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ololo11 [35]
3 years ago
14

A teacher wants to perform a classroom demonstration that illustrates both chemical and physical changes. Which would be the bes

t demonstration that she could use?
A) bending a piece of aluminum
B)breaking a matchstick
C) dissolving sugar in water
D) burning a candle
Physics
2 answers:
anygoal [31]3 years ago
7 0

Answer

D) burning a candle

Explanation

When burning a candle no new substance is form.

We have both physical and chemical change occuring.

Physical part: Melting of the solid wax and evaporation of the liquid forms the physical change.

Chemical part: burning of the wax vapour forms the chemical change.

puteri [66]3 years ago
3 0
Burning a candle because the candle gets smaller and causes a chemical change
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Give 10 examples of units you might use or see in any given day
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Cups
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A technological device that can be used to see sound waves is an
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The answer is oscilloscope
4 0
3 years ago
A runner taking part in the 200 m dash must run around the end of a track that has a circular arc with a radius of curvature of
34kurt

Answer:

The centripetal acceleration of the runner is 1.73\ m/s^2.

Explanation:

Given that,

A runner completes the 200 m dash in 24.0 s and runs at constant speed throughout the race. We need to find the centripetal acceleration as he runs the curved portion of the track. We know that the centripetal acceleration is given by :

a=\dfrac{v^2}{r}

v is the velocity of runner

v=\dfrac{200\ m}{24\ s}\\\\v=8.34\ m/s

Centripetal acceleration,

a=\dfrac{(8.34)^2}{40}\\\\a=1.73\ m/s^2

So, the centripetal acceleration of the runner is 1.73\ m/s^2. Hence, this is the required solution.

5 0
4 years ago
An electron is in a vacuum near Earth's surface and located at y = 0 m on a vertical y axis. At what value of y should a group o
Bingel [31]

Answer:

the bunch of 23 electrons must be placed on y axis at coordinate y = - 24.35 m.

Explanation:

As we know that the gravitational force on electron at y = 0 is counter-balanced by the weight of the electron

So we have

\frac{kq_1q_2}{r^2} = mg

here we have

q_1 = e

q_2 = 23 e

m = 9.11 \times 10^{-31} kg

also we know that

e = 1.6 \times 10^{-19} C

so we will have

\frac{(9\times 10^9)(1.6 \times 10^{-19})(23\times 1.6 \times 10^{-19})}{r^2} = (9.11 \times 10^{-31})(9.81)

\frac{5.3 \times 10^{-27}}{r^2} = 8.94 \times 10^{-30}

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7 0
3 years ago
Denial cycled at 6 mph for 0.8 h. Then he jogged at 2 mph for
dexar [7]
<h3>Deniel's average speed for the total</h3><h3>distance is 3.6 mph</h3>

Average speed is difined as the total distance travelled divided by the total time required to cover the distance. Mathematically, it is expressed as:

Ave. Speed = \frac{Total distance }{Total time}

To obtain the average speed of Daniel, we shall determine the distance travelled in each case. This is illustrated below:

<h3>Case 1:</h3>

Speed 1 (S₁) = 6 mph

Time 1 (t₁) = 0.8 h.

<h3>Distance 1 (D₁) =? </h3>

Speed = \frac{Distance}{Time} \\\\S_{1}  = \frac{D_{1}}{t_{1}} \\\\6 = \frac{D_{1}}{0.8}

Cross multiply

D₁ = 6 × 0.8

<h3>D₁ = 4.8 mile</h3>

<h3>Case 2:</h3>

Speed 2 (S₂) = 2 mph

Time 2 (t₂) = 1.2 h.

<h3>Distance 2 (D₂) =? </h3>

S_{2} = \frac{D_{2}}{t_{2}}\\\\2 = \frac{D_{2}}{1.2}

Cross multiply

D₂ = 2 × 1.2

<h3>D₂ = 2.4 mile </h3>

Next, we shall determine the total time. This can be obtained as follow:

Time 1 (t₁) = 0.8 h.

Time 2 (t₂) = 1.2 h.

<h3>Total time (T) =? </h3>

T = t₁ + t₂

T = 0.8 + 1.2

<h3>T = 2 h</h3>

Next, we shall determine the total distance. This can be obtained as follow:

Distance 1 (D₁) = 4.8 mile

Distance 2 (D₂) = 2.4 mile

<h3>Total distance (D) =?</h3>

D = D₁ + D₂

D = 4.8 + 2.4

<h3>D = 7.2 mile</h3>

Finally, we shall determine the average speed of Daniel. This can be obtained as follow:

Total time = 2 h

Total distance  = 7.2 mile

<h3>Average speed =?</h3>

Ave. speed = \frac{Total distance }{Total time} \\\\Ave. speed = \frac{7.2 }{2}

<h3>Average speed = 3.6 mph</h3><h3 />

Therefore, the average speed of Daniel is 3.6 mph

Learn more: brainly.com/question/680492

5 0
3 years ago
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