How many times is larger than a centigram is a dekagram

To the speaker makes the sound louder.

Damm [24]

Answer:

the right control panel.

Place the listener at a certain distance away from the speakers. Slowly move the listener up and down. What do y

Explanation:

4 0

2 years ago

A horizontal circular wire loop of radius 0.7 m lies in a plane perpendicular to a uniform magnetic field that is pointing down

Maru [420]

Answer:

$\epsilon=1.10\ V$

Explanation:

It is given that,

Radius of the circular loop, r = 0.7 m

Magnetic field, B = 0.44 T

In 0.14 s the wire is reshaped from a circle into a square, but remains in the same plane.

Area of the circular wire,

$A_1=\pi r^2$

$A_1=\pi (0.7)^2=1.539\ m^2$

For the area of square,

The circumference of wire, $C=2\pi r=2\pi \times 0.7=4.39\ m$

Side of square, $l=\dfrac{4.39}{4}=1.09\ m$

Area of square, $A_2=1.09^2=1.188\ m^2$

An emf is induced in the loop due to change in its area. The induced emf is given by :

$\epsilon=-B\dfrac{dA}{dt}$

$\epsilon=-B\dfrac{A_2-A_1}{t}$

$\epsilon=-0.44\times \dfrac{1.188-1.539}{0.14}$

$\epsilon=1.10\ V$

So, the magnitude of the average induced emf is 1.10 volts. Hence, this is the required solution.

3 0

2 years ago

Sand falls from an overhead bin and accumulates in a conical pile with a radius that is always threethree times its height. Supp

hammer [34]

Answer:

159241.048 cm³/s

Explanation:

r = Radius = 3×height = 3h

h = height = 16 cm

Height of the pile increases at a rate = $\frac{dh}{dt}=22\ cm/s$

$\text{Volume of cone}=\frac{1}{3}\pi r^2h\\\Rightarrow V=\frac{1}{3}\pi (3h)^2h\\\Rightarrow V=3\pi h^3$

Differentiating with respect to time

$\frac{dv}{dt}=9\pi h^2\frac{dh}{dt}\\\Rightarrow \frac{dv}{dt}=9\pi 16^2\times 22\\\Rightarrow \frac{dv}{dt}=159241.048\ cm^3/s$

∴ Rate is the sand leaving the bin at that instant is 159241.048 cm³/s

5 0

2 years ago

What is the definition of displacement

Savatey [412]

<span>the action of moving something from its place or position.</span>

6 0

2 years ago

Read 2 more answers

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

2 years ago