The answer is B.
She can measure the mass of the water, marble and the graduated cylinder with the balance.
The volume of the water can be shown on the marked graduated cylinder, the volume of the marble can be measured by the volume difference of the water before and after the marble is put in.
A motion where gravity is the only force acting upon it
The figure is showing a volume of 2.4 mL becuase it's feel 4 little segments.
Therefore, the answer is 2.4 mL.
Refer to the diagram shown below.
m = the mass of the object
x = the distance of the object from the equilibrium position at time t.
v = the velocity of the object at time t
a = the acceleration of the object at time t
A = the amplitude ( the maximum distance) of the mass from the equilibrium
position
The oscillatory motion of the object (without damping) is given by
x(t) = A sin(ωt)
where
ω = the circular frequency of the motion
T = the period of the motion so that ω = (2π)/T
The velocity and acceleration are respectively
v(t) = ωA cos(ωt)
a(t) = -ω²A sin(ωt)
In the equilibrium position,
x is zero;
v is maximum;
a is zero.
At the farthest distance (A) from the equilibrium position,
x is maximum;
v is zero;
a is zero.
In the graphs shown, it is assumed (for illustrative purposes) that
A = 1 and T = 1.
Answer:
1. 12 V
2a. R₁ = 4 Ω
2b. V₁ = 4 V
3a. A = 1.5 A
3b. R₂ = 4 Ω
4. Diagram is not complete
Explanation:
1. Determination of V
Current (I) = 2 A
Resistor (R) = 6 Ω
Voltage (V) =?
V = IR
V = 2 × 6
V = 12 V
2. We'll begin by calculating the equivalent resistance. This can be obtained as follow:
Voltage (V) = 12 V
Current (I) = 1 A
Equivalent resistance (R) =?
V = IR
12 = 1 × R
R = 12 Ω
a. Determination of R₁
Equivalent resistance (R) = 12 Ω
Resistor 2 (R₂) = 8 Ω
Resistor 1 (R₁) =?
R = R₁ + R₂ (series arrangement)
12 = R₁ + 8
Collect like terms
12 – 8 =
4 = R₁
R₁ = 4 Ω
b. Determination of V₁
Current (I) = 1 A
Resistor 1 (R₁) = 4 Ω
Voltage 1 (V₁) =?
V₁ = IR₁
V₁ = 1 × 4
V₁ = 4 V
3a. Determination of the current.
Since the connections are in series arrangement, the same current will flow through each resistor. Thus, the ammeter reading can be obtained as follow:
Resistor 1 (R₁) = 4 Ω
Voltage 1 (V₁) = 6 V
Current (I) =?
V₁ = IR₁
6 = 4 × I
Divide both side by 4
I = 6 / 4
I = 1.5 A
Thus, the ammeter (A) reading is 1.5 A
b. Determination of R₂
We'll begin by calculating the voltage cross R₂. This can be obtained as follow:
Total voltage (V) = 12 V
Voltage 1 (V₁) = 6 V
Voltage 2 (V₂) =?
V = V₁ + V₂ (series arrangement)
12 = 6 + V₂
Collect like terms
12 – 6 = V₂
6 = V₂
V₂ = 6 V
Finally, we shall determine R₂. This can be obtained as follow:
Voltage 2 (V₂) = 6 V
Current (I) = 1.5 A
Resistor 2 (R₂) =?
V₂ = IR₂
6 = 1.5 × R₂
Divide both side by 1.5
R₂ = 6 / 1.5
R₂ = 4 Ω
4. The diagram is not complete