Answer:
100 Ω
Explanation:
Given that
Open circuit voltage, V = 1 V
Short circuit current, I = 10 mA
Source resistance R, = ?
This is rather a straight forward question. Remember Ohms Law? Current being directly proportional to the voltage and inversely proportional to the resistance?
Yeah, that's the formula we'd be using.
Ohms Law states that V = IR, and thus, if we make R subject of the formula, we have
R = V / I, on substituting the values, we have
R = 1 / 10*10^-3
R = 1 / 0.01
R = 100 Ω
Answer:
Q = 836.4 Joules.
Explanation:
Given the following data;
Mass = 100 grams
Initial temperature = 25°C
Final temperature = 45°C
We know that the specific heat capacity of water is equal to 4.182 J/g°C.
To find the quantity of heat;
Heat capacity is given by the formula;
Where;
Q represents the heat capacity or quantity of heat.
m represents the mass of an object.
c represents the specific heat capacity of water.
dt represents the change in temperature.
dt = T2 - T1
dt = 45 - 25
dt = 20°C
Substituting the values into the equation, we have;
Q = 836.4 Joules.
Answer:
C_{y} = 4.96 and θ' = 104,5º
Explanation:
To add several vectors we can decompose each one of them, perform the sum on each axis, to find the components of the resultant and then find the module and direction.
Let's start by decomposing the two vectors.
Vector A
sin θ =
/ A
cos θ = Aₓ / A
A_{y} = A sin θ
Ax = A cos θ
A_{y} = 4.9 sin 31 = 2.52
Ax = 4.9 cos 31 = 4.20
Vector B
B_{y} = B sin θ
Bx = B cos θ
B_{y} = 6 sin 156 = 2.44
Bx = 6 cos 156 = -5.48
The components of the resulting vector are
X axis
Cx = Ax + B x
Cx = 4.20 -5.48
Cx = -1.28
Axis y
C_{y} = Ay + By
C_{y} = 2.52 + 2.44
C_{y} = 4.96
Let's use the Pythagorean theorem to find modulo
C = √ (Cₙ²x2 + Cy2)
C = Ra (1.28 2 + 4.96 2)
C = 5.12
We use trigonemetry to find the angle
tan θ = C_{y} / Cₓ
θ’ = tan⁻¹ (4.96 / (1.28))
θ’ = 75.5
como el valor de Cy es positivo y Cx es negativo el angulo este en el segundo cuadrante, por lo cual el angulo medido respecto de eje x positivo es
θ’ = 180 – tes
θ‘= 180 – 75,5
θ' = 104,5º