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Sagot :
Sure, let's break this down step by step to show that both equations are homogeneous in terms of their respective units.
### Question 7
For the equation [tex]\( Q = 1 \times t \)[/tex], we need to prove that the units on both sides are consistent. Here, [tex]\( Q \)[/tex] is electric charge, and [tex]\( t \)[/tex] is time, with electric current given as 1.
1. Units of Electric Charge [tex]\( Q \)[/tex]:
- Electric charge [tex]\( Q \)[/tex] is measured in Coulombs (C).
2. Units of Electric Current [tex]\( I \)[/tex]:
- Electric current [tex]\( I \)[/tex] is measured in Amperes (A).
- Given [tex]\( I = 1 \)[/tex] A.
3. Units of Time [tex]\( t \)[/tex]:
- Time [tex]\( t \)[/tex] is measured in seconds (s).
From the basic relationship in electrical formulas:
[tex]\[ I = \frac{Q}{t} \][/tex]
Rearranging this to solve for [tex]\( Q \)[/tex]:
[tex]\[ Q = I \times t \][/tex]
Now, replacing the given values:
[tex]\[ Q = 1 \times t \][/tex]
Next, we confirm the homogeneity of the units:
- Electric charge [tex]\( Q \)[/tex] has units of Coulombs (C).
- Time [tex]\( t \)[/tex] has units of seconds (s).
- Since [tex]\( I \)[/tex] is electric current, it has units of Amperes (A).
Using the fundamental equation [tex]\( I = \frac{Q}{t} \)[/tex]:
[tex]\[ \text{1 Coulomb (C)} = \text{1 Ampere (A)} \times \text{1 second (s)} \][/tex]
This shows that:
[tex]\[ \text{C} = \text{A} \times \text{s} \][/tex]
Thus, the units on both sides of [tex]\( Q = 1 \times t \)[/tex] are homogeneous, as Coulombs are equivalent to [tex]\( \text{Ampere} \times \text{seconds} \)[/tex].
### Question 8
For the equation [tex]\( P = IV \)[/tex], we need to demonstrate that the units on both sides are consistent. Here, [tex]\( P \)[/tex] is power, [tex]\( I \)[/tex] is electric current, and [tex]\( V \)[/tex] is voltage.
1. Units of Power [tex]\( P \)[/tex]:
- Power [tex]\( P \)[/tex] is measured in Watts (W).
2. Units of Electric Current [tex]\( I \)[/tex]:
- Electric current [tex]\( I \)[/tex] is measured in Amperes (A).
3. Units of Voltage [tex]\( V \)[/tex]:
- Voltage [tex]\( V \)[/tex] is measured in Volts (V).
Given:
[tex]\[ P = I \times V \][/tex]
Next, we confirm the homogeneity of the units:
- Power [tex]\( P \)[/tex] has units of Watts (W).
- Electric current [tex]\( I \)[/tex] has units of Amperes (A).
- Voltage [tex]\( V \)[/tex] has units of Volts (V).
Using the known relationship:
[tex]\[ 1 \text{ Watt (W)} = 1 \text{ Ampere (A)} \times 1 \text{ Volt (V)} \][/tex]
Therefore:
[tex]\[ \text{W} = \text{A} \times \text{V} \][/tex]
This shows that the units on both sides of [tex]\( P = IV \)[/tex] are homogeneous, as Watts are equivalent to [tex]\( \text{Ampere} \times \text{Volts} \)[/tex].
In conclusion, we have shown that both equations [tex]\( Q = 1 \times t \)[/tex] and [tex]\( P = IV \)[/tex] are homogeneous with respect to their units.
### Question 7
For the equation [tex]\( Q = 1 \times t \)[/tex], we need to prove that the units on both sides are consistent. Here, [tex]\( Q \)[/tex] is electric charge, and [tex]\( t \)[/tex] is time, with electric current given as 1.
1. Units of Electric Charge [tex]\( Q \)[/tex]:
- Electric charge [tex]\( Q \)[/tex] is measured in Coulombs (C).
2. Units of Electric Current [tex]\( I \)[/tex]:
- Electric current [tex]\( I \)[/tex] is measured in Amperes (A).
- Given [tex]\( I = 1 \)[/tex] A.
3. Units of Time [tex]\( t \)[/tex]:
- Time [tex]\( t \)[/tex] is measured in seconds (s).
From the basic relationship in electrical formulas:
[tex]\[ I = \frac{Q}{t} \][/tex]
Rearranging this to solve for [tex]\( Q \)[/tex]:
[tex]\[ Q = I \times t \][/tex]
Now, replacing the given values:
[tex]\[ Q = 1 \times t \][/tex]
Next, we confirm the homogeneity of the units:
- Electric charge [tex]\( Q \)[/tex] has units of Coulombs (C).
- Time [tex]\( t \)[/tex] has units of seconds (s).
- Since [tex]\( I \)[/tex] is electric current, it has units of Amperes (A).
Using the fundamental equation [tex]\( I = \frac{Q}{t} \)[/tex]:
[tex]\[ \text{1 Coulomb (C)} = \text{1 Ampere (A)} \times \text{1 second (s)} \][/tex]
This shows that:
[tex]\[ \text{C} = \text{A} \times \text{s} \][/tex]
Thus, the units on both sides of [tex]\( Q = 1 \times t \)[/tex] are homogeneous, as Coulombs are equivalent to [tex]\( \text{Ampere} \times \text{seconds} \)[/tex].
### Question 8
For the equation [tex]\( P = IV \)[/tex], we need to demonstrate that the units on both sides are consistent. Here, [tex]\( P \)[/tex] is power, [tex]\( I \)[/tex] is electric current, and [tex]\( V \)[/tex] is voltage.
1. Units of Power [tex]\( P \)[/tex]:
- Power [tex]\( P \)[/tex] is measured in Watts (W).
2. Units of Electric Current [tex]\( I \)[/tex]:
- Electric current [tex]\( I \)[/tex] is measured in Amperes (A).
3. Units of Voltage [tex]\( V \)[/tex]:
- Voltage [tex]\( V \)[/tex] is measured in Volts (V).
Given:
[tex]\[ P = I \times V \][/tex]
Next, we confirm the homogeneity of the units:
- Power [tex]\( P \)[/tex] has units of Watts (W).
- Electric current [tex]\( I \)[/tex] has units of Amperes (A).
- Voltage [tex]\( V \)[/tex] has units of Volts (V).
Using the known relationship:
[tex]\[ 1 \text{ Watt (W)} = 1 \text{ Ampere (A)} \times 1 \text{ Volt (V)} \][/tex]
Therefore:
[tex]\[ \text{W} = \text{A} \times \text{V} \][/tex]
This shows that the units on both sides of [tex]\( P = IV \)[/tex] are homogeneous, as Watts are equivalent to [tex]\( \text{Ampere} \times \text{Volts} \)[/tex].
In conclusion, we have shown that both equations [tex]\( Q = 1 \times t \)[/tex] and [tex]\( P = IV \)[/tex] are homogeneous with respect to their units.
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