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1 year ago

Alan wrote the following examples of changes in substance.

Physics

2 answers:

Anastaziya [24]1 year ago

6 0

Answer:

The correct option is B

Explanation:

A physical change is characterized by the following

no new substance is formed
easily reversible
cannot be represented by a chemical reaction

A chemical change is characterized by the following

new substances are formed
easily reversible
can be represented by a chemical reaction

A camp fire is usually made by burning woods. This process is called combustion, it involves the burning of a substance in oxygen to produce carbon dioxide (CO₂). This is a chemical change because new substance (CO₂) is formed and the ashes formed (after burning) cannot be reversed back to wood.

Plants convert carbon dioxide and water into glucose and oxygen through photosynthesis. This process is also a chemical change because new substances (glucose and oxygen) are formed and the reaction cannot also be easily reversed.

The reaction involved can be represented as

6CO₂ + 6H₂O ⇒ C₆H₁₂O₆ + 6O₂

Yuliya22 [10]1 year ago

5 0

A campfire being lighted and plants converting carbon-dioxide and water into glucose and oxygen are both forms of chemical change.

Therefore, the answer is:

B. Both are examples of chemical change.

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A point charge with charge q1 is held stationary at the origin. A second point charge with charge q2 moves from the point (x1, 0

Scilla [17]

Answer:

$W=kq_1q_2(\dfrac{1}{x_1}-\dfrac{1}{\sqrt{x_2^2+y_2^2}})$

Explanation:

Position of charge q₁ is (0,0)

Position of charge q₂ is (x₁,0)

So, the electric potential energy between the charges is given by :

$U_1=k\dfrac{q_1q_2}{x_1}$

Now the position of charge q₂ has been changes from (x₁,0) to (x₂,y₂). Now, electric potential energy between the charges is :

$U_2=k\dfrac{q_1q_2}{\sqrt{x_2^2+y_2^2}}$

We know form the work energy theorem that, the change in potential energy is equal to the work done. Mathematically, it is given by :

$W=-\Delta U$

$W=-(U_2-U_1)$

$W=(U_1-U_2)$

$W=(k\dfrac{q_1q_2}{x_1}-k\dfrac{q_1q_2}{\sqrt{x_2^2+y_2^2}})$

$W=kq_1q_2(\dfrac{1}{x_1}-\dfrac{1}{\sqrt{x_2^2+y_2^2}})$

Hence, the work done by the electrostatic force on the moving point charge is $kq_1q_2(\dfrac{1}{x_1}-\dfrac{1}{\sqrt{x_2^2+y_2^2}})$ . Hence, this is the required solution.

3 0

1 year ago

Megan rode the bus to school, which is located 8 kilometers from her home. If Megan's frame of reference is her house, and it to

Dmitriy789 [7]

Answer:

Explanation: idk sry

7 0

1 year ago

A company designed and sells an ultrasonic receiver, which detects sounds unable to be heard by the human ear. The receiver can

Ksenya-84 [330]

Answer:

28√3 m

Explanation:

A = vertex where receiver is placed

S = focus

Bp = r = radius of the outside edge

Bc = 2r = diameter

The full explanation is shown in the picture attached herewith. Thank you and i hope it helps.

7 0

1 year ago

A 450g mass on a spring is oscillating at 1.2Hz. The totalenergy of the oscillation is 0.51J. What is the amplitude.

Volgvan

Answer:

A=0.199

Explanation:

We are given that

Mass of spring=m=450 g= $=\frac{450}{1000}=0.45 kg$

Where 1 kg=1000 g

Frequency of oscillation= $\nu=1.2Hz$

Total energy of the oscillation=0.51 J

We have to find the amplitude of oscillations.

Energy of oscillator= $E=\frac{1}{2}m\omega^2A^2$

Where $\omega=2\pi\nu$ =Angular frequency

A=Amplitude

$\pi=\frac{22}{7}$

Using the formula

$0.51=\frac{1}{2}\times 0.45(2\times \frac{22}{7}\times 1.2)^2A^2$

$A^2=\frac{2\times 0.51}{0.45\times (2\times \frac{22}{7}\times 1.2)^2}=0.0398$

$A=\sqrt{0.0398}=0.199$

Hence, the amplitude of oscillation=A=0.199

4 0

1 year ago

a torch bulb is rated 2.5V and 750mA. Calculate its power,its resistance and the energy consumed if this bulb lighted for 4 hour

Hatshy [7]

Using Ohm's Law, we can derived from this the value of resistance. If I=V/R, therefore, R = V/I
Substituting the values to the given,
P = Power = ?
R = Resistance = ?
V = Voltage = 2.5 V
I = Current = 750 mA

R = V/I = 2.5/ (750 x 10^-3)
R = 3.33 ohms

Calculating the power, we have P = IV

P = (750 x 10^-3)(2.5)
P = 1.875 W

The power consumption is the power consumed multiply by the number of hours. In here, we have;
1.875W x 4 hours = 7.5 watt-hours

3 0

1 year ago