8 0

3 years ago

Two loudspeakers emit sound waves of the same frequency along the x-axis. The amplitude of each wave is a. The sound intensity i

leonid [27]

Answer:

Explanation:

To find the amplitude of the sound, we must first determine the wavelength and the phase difference between the two speakers.

For the wavelength;

Recall that, the separation between two successive max. and min. intensity points are $\dfrac{\lambda}{2}$

Thus; for both speakers; the wavelength of the sound is:

$\dfrac{\lambda}{2} = (10+30) cm$

$\dfrac{\lambda}{2} = (40) cm$

λ = 80 cm

The relation between the path difference(Δx) and the phase difference(Δ∅) is:

$\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o$

where;

Δx = 10 cm

λ = 80 cm

Δ∅ = π rad

∴

$\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o$

$\pi \ rad = \dfrac{2 \pi}{80 \ cm}(10 \ cm) + \Delta \phi_o$

$\pi \ rad = \dfrac{2 \pi}{8}+ \Delta \phi_o$

$\pi \ rad = \dfrac{ \pi}{4}+ \Delta \phi_o$

$\Delta \phi_o = \pi -\dfrac{ \pi}{4}$

$\Delta \phi_o = \dfrac{ 4\pi - \pi}{4}$

$\Delta \phi_o = \dfrac{ 3\pi}{4} \ rad$

Suppose both speakers are placed side-by-side, then the path difference between the two speakers is: Δx = 0 cm

Thus, we have:

$\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o$

$\Delta \phi = \dfrac{2 \pi}{\lambda}(0 \ cm ) + \dfrac{3 \pi}{4} \ rad$

$\Delta \phi = \dfrac{3 \pi}{4} \ rad$

∴

The amplitude of the sound wave if the two speakers are placed side-by-side is:

$A = 2a \ cos \bigg (\dfrac{\Delta \phi }{2} \bigg)$

$A = 2a \ cos \bigg (\dfrac{\dfrac{3 \pi}{4} }{2} \bigg)$

$A = 2a \ cos \bigg ({\dfrac{3 \pi}{8} } \bigg)$

A = 0.765a

7 0

3 years ago

In the picture below, label which particle is the solute and which is the solvent

klasskru [66]

The black circles are the solvent and the open circles are the solute

8 0

3 years ago