Could someone answer this?
1 answer:
Answer:
Here the circuit in which a 4Ω resistor resistor is connected in series and two 8Ω resistor resistors are connected in parallel. Also, ammeter and voltmeter connected in series and parallel circuit respectively.
Now,
The maximum power of each resistance is 16 W
The 4Ω resistor is linked in series with the circuit.
so, P o w e r = I
two
R, here i is the current through the resistor resistor R
1 6 = I
two
∗ 4 Ω
i = 2A
Now 2A passes through parallel resistors of 8Ω resistance.
we know that, in parallel, the potential difference must be constant,
the current is divided into two parts, because the same resistance current in each resistance will be half. then the current through each resistor in parallel is
2 A
two
.
= 1 A
So finally the current through the 4Ω resistor = 2 A
current through each 8Ω resistor = 1 A
Explanation:
I hope this answer has helped you
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The motion described here is a projectile motion which is characterized by an arc-shaped direction of motion. There are already derived equations for this type of motions as listed:
Hmax = v₀²sin²θ/2g
t = 2v₀sinθ/g
y = xtanθ + gx²/(2v₀²cos²θ)
where
Hmax = max. height reached by the object in a projectile motion
θ=angle of inclination
v₀= initial velocity
t = time of flight
x = horizontal range
y = vertical height
Part A.
Hmax = v₀²sin²θ/2g = (30²)(sin 33°)²/2(9.81)
Hmax = 13.61 m
Part B. In this part, we solve the velocity when it almost reaches the ground. Approximately, this is equal to y = 28.61 m and x = 31.91 m. In projectile motion, it is important to note that there are two component vectors of motion: the vertical and horizontal components. In the horizontal component, the motion is in constant speed or zero acceleration. On the other hand, the vertical component is acting under constant acceleration. So, we use the two equations of rectilinear motion:
y = v₀t + 1/2 at²
28.61 = 30(t) + 1/2 (9.81)(t²)
t = 0.839 seconds
a = (v₁-v₀)/t
9.81 = (v₁ - 30)/0.839
v₁ = 38.23 m/s
Part C.
y = xtanθ + gx²/(2v₀²cos²θ)
Hmax + 15 = xtanθ + gx²/(2v₀²cos²θ)
13.61 + 15 = xtan33° + (9.81)x²/[2(30)²(cos33°)²]
Solving using a scientific calculator,
x = 31.91 m
Hmax = v₀²sin²θ/2g
t = 2v₀sinθ/g
y = xtanθ + gx²/(2v₀²cos²θ)
where
Hmax = max. height reached by the object in a projectile motion
θ=angle of inclination
v₀= initial velocity
t = time of flight
x = horizontal range
y = vertical height
Part A.
Hmax = v₀²sin²θ/2g = (30²)(sin 33°)²/2(9.81)
Hmax = 13.61 m
Part B. In this part, we solve the velocity when it almost reaches the ground. Approximately, this is equal to y = 28.61 m and x = 31.91 m. In projectile motion, it is important to note that there are two component vectors of motion: the vertical and horizontal components. In the horizontal component, the motion is in constant speed or zero acceleration. On the other hand, the vertical component is acting under constant acceleration. So, we use the two equations of rectilinear motion:
y = v₀t + 1/2 at²
28.61 = 30(t) + 1/2 (9.81)(t²)
t = 0.839 seconds
a = (v₁-v₀)/t
9.81 = (v₁ - 30)/0.839
v₁ = 38.23 m/s
Part C.
y = xtanθ + gx²/(2v₀²cos²θ)
Hmax + 15 = xtanθ + gx²/(2v₀²cos²θ)
13.61 + 15 = xtan33° + (9.81)x²/[2(30)²(cos33°)²]
Solving using a scientific calculator,
x = 31.91 m
Gravitational potential energy can be calculated using the formula:
Where:
PEgrav = Gravitational potential energy
m= mass
g = acceleration due to gravity
h = height
On Earth acceleration due to gravity is a constant 9.8 but since the scenario is on Mars, the pull of gravity is different. In this case, it is 3.7, so we will use that for g.
So put in what you know and solve for what you don't know.
m = 10kg
g = 3.7m/s^2
h = 1m
So we put that in and solve it.
Where:
PEgrav = Gravitational potential energy
m= mass
g = acceleration due to gravity
h = height
On Earth acceleration due to gravity is a constant 9.8 but since the scenario is on Mars, the pull of gravity is different. In this case, it is 3.7, so we will use that for g.
So put in what you know and solve for what you don't know.
m = 10kg
g = 3.7m/s^2
h = 1m
So we put that in and solve it.