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andrezito [222]
3 years ago
8

True or false Water is a solute in moist air

Physics
2 answers:
Advocard [28]3 years ago
8 0

Answer:

True

Explanation:

Oxana [17]3 years ago
3 0

Answer:

its true water is a solute in moist air

Explanation:

i hope i helped

a crown is always appreciated

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A circuit carries a current of 0.64 A when there is a resistance of 50.0 Ω. The voltage applied to the circuit, to the nearest w
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1. Take the speed at which the body moves as 10 m/s 2. Calculate the distance travelled by the body every second 3. Take the tim
exis [7]

Answer:

a)this graph is also a line     b) in both cases we have a uniform movement

Explanation:

In this exercise we have a uniform movement

     v = d / t

     d = v t

in the table we give some values ​​to make the graph

       t (s)    d (m)

        1         10

        2        20

        3         30

In the attached we can see the graph that is a straight line

we have another vehicle at v = 50 me / S

t (s)     d (m)

1         50

2        100

3         150

this graph is also a line

b) in both cases we have a uniform movement

3 0
3 years ago
chanice drives her scooter 7 km north she stops for lunch and then drives 5 km east what distance did she cover. what was her di
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the diameter of the wheels on your car ( including the tires) is 25 inches. you are going to drive 250 miles today. each of your
timofeeve [1]
So the answer is 324.

Explanation
Because I took that test and I got right if u don’t believe me u can brainy it
3 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
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