Answer:
see solution below
Explanation:
The given resistors are connected in series.
Equivalent resistance in series = 30 + 55 + 15
Equivalent resistance in series Rt = 100 ohms
Since the potential difference in the circuit = 36V
Get the current in the circuit first
I = V/Rt
I = 36/100
I = 0.36A
Get the voltage across 30ohms resistor;
V30 = 0.36 * 30
V30 = 10.8volts
Hence the voltage across the 30ohms resistor is 10.8volts
Get the voltage across 55ohms resistor;
V55 = 0.36 * 55
V55 = 19.8volts
Hence the voltage across the 55ohms resistor is 19.8volts
Get the voltage across 15ohms resistor;
V15 = 0.36 * 15
V15 = 5.4volts
Hence the voltage across the 15ohms resistor is 5.4volts
I think that the answer to that is true hope that helps
Incomplete Question.The Complete question is
The Earth spins on its axis and also orbits around the Sun. For this problem use the following constants. Mass of the Earth: 5.97 × 10^24 kg (assume a uniform mass distribution) Radius of the Earth: 6371 km Distance of Earth from Sun: 149,600,000 km
(i)Calculate the rotational kinetic energy of the Earth due to rotation about its axis, in joules.
(ii)What is the rotational kinetic energy of the Earth due to its orbit around the Sun, in joules?
Answer:
(i) KE= 2.56e29 J
(ii) KE= 2.65e33 J
Explanation:
i) Treating the Earth as a solid sphere, its moment of inertia about its axis is
I = (2/5)mr² = (2/5) * 5.97e24kg * (6.371e6m)²
I = 9.69e37 kg·m²
About its axis,
ω = 2π rads/day * 1day/24h * 1h/3600s
ω= 7.27e-5 rad/s,
so its rotational kinetic energy
KE = ½Iω² = ½ * 9.69e37kg·m² * (7.27e-5rad/s)²
KE= 2.56e29 J
(ii) About the sun,
I = mR²
I= 5.97e24kg * (1.496e11m)²
I= 1.336e47 kg·m²
and the angular velocity
ω = 2π rad/yr * 1yr/365.25day * 1day/24h * 1h/3600s
ω= 1.99e-7 rad/s
so
KE = ½ * 1.336e47kg·m² * (1.99e-7rad/s)²
KE= 2.65e33 J
<span>This problem can be solved by the formula used to find resistance. The formula is R=V/I which basically means divide the Voltage by the Current to find the Resistance in an object. Ohm's law.</span>
Answer:
140265.8 C = 1.403 × 10⁵ C
Explanation:
The battery's electric potential energy is used to account for the kinetic and potential work done in moving the car up this hill.
Potential work required to move the 757 kg car up a vertical height of 195 m = mgh
P.E = 757 × 9.8 × 195 = 1446627 J
Kinetic work done = (1/2)(m)(v²)
K.E = (1/2)(757)(25²) = 236562.5 J
Total work done in moving the car up that height = 1446627 + 236562.5 = 1683189.5 J
And this would be equal to the potential of the battery.
For the battery, potential difference = (electric potential energy)/(charges moved)
ΔV = ΔU/q
q = ΔU/ΔV
ΔU = 1683189.5 J
ΔV = 12.0 V
q = 1683189.5/12 = 140265.8 C
