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Molodets [167]
2 years ago
13

What's the difference between a direct relationship and a positive one?

Physics
1 answer:
borishaifa [10]2 years ago
5 0
A positive or direct relationship is one in which the two variables (we will generally call them x and y) move together, that is, they either increase or decrease together. In a negative or indirect relationship, the two variables move in opposite directions, that is, as one increases, the other descremases
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You need 5 blocks of the smaller object to contain the same amount of volume of the bigger object

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3 years ago
A sound is recorded at 19 decibels. What is the intensity of the sound?
sp2606 [1]

1 \times 10^{-10.1} \mathrm{Wm}^{-2} is the intensity of the sound.

Answer: Option B

<u>Explanation:</u>

The range of sound intensity that people can recognize is so large (including 13 magnitude levels). The intensity of the weakest audible noise is called the hearing threshold. (intensity about 1 \times 10^{-12} \mathrm{Wm}^{-2}). Because it is difficult to imagine numbers in such a large range, it is advisable to use a scale from 0 to 100.

This is the goal of the decibel scale (dB).  Because logarithm has the property of recording a large number and returning a small number, the dB scale is based on a logarithmic scale. The scale is defined so that the hearing threshold has intensity level of sound as 0.

                     \text { Intensity }(d B)=(10 d B) \times \log _{10}\left(\frac{I}{I_{0}}\right)

Where,

I = Intensity of the sound produced

I_{0} = Standard Intensity of sound of 60 decibels = 1 \times 10^{-12} \mathrm{Wm}^{-2}

So for 19 decibels, determine I as follows,

                   19 d B=(10 d B) \times \log _{10}\left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)

                  \log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=\frac{19}{10}

                  \log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=1.9

When log goes to other side, express in 10 to the power of that side value,

                  \left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)=10^{1.9}

                  I=1 \times 10^{-12} \mathrm{Wm}^{-2} \times 10^{1.9}=1 \times 10^{-12-1.9}=1 \times 10^{-10.1} \mathrm{Wm}^{-2}

5 0
3 years ago
What characteristic of solitaire did Mendeleev<br> use as a model for his periodic table?
forsale [732]
That the player must produce an arrangement in which each suit is ordered from ace to king
7 0
3 years ago
When are you most aware of your motion in a moving vehicle: when it is moving steadily in a straight line or when it is accelera
Ghella [55]

Answer:

Acceleration is percieved, not constant velocity.

Explanation:

You are most aware when the vehicle is accelerating. At constant velocity you would not be aware of the motion. Only if the system is accelerated the dynamics must be solved considering a pseudo-force (of inertial origin) acting.

It's because of this that:

(A) False. The acceleration can be detected from the inside of a closed car.

(B) False. You would be aware of the motion, but not because humans can sense speed but acceleration.

(C) False. Constant velocity cannot be felt in a closed car.

(D) False. Again, you can't feel constant speed.

6 0
2 years ago
A car is moving in uniform circular motion. If the cars speed were to double to keep the car moving with the same radius the acc
Stells [14]

Answer:

<em>The centripetal acceleration would increase by a factor of 4</em>

<em>Correct choice: B.</em>

Explanation:

<u>Circular Motion</u>

The circular motion is described when an object rotates about a fixed point called center. The distance from the object to the center is the radius. There are other magnitudes in the circular motion like the angular speed, tangent speed, and centripetal acceleration. The formulas are:

v_t=w\ r

\displaystyle a_c=\frac{v_t^2}{r}

If the speed is doubled and the radius is the same, then

\displaystyle a_c=\frac{(2v_t)^2}{r}

\displaystyle a_c=4\frac{v_t^2}{r}

The centripetal acceleration would increase by a factor of 4

Correct choice: B.

5 0
3 years ago
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