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kotykmax [81]
3 years ago
14

Which of the following decreases the dissolving rate of a solid in water?

Physics
1 answer:
xenn [34]3 years ago
6 0

Answer:

Using larger pieces of solids

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As the velocity of the sound source approaches the speed of sound, the wave fronts ahead of the object begin to appear as:
sineoko [7]

Answer:

The correct answer is D

shock waves

Explanation:

When a body approaches the speed of sound in the media, the waves it emits are getting closer and closer to each other, the envelope (sum) results in a kind of cylindrical cone that is similar to that formed in a supersonic process

The correct answer is D

When the speed of sound is exceeded, the body goes faster than the wave, so the so-called shock waves are formed.

  They carry a lot of energy

4 0
3 years ago
If you adjusted the frequency to be lower than the resonant frequency, would the voltage have been leading ahead of or lagging b
tatyana61 [14]

Answer:

current will lead the voltage

Explanation:

From the bellow figure we can see the plot between the frequency and impedance

From the figure it is clear that when the frequency is less than the resonance frequency then the impedance decreases as the frequency increases it is possible only when impedance is capacitive in nature.

As the impedance is capacitive in nature so current will lead than the voltage

4 0
3 years ago
A proton with an initial speed of 8.20×105 m/s is brought to rest by an electric field. Part APart complete Did the proton move
Anastaziya [24]

Answer:

(a) Region of higher potential

(b) ΔV=1.911V

Explanation:

For Part (a)

Proton moves into a region of higher potential because it must goes in opposite direction of force to get rest

For Part (b)

Potential difference we can calculate bu definition of kinetic energy

ΔK=mv²/2= -qΔV

express in ΔV

ΔV= -mv²/2q

Substitute the given values

So

ΔV= -mv²/2q

=-\frac{mv^2}{2e}\\ =\frac{9.11*10^{-31}*(8.20*10^{5}m/s)^2}{2*1.602*10^{-19}}\\ =1.911V

ΔV=1.911V

5 0
3 years ago
B. The coefficient of friction between the tires and the road is 0.850 and the mass of the car is
Mrac [35]

Answer:

156.26N

Explanation:

The data needed are incomplete. Let the acceleration of the body be 3.5m/s²

Other given parameters

Mass = 1.35×10^1 = 13.5kg

coefficient of friction between the tires and the road = 0.850

Acceleration due to gravity = 9.8m/s²

According to Newton's second law:

Fnet = ma

Fnet = Fapp - Ff

Fapp is the applied force

Ff is the frictional force = umg

The equation becomes:

Fapp - Ff = ma

Fapp-umg = ma

Fapp - 0.85(13.5)(9.8) = 13.5(3.5)

Fapp - 109.0125 = 47.25

Fapp = 47.25+109.0125

Fapp = 156.2625N

Hence the applied force that caused the acceleration is 156.26N

Note that the acceleration of the car was assumed. Any value of acceleration can be used for the calculation.

8 0
2 years ago
You have two springs. one has a greater spring constant than the other. you also have two objects, one with a greater mass than
neonofarm [45]
The answer should be "The object with the greater mass should be attached to the spring with the smaller <span>spring constant."

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
</span>
4 0
3 years ago
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