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xxMikexx [17]
3 years ago
8

Your teacher wants you to perform an experiment that will demonstrate a chemical property. Which experiment should you choose? A

) Measure the pH of different household chemicals B) Form a density column using common household materials. C) Measure the melting point of common household chemicals. D) Measure the solubility of different household substances.
Physics
2 answers:
son4ous [18]3 years ago
7 0

Answer:

A

Explanation:

because it say chemical

Dovator [93]3 years ago
4 0

A) Measure the pH of different household chemicals

Explanation:

To demonstrate a chemical property using an experiment, measuring the pH of different household chemicals will be the best way. The pH is the degree of hydrogen or hydroxyl ion concentration in a solution.

  • Chemical properties tell us about what a substance can do.
  • It shows if a substance will react with other substances or not.
  • Examples are flammability, rusting of iron, precipitation, decomposition of water by electric current e.t.c.
  • Measuring the pH is a chemical property determination procedure.
  • The pH points to the degree of acidity or alkalinity of a solution.

Learn more:

Chemical change brainly.com/question/9388643

#learnwithBrainly

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ANSWER:

(a) 1036 N

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(c) 2590 N

STEP-BY-STEP EXPLANATION:

Given:

Mc = 1400 kg

Mt = 560 kg

a = 1.85 m/s^2

(a)

Force by car on trailer:

\begin{gathered} F_c=m\cdot a \\ F_c=560\cdot1.85 \\ F_c=1036\text{ N} \end{gathered}

(b)

\begin{gathered} F_t=-F_c \\ F_t=-1036\text{ N} \end{gathered}

(c)

\begin{gathered} F_n=1400\cdot1.85 \\ F_n=2590\text{ N} \end{gathered}

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1 year ago
Calculate the work done when a 100-W lightbulb is lit for 30 seconds.
Mila [183]

Answer:

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Explanation:

8 0
3 years ago
A 0.29 kg particle moves in an xy plane according to x(t) = - 19 + 1 t - 3 t3 and y(t) = 20 + 7 t - 9 t2, with x and y in meters
Artist 52 [7]

Answer:

Part a)

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Part b)

\theta = -137.7 degree

Part c)

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Explanation:

As we know that acceleration is rate of change in velocity of the object

So here we know that

x = -19 + t - 3t^3

y = 20 + 7t - 9t^2

Part a)

differentiate x and y two times with respect to time to find the acceleration

a_x = \frac{d^2}{dt^2}(-19 + t - 3t^3)

a_x = \frac{d}{dt}(0 +1 - 9t^2)

a_x = -18t

a_y = \frac{d^2}{dt^2}(20 + 7t - 9t^2)

a_y = \frac{d}{dt}(0 +7 - 18t)

a_y = -18

Now the acceleration of the object is given as

\vec a = (-18t)\hat i + (-18)\hat j

at t= 1.1 s we have

\vec a = -19.8 \hat i - 18 \hat j

now the net force of the object is given as

\vec F = m\vec a

\vec F = (0.29 kg)(-19.8 \hat i - 18 \hat j)

\vec F = -5.74 \hat i - 5.22 \hat j

now magnitude of the force will be

F = \sqrt{5.74^2 + 5.22^2} = 7.76 N

Part b)

Direction of the force is given as

tan\theta = \frac{F_y}{F_x}

tan\theta = \frac{-5.22}{-5.74}

\theta = -137.7 degree

Part c)

For velocity of the particle we have

v_x = \frac{dx}[dt}

v_x = (0 +1 - 9t^2)

v_y = \frac{dy}{dt}

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\vec v = -9.89\hat i - 12.8 \hat j

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\theta = tan^{-1}(\frac{-12.8}{-9.89})

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Answer:

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Explanation:

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slamgirl [31]

Answer:

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Explanation:

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make I the subject of the equation

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Substitute into equation 2

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Also,

Q = It............... Equation 3

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make t the subject of the equation

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t =  7.899×10⁻¹⁹/0.276

t = 2.86×10⁻¹⁸ seconds

5 0
3 years ago
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