Sam is moving house and is carrying a 300N box of books up a flight of steps 5m high, it takes her 30 seconds. Gary follows her
1 answer:
Answer: Sam provides the biggest power
Explanation: <u>Power</u> is defined, in Physics, as the rate of work done. <u>Work</u> is energy transferred to an object due to the force causing the displacement of the object.
So, to determine Power:
For Sam:
P =
P = 50 Watts
The unit for Power is [P] = J/s = Watt
For Gary:
P =
P = 40 Watts
Comparing power of Sam and Gary, we can conclude that Sam provides the biggest power of 50 Watts.
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Answer:
υ = 345.82 m/s
Explanation:
The formula used to find the speed of sound in air, at different temperatures is given as follows:
where,
υ = speed of sound at given temperature = ?
υ₀ = speed of sound at 0°C = 331 m/s
T = temperature in K = 15°C + 273 = 298 k
Therefore, using these values in the equation, we get:
<u>υ = 345.82 m/s</u>
Answer:
A) C_{eq} = 15 10⁻⁶ F, B) U₃ = 3 J, U₄ = 0.5 J
Explanation:
In a complicated circuit, the method of solving them is to work the circuit in pairs, finding the equivalent capacitance to reduce the circuit to simpler forms.
In this case let's start by finding the equivalent capacitance.
A) Let's solve the part where C1 and C3 are. These two capacitors are in serious
(you has an mistake in the formula)
= 0.1 10⁶
= 10 10⁻⁶ F
capacitors C₂, C₄ and C₅ are in series
= 0.2 10⁶
= 5 10⁻⁶ F
the two equivalent capacitors are in parallel therefore
C_{eq} = C_{eq1} + C_{eq2}
C_{eq} = (10 + 5) 10⁻⁶
C_{eq} = 15 10⁻⁶ F
B) the energy stored in C₃
The charge on the parallel voltage is constant
is the sum of the charge on each branch
Q = C_{eq} V
Q = 15 10⁻⁶ 6
Q = 90 10⁻⁶ C
the charge on each branch is
Q₁ = Ceq1 V
Q₁ = 10 10⁻⁶ 6
Q₁ = 60 10⁻⁶ C
Q₂ = C_{eq2} V
Q₂ = 5 10⁻⁶ 6
Q₂ = 30 10⁻⁶ C
now let's analyze the load on each branch
Branch C₁ and C₃
In series combination the charge is constant Q = Q₁ = Q₃
U₃ =
U₃ =
U₃ = 3 J
In Branch C₂, C₄, C₅
since the capacitors are in series the charge is constant Q = Q₂ = Q₄ = Q₅
U₄ =
U₄ = 0.5 J