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3 years ago

When velocity is graphed with respect to time, what does the area under

Physics

2 answers:

denpristay [2]3 years ago

6 0

The Answer

Is Displacement

Juliette [100K]3 years ago

4 0

Answer:

the answer is D

Explanation:

Since the velocity of the object is the derivative of the position graph, the area under the line in the velocity vs. time graph is the displacement of the object.

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Determining the pH of substances such as purple grape juice and catsup using test strips can be difficult. Why?

Charra [1.4K]

Answer:

Because they would naturally dye the test strips in the colors violet and red, regardless of their pH values

(would really appreciate the brainliest)

4 0

4 years ago

A firework is designed so that when it is fired directly upwards, it explodes and splits into three equal-massed parts at its pe

Arturiano [62]

Answer:

the third piece landed $2\,\sqrt{2}$ meters in a direction 45 degrees from the observer's right towards the front

Explanation:

Due to conservation of momentum, the momentum of the third piece must add to the other two momenta in vector from to render zero (initial momentum of the firework on a plane parallel to ground. Therefore the momentum components of the third piece must have a component to the right equal in magnitude to the momentum of the piece that traveled to the left, and a momentum component pointing to the front of the observer of equal magnitude to that of the piece that traveled behind the observer. Then the direction of the third momentum must be at 45 degrees from the observer's right towards the observer's front.

Also, the magnitude of a momentum vector of such components, is given by the Pythagorean theorem. Recall also that the magnitudes of the momentum vectors of the first two pieces must be equal since they traveled equal distances.

$|P_3|=\sqrt{P_1^2+P_2^2} \\|P_3|=\sqrt{P^2+P^2}\\|P_3|=\sqrt{2P^2}\\|P_3|=P\sqrt{2}$

Then the distance traveled by the third piece must also keep this proportionality:

$D_3=\sqrt{2} \,\,D\\D_3=\sqrt{2} \,*\,2\,\,meters\\D_3=2\,\sqrt{2} \,\,meters$

4 0

3 years ago

What is the intensity of a sound with a sound intensity level (SIL) 67 dB, in units of W/m^2?

vfiekz [6]

Answer:

The intensity of sound (I) = 3.16 x 10⁻⁶ W/m²

Explanation:

We have expression for sound intensity level (SIL),

$L=10log_{10}\left ( \frac{I}{I_0}\right )$

Here we need to find the intensity of sound (I).

$L=10log_{10}\left ( \frac{I}{I_0}\right )\\\\log_{10}\left ( \frac{I}{I_0}\right )=0.1L\\\\\frac{I}{I_0}=10^{0.1L}\\\\I=I_010^{0.1L}$

Substituting

L = 67 dB and I₀ = 10⁻¹² W/m² in the equation

$I=I_010^{0.1L}=10^{-12}\times 10^{0.1\times 65}\\\\I_0=10^{-12}\times 10^{6.5}=10^{-5.5}=3.16\times 10^{-6}W/m^2$

The intensity of sound (I) = 3.16 x 10⁻⁶ W/m²

8 0

3 years ago

. An experimental rocket plane lands on skids on a dry lake bed. If it’s traveling at 80.0 m/s when it touches down, how far doe

kobusy [5.1K]

Answer:

1. A substance that can be separated into two or more substances only by a chemical change is a(n) _____.

solution

element

mixture

compound

Which of the following materials is a substance?

air

gasoline

stainless steel

silver

The first figure in a properly written chemical symbol always is _____.

boldfaced

capitalized

italicized

underlined

Which of the following represents a compound?

H-3

H2O

O-16

5. What do chemical symbols and formulas represent, respectively?

elements and compounds

atoms and mixtures

compounds and mixtures

elements and ions

Explanation:

5 0

2 years ago

13.An airplane starts from rest at the end of a runway and accelerates at a constant rate. In the first second, the airplane tra

Murrr4er [49]

Answer:

The answer is 3.33m

Explanation:

The acceleration "a" is constant.

Acceleration is the variation of velocity over time,

$\frac{dv}{dt} = a$ .

solving the last equation

$\int_{v_0}^v dv = a\int_0^t dt \rightarrow v-v_0 = at$ ,

where $v_0=0$ because the airplane starts from rest.

Once again, velocity is the variation of distance over time.

$\frac{dx}{dt} = at \rightarrow \int_{x_0}^x dx = a\int_0^t t\ dt$

then

$x- x_0 = \frac{1}{2}at^2$

where $x_0=0$ if we consider the end of the runway as the initial point (this step is for simplicity but you can let it expressed, it's going to cancel anyway).

If $x=1.11\ m$ at $t=1s$ , then

$a = \frac{2x}{t^2} = 2.22\ m/s^2$

and the final expression for the distance is

$x = 1.11 t^2$ .

If t = 2s, x = 4.44 m. Which means thad the additional distance is

$x(2s) - x(1s) = 4.44 - 1.11 = 3.33\ m$

8 0

4 years ago