Answer:
A) the ammeter is x
B)
- voltage across R₁ (left resistor) = 0.75 V
- voltage across the right one = 0.3 V
C) 1.05 V
Explanation:
From the diagram attached below;
A) Assuming the homes were wired in series, and one of the homes face short circuit then all the houses would face power cut but it doesn't happen. So they must be connected in parallel.
Therefore; The ammeter is connected in series, Hence, the ammeter is x and the voltmeter must be z.
B)
Given that:
x = 0.15 A
z = 0.3 V
Resistor (R) on the left = 5 ohms
Then, voltage across R₁ (left resistor) = 5×(x)
= 5×0.15
= 0.75 V
voltage across the right one = z = 0.3 V
C)
The total voltage of battery = 0.75+0.3 = 1.05 V
To solve this problem we will apply the concepts related to Ohm's law and Electric Power. By Ohm's law we know that resistance is equivalent to,
Here,
V = Voltage
I = Current
While the power is equivalent to the product between the current and the voltage, thus solving for the current we have,
Applying Ohm's law
Therefore the equivalent resistance of the light string is 
The answer for this question is letter "B.Fission releases energy, and its products have greater stability."
Fission and Fusion are both nuclear reactions that when they release energy, they make the nuclei more stable. So among the choices, option B is the most fitting for the definition.
You multiply the high length and width and if your using centimeters then divide by 500 and then there's your answer.hoped this helped.
Answer:
(a) 2.85 m
(b) 16.5 m
(c) 21.7 m
(d) 22.7 m
Explanation:
Given:
v₀ₓ = 19 cos 71° m/s
v₀ᵧ = 19 sin 71° m/s
aₓ = 0 m/s²
aᵧ = -9.8 m/s²
(a) Find Δy when t = 3.5 s.
Δy = v₀ᵧ t + ½ aᵧ t²
Δy = (19 sin 71° m/s) (3.5 s) + ½ (-9.8 m/s²) (3.5 s)²
Δy = 2.85 m
(b) Find Δy when vᵧ = 0 m/s.
vᵧ² = v₀ᵧ² + 2 aᵧ Δy
(0 m/s)² = (19 sin 71° m/s)² + 2 (-9.8 m/s²) Δy
Δy = 16.5 m
(c) Find Δx when t = 3.5 s.
Δx = v₀ₓ t + ½ aₓ t²
Δx = (19 cos 71° m/s) (3.5 s) + ½ (0 m/s²) (3.5 s)²
Δx = 21.7 m
(d) Find Δx when Δy = 0 m.
First, find t when Δy = 0 m.
Δy = v₀ᵧ t + ½ aᵧ t²
(0 m) = (19 sin 71° m/s) t + ½ (-9.8 m/s²) t²
0 = t (18.0 − 4.9 t)
t = 3.67
Next, find Δx when t = 3.67 s.
Δx = v₀ₓ t + ½ aₓ t²
Δx = (19 cos 71° m/s) (3.67 s) + ½ (0 m/s²) (3.67 s)²
Δx = 22.7 m
